Wave motion

What is a wave?

A wave is essentially a disturbance that travels through a medium, transferring energy from one point to another without physically moving the matter of the medium itself (the energy is transferred from one place to another without any actual transfer of the particles of the medium); common examples include water waves, sound waves, and light waves, where the disturbance propagates away from its source while the medium particles only vibrate around their equilibrium positions. 

In other words, “a wave is any disturbance from a normal or equilibrium condition that propagates without the transport of matter. In general, a wave transports both energy and momentum.”

The most intuitive and easiest wave to imagine is the familiar water wave. For water waves, the disturbance is in the surface of the water, perhaps created by a rock thrown into a pond or by a swimmer splashing the surface repeatedly. For sound waves, the disturbance is a change in air pressure, perhaps created by the oscillating cone inside a speaker. Water waves exhibit characteristics common to all waves, such as amplitude, period, frequency, and energy. All wave characteristics can be described by a small set of underlying principles.

Misconception: Many people think that water waves push water from one direction to another. In fact, the particles of water tend to stay in one location, save for moving up and down due to the energy in the wave. The energy moves forward through the water, but the water stays in one place. If you feel yourself pushed in an ocean, what you feel is the energy of the wave, not a rush of water.

What is wave energy?

The energy in water waves primarily comes from wind blowing across the surface of the water; as the wind interacts with the water, it transfers its kinetic energy, creating waves that propagate across the ocean surface. Wave energy (or wave power) harnesses the ocean’s waves to generate energy by converting a wave’s kinetic energy into electricity. Wave power is a form of renewable and sustainable energy that’s often overlooked but has immense potential.

Key points about water wave energy:

Source of wind energy: The wind itself is driven by the sun’s heat, making wave energy an indirect form of solar energy.

Mechanism: Friction between the wind and water surface causes the water to move in a circular motion, forming waves.

Wave characteristics: The strength of the wind and the duration it blows over the water determine the size and energy of the waves.

Key points about waves:

Energy transfer: The primary function of a wave is to transport energy, not matter (means of transferring energy and momentum from one point to another without actual transport of matter between two points). 

Medium: Waves require a medium to travel through, like water for water waves or air for sound waves.

Disturbance: A wave is created by a disturbance that disrupts the equilibrium of the medium.

Propagation: The disturbance travels through the medium, transferring energy from one point to another.

What is wave motion?

A wave motion is a means of transferring energy and momentum from one point to another without the actual transport of matter between two points. Transmission of energy over considerable distances is possible through wave motion. 

The simplest waves repeat themselves for several cycles and are associated with simple harmonic motion. Let us start by considering the simplified water wave in Figure 2. The wave is an up-and-down disturbance of the water surface. It causes a seagull to move up and down in simple harmonic motion as the wave crests and troughs (peaks and valleys) pass under the bird. The time for one complete up and down motion is the wave’s period \(T\). The wave’s frequency is
\(
f=\frac{1}{T}
\)
, as usual. The wave itself moves to the right in Figure 2. This movement of the wave is actually the disturbance moving to the right, not the water itself (or the bird would move to the right). We define wave velocity \(v_w\) to be the speed at which the disturbance moves. Wave velocity is sometimes also called the propagation velocity or propagation speed because the disturbance propagates from one location to another.

The water wave in the figure also has a length associated with it, called its wavelength \(\lambda\), the distance between adjacent identical parts of a wave. ( \(\lambda\) is the distance parallel to the direction of propagation.) The speed of propagation \(v_{\mathrm{w}}\) is the distance the wave travels in a given time, which is one wavelength in the time of one period. In equation form, that is
\(
v_{\mathrm{w}}=\frac{\lambda}{T}
\)
\(
\text { or } v_{\mathrm{w}}=f \lambda \quad [f=\frac{1}{T}]
\)
This fundamental relationship holds for all types of waves. For water waves, \(v_{\mathrm{w}}\) is the speed of a surface wave; for sound, \(v_{\mathrm{w}}\) is the speed of sound; and for visible light, \(v_{\mathrm{w}}\) is the speed of light.

Transverse and Longitudinal Waves

A simple wave consists of a periodic disturbance that propagates from one place to another. The wave in Figure 3 propagates in the horizontal direction while the surface is disturbed in the vertical direction. Such a wave is called a transverse wave or shear wave; in such a wave, the disturbance is perpendicular to the direction of propagation.

In contrast, in a longitudinal wave or compressional wave, the disturbance is parallel to the direction of propagation. Figure 4 shows an example of a longitudinal wave. The size of the disturbance is its amplitude \(X\) and is completely independent of the speed of propagation \(v_{\mathrm{w}}\).

Waves may be transverse, longitudinal, or a combination of the two. (Water waves are actually a combination of transverse and longitudinal. The simplified water wave illustrated in Figure 2 shows no longitudinal motion of the bird.) The waves on the strings of musical instruments are transverse, so are electromagnetic waves, such as visible light.

Sound waves in air and water are longitudinal. Their disturbances are periodic variations in pressure that are transmitted in fluids. Fluids do not have appreciable shear strength, and thus the sound waves in them must be longitudinal or compressional. Sound in solids can be both longitudinal and transverse.

Classification Based on Motion of Wave in Space

The waves are classified into three types based on the motion of wave in a space. They are as follows:

• One-Dimensional waves
• Two-Dimensional waves
• Three-Dimensional waves

\(
\begin{array}{|c|c|c|}
\hline \text { Wave Type } & \text { Definition } & \text { Example } \\
\hline \begin{array}{c}
\text { One- } \\
\text { Dimensional } \\
\text { waves }
\end{array} & \begin{array}{c}
\text { The wave that moves } \\
\text { along one dimension only } \\
\text { is called a one- } \\
\text { dimensional wave. }
\end{array} & \begin{array}{c}
\text { Waves produced } \\
\text { in a string. }
\end{array} \\
\hline \begin{array}{c}
\text { Two- } \\
\text { Dimensional } \\
\text { waves }
\end{array} & \begin{array}{c}
\text { The wave that moves in a } \\
\text { plane is called a two- } \\
\text { dimensional wave. }
\end{array} & \text { Ripple in water. } \\
\hline \begin{array}{c}
\text { Three- } \\
\text { Dimensional } \\
\text { waves }
\end{array} & \begin{array}{c}
\text { The wave that moves in a } \\
\text { three-dimensional space } \\
\text { is called a three- } \\
\text { dimensional wave. }
\end{array} & \begin{array}{c}
\text { Propagation of } \\
\text { light and sound } \\
\text { waves. }
\end{array} \\
\hline
\end{array}
\)

Types of waves

There are mainly three types of waves

• Mechanical waves: Mechanical waves are those waves that require a medium for their propagation. They cannot travel through vacuum. e.g. Sound waves, water waves, etc. They are governed by Newton’s laws, and they can exist only within a material medium.
• Electromagnetic waves:  Electromagnetic waves are those waves that do not require a medium for their propagation. These waves travel in the form of oscillating electric and magnetic fields. e.g. X-rays, radio waves, etc. Light waves from stars, for example, travel through the vacuum of space to reach us. All electromagnetic waves travel through a vacuum at the same speed \(c=299792458 \mathrm{~m} / \mathrm{s}\).
• Matter waves: Moving microscopic particles such as electrons, protons, neutrons, atoms, molecules, etc., sometimes behave like waves. These are called matter waves. molecules. We commonly think of these particles as constituting matter, such waves are called matter waves.

Definitions related to the wave motion

A wave travels in the form of a sine wave as shown below. Amplitude, crest, trough, and wavelength are a part of the wave.

Amplitude: The maximum displacement suffered by the particles of the medium from their mean position is called amplitude. It is denoted by \(A\).

Time period: The time taken for one complete cycle of a wave to pass a given point is called the period. It is the reciprocal of frequency. It is denoted by \(T\). It is the reciprocal of frequency.

Frequency: The number of waves produced per unit time in the given medium is known as frequency of a wave. It is denoted by \(f=1/T\).  It is measured in Hertz (Hz).

Wavelength: The distance covered by a wave during the time in which a particle of medium completes one vibration about its mean position is known as the wavelength of a wave. It is denoted by \(\lambda\).

Wave Number: The number of waves present in a unit distance along the direction of propagation is known as wave number. It is equal to the reciprocal of wavelength \((\lambda)\). It is denoted by \(\bar{\nu}\).
\(
\therefore \quad \bar{\nu}=\frac{1}{\lambda}
\)
The SI unit of wave number is \(\mathrm{m}^{-1}\).

Phase: The position of a point in time on a waveform is known as the phase of a wave. Phase can also be expressed as relative displacement between two corresponding peaks of a waveform.

The two waves shown above (A versus B) are of the same amplitude and frequency, but they are out of step with each other. In technical terms, this is called a phase shift. 

A sampling of different phase shifts is given in the following graphs to better illustrate this concept: Figure below

Path Difference: The difference in the path traversed by the two waves measured in terms of the wavelength of the associated waves is called path difference.

In Figures 6 and 7 you can see that at the different points on the screen the waves from \(\mathrm{S}_1\) have travelled a different distance from those from \(S_2\). In Figure 6 the path difference is zero, in Figure 7 it is half a wavelength.

Speed of a Wave: The speed of a wave is the measure of how fast the wave travels. It is calculated as the ratio of how far a wave travels to the time taken by the wave to travel that distance.

\(
\text { speed }=\frac{\text { distance }}{\text { time }}
\)

Wave Pulse: It is a short wave produced in a medium when the disturbance is created for a short time. If we give an upward jerk to one end of a long rope whose opposite end is fixed, a single wave pulse is formed as shown in the figure and it travels along the rope with a fixed speed.

Relation between frequency, speed and wavelength

When a particle of the medium completes one oscillation about its mean position in periodic time \((T)\), then the wave travels a distance equal to the wavelength ( \(\lambda\) ). So,
\(
\begin{array}{ll}
& \text { Wave speed }=\frac{\text { Distance }}{\text { Time }} \\
\Rightarrow & v=\frac{\lambda}{T} \Rightarrow v=f \times \lambda \left(\because f=\frac{1}{T}\right) \\
\text { i.e. } & \text { Wave speed }=\text { Frequency } \times \text { Wavelength }
\end{array}
\)

Example 1: The speed of a wave in a medium is \(960 \mathrm{~ms}^{-1}\). If 3600 waves pass through a point in the medium in 1 minute, then determine its wavelength.

Solution: Given, the speed of the wave, \(v=960 \mathrm{~ms}^{-1}\)
Frequency of the wave, \(f=3600 \mathrm{~min}^{-1}\)
\(
=\frac{3600}{60}=60 \mathrm{~s}^{-1}
\)
\(\therefore\) Wavelength, \(\lambda=\frac{v}{f}=\frac{960}{60}=16 \mathrm{~m} \Rightarrow \lambda=16 \mathrm{~m}\)

Progressive wave

A wave that travels from one point of the medium to another is called a progressive wave. It can be transverse or longitudinal. A wave which travels continuously in a medium in the same direction without a change in its amplitude is called a travelling wave or a progressive wave.

For convenience, we shall take the wave to be transverse so that if the position of the constituents of the medium is denoted by \(x\), the displacement from the equilibrium position may be denoted by \(y\). A sinusoidal travelling wave is then described by:
\(
y(x, t)=a \sin (k x-\omega t+\phi) \dots(i)
\)
The term \(\phi\) in the argument of sine function means equivalently that we are considering a linear combination of sine and cosine functions:
\(
y(x, t)=A \sin (k x-\omega t)+B \cos (k x-\omega t) \dots(ii)
\)
From Equations (i) and (ii),
\(
a=\sqrt{A^2+B^2} \text { and } \phi=\tan ^{-1}\left(\frac{B}{A}\right)
\)
To understand why Equation (i) represents a sinusoidal travelling wave, take a fixed instant, say \(t=t_0\). Then, the argument of the sine function in Equation (i) is simply \(k x+\) constant. Thus, the shape of the wave (at any fixed instant) as a function of \(x\) is a sine wave. Similarly, take a fixed location, say \(x=X_0\). Then, the argument of the sine function in Equation (i) is constant \(-\omega t\). The displacement \(y\), at a fixed location, thus, varies sinusoidally with time. That is, the constituents of the medium at different positions execute simple harmonic motion. Finally, as \(t\) increases, \(x\) must increase in the positive direction to keep \(k x-\omega t+\phi\) constant. Thus, Eq. (i) represents a sinusiodal (harmonic) wave travelling along the positive direction of the \(x\)-axis. On the other hand, a function
\(
y(x, t)=a \sin (k x+\omega t+\phi)
\)
represents a wave travelling in the negative direction of \(x\)-axis. Fig 15.6. below gives the names of the various physical quantities appearing in Eq. (i) that we now interpret.

Displacement Relation in a Progressive Wave

During the propagation of a wave through a medium, if the particles of the medium vibrate simple harmonically about their mean positions, then the wave is said to be plane progressive harmonic wave.

In a progressive wave, the displacement relation describes how the position of a particle changes over time as the wave passes through. It shows the relationship between the displacement of the particle from its equilibrium position and both the position along the wave (represented by \(x\) ) and the time (represented by \(t\) ).
Mathematically, the displacement relation for a sinusoidal (harmonic)  wave travelling in \(+x\)-direction is given by
\(
y(x, t)=A \sin (\omega t-k x+\phi) \dots(1)
\)
On the other hand, a wave travelling in \(-x\)-direction is given by
\(
y(x, t)=A \sin (\omega t+k x+\phi) \dots(2)
\)
Where, \(y(x, t)\) is the displacement of the particle at position \(x\) and time \(t\),
\(A\) is the amplitude, representing the maximum displacement from equilibrium,
\(k\) is the angular wave number, determining the spatial frequency of the wave, \(k=2 \pi / \lambda\)
\(\omega\) is the angular frequency, indicating how quickly the wave oscillates in time,
\(\phi\) is the phase constant, determining the initial position of the wave.
This relation shows how the displacement of a particle varies with both position and time along the wave. As the wave propagates, particles oscillate back and forth around their equilibrium positions according to this relation. It helps understand the behavior of waves and their interaction with particles in a medium.

Amplitude and Phase

In Eq. (1), since the sine function varies between 1 and -1, the displacement \(y(x, t)\) varies between \(A\) and \(-A\). We can take \(A\) to be a positive constant, without any loss of generality. Then, \(A\) represents the maximum displacement of the constituents of the medium from their equilibrium position. Note that the displacement \(y\) may be positive or negative, but \(A\) is positive. It is called the amplitude of the wave.

The quantity ( \(\omega t-k x+\phi\) ) appearing as the argument of the sine function in Eq. (1) is called the phase of the wave. Given the amplitude \(A\), the phase determines the displacement of the wave at any position and at any instant. Clearly \(\phi\) is the phase at \(x=0\) and \(t=0\). Hence, \(\phi\) is called the initial phase angle. By suitable choice of origin on the \(x\)-axis and the initial time, it is possible to have \(\phi=0\). Thus there is no loss of generality in dropping \(\phi\), i.e., in taking Eq. (1) with \(\phi=0\).

Wavelength and Angular Wave Number

The minimum distance between two points having the same phase is called the wavelength of the wave, usually denoted by \(\lambda\). For simplicity, we can choose points of the same phase to be crests or troughs. The wavelength is then the distance between two consecutive crests or troughs in a wave. Taking \(\phi=0\) in Eq. (1), the displacement at \(t=0\) is given by
\(
y(x, 0)=A \sin k x \dots(3)
\)
Since the sine function repeats its value after every \(2 \pi\) change in angle,
\(
\sin k x=\sin (k x+2 n \pi)=\sin k\left(x+\frac{2 n \pi}{k}\right)
\)
That is the displacements at points \(x\) and at
\(
x+\frac{2 n \pi}{k}
\)
are the same, where \(n=1,2,3, \ldots\) The least distance between points with the same displacement (at any given instant of time) is obtained by taking \(n=1. \lambda\) is then given by
\(
\lambda=\frac{2 \pi}{k} \quad \text { or } \quad k=\frac{2 \pi}{\lambda} \dots(4)
\)
\(k\) is the angular wave number or propagation constant; its SI unit is radian per metre or \(\mathrm{rad}  \mathrm{~m}^{-1 }\)

Period, Angular Frequency and Frequency

\(\omega\) is called the angular frequency. Here,
\(
\omega=2 \pi f=\frac{2 \pi}{T} \quad \text { and } \quad f=\frac{1}{T}
\)
where, \(f\) is the normal frequency (natural frequency) of oscillations and \(T\) is the time period of oscillations. SI unit of \(\omega\) is radian/sec while that of \(f\) is Hz or \(\mathrm{s}^{-1}\).

Wave Speed (\(v\))

Wave speed \(v=\frac{\text { coefficient of } t}{\text { coefficient of } x}=\frac{\omega}{k}\)
But \(\omega=2 \pi f\) and \(k=\frac{2 \pi}{\lambda}\)
\(v=f \lambda=\frac{\lambda}{T}\)

Notes: (i) Equation of the form \(y(x, t)=f\) (at \(\pm b x\) ) represents progressive or travelling waves. The plus sign denotes a wave traveling in the negative direction of the \(x\) axis, and the minus sign a wave traveling in the positive direction.
Some common progressive wave equations are
\(y=A \log (a t+b x)\)
\(y=\sqrt{(a x+b t)}\)
\(y=(a x-b t)^2\)
\(y=A \sin (a x-b t)^2\)
\(y=a \cos ^2(\omega t-k x)\)
\(y=a \cos \Delta \omega t \sin (\omega t-k x)\)

Sound Waves

Sound is a form of energy that produces a sensation of hearing in our ears. Sound waves are longitudinal in nature. Sound waves can be classified in three groups according to their range of frequencies

• Infrasonic waves: Longitudinal waves having frequencies below 20 Hz are called infrasonic waves. They cannot be heard by human beings.
• Audible waves: Longitudinal waves having frequencies lying between 20-20000 Hz} are called audible waves.
• Ultrasonic waves: Longitudinal waves having frequencies above 20000 Hz are called ultrasonic waves. They are produced and heard by bats. They have a large energy content.

Particle velocity and acceleration in plane progressive harmonic wave

As particle displacement, \(y=A \sin (\omega t-k x)\)
\(\therefore\) Particle velocity, \(v_p=\frac{d y}{d t}=A \omega \cos (\omega t-k x) \dots(1)\)
Again, by differentiating the above equation, we get Particle acceleration,
\(
a_p=\frac{d^2 y}{d t^2}=-A \omega^2 \sin (\omega t-k x)=-\omega^2 y
\)
Consider the motion of a string along \(x\)-axis. Its one end is connected to a source executing SHM of amplitude \(A\), time period, \(T=2 \pi / \omega\) and frequency of oscillation, \(f=1 / T\). The wave produced by such a vibrating source is a sine wave.
Now, slope of string, \(\frac{d y}{d x}=-A k \cos (\omega t-k x) \dots(2)\)
\(
\therefore \quad \frac{d y / d t}{d y / d x}=\frac{A \omega \cos (\omega t-k x)}{-A k \cos (\omega t-k x)}=-\frac{\omega}{k} \Rightarrow \frac{d y}{d t}=-v_w \frac{d y}{d x}
\)
\(\Rightarrow\) Particle velocity \(=-\) Wave velocity \(\times\) Slope
Similarly, \(\frac{d^2 y}{d t^2} \times \frac{d x^2}{d^2 y}=\frac{\omega^2}{k^2}=v^2\)
\(
\Rightarrow \quad \frac{d^2 y}{d t^2}=v^2 \frac{d^2 y}{d x^2} \text { (Wave equation) }
\)

Speed of Different Waves

Normally, two wave speeds are required at this stage.

• Transverse wave speed on a string.
• Longitudinal wave speed in all three states: solid, liquid, and gas.

Transverse Wave Speed on a Stretched String

The speed of transverse wave on a string is given by \(v=\sqrt{\frac{T}{\mu}}\)
Here, \(\mu=\) mass per unit length of the string \(=\frac{m}{l}=\frac{m A}{l A} \quad(A=\) area of cross-section of the string \()\), (\(T\)) is the tension in the string (the stronger the tension in the string, the faster the wave travels).
\(=\left(\frac{m}{V}\right) A \quad(V=\) volume of string \()\)
\(=\rho A \quad(\rho=\) density of string \()\)
Hence, the above expression can also be written as \(v=\sqrt{\frac{T}{\rho A}}\)

If radius of the string is \(r\) having volume \(V\) and the density of the material of the string is \(\rho\), then
\(
\begin{aligned}
& \mu=\frac{V}{L} \times \rho=\pi r^2 \rho \\
& v=\sqrt{\frac{T}{\pi r^2 \rho}}
\end{aligned}
\)
When a block of mass \(M\) is suspended from the string, then
\(
T=M g
\)
The speed of the wave along a stretched ideal string depends only on the tension and the linear mass density of the string and does not depend on the frequency of the wave.
The frequency of the wave is determined by the source that generates the wave. The wavelength is given by
\(
\lambda=\frac{v}{f}
\)

Derivation: Speed \(v\) of a transverse wave depends upon tension in the string and linear mass density \(\mu\) (mass/length)
\(
\begin{aligned}
& v \propto T^a \mu^b \\
& \Rightarrow v=C T^a \mu^b \\
& \Rightarrow\left[L T^{-1}\right]=C\left[M L T^{-2}\right]^a\left[M L^{-1}\right]^b \\
& \Rightarrow\left[L T^{-1}\right]=C\left[M^{a+b} L^{a-b} T^{-2 a}\right] \\
& a+b=0, a-b=1, \\
& \therefore a=\frac{1}{2}, b=-\frac{1}{2}
\end{aligned}
\)
\(
\begin{aligned}
& v=C T^{1 / 2} \mu^{-1 / 2} \\
& v=C \frac{T^{1 / 2}}{\mu^{1 / 2}} \\
& v=C \sqrt{\frac{T}{\mu}}
\end{aligned}
\)
\(C\) is a dimensionless constant that cannot be determined with dimensional analysis (\(C=1\)).
\(
v=\sqrt{\frac{T}{\mu}}
\)

Example 2: A steel wire 0.72 m long has a mass of \(5.0 \times 10^{-3} \mathrm{~kg}\). If the wire is under a tension of 60 N , what is the speed of transverse waves on the wire?

Solution: Mass per unit length of the wire,
\(
\begin{aligned}
\mu & =\frac{5.0 \times 10^{-3} \mathrm{~kg}}{0.72 \mathrm{~m}} \\
& =6.9 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}
\end{aligned}
\)
Tension, \(T=60 \mathrm{~N}\)
The speed of wave on the wire is given by
\(
v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{60 \mathrm{~N}}{6.9 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}}}=93 \mathrm{~m} \mathrm{~s}^{-1}
\)

Speed of a longitudinal wave (sound wave) in a medium

In a longitudinal wave, the constituents of the medium oscillate forward and backward in the direction of wave propagation. Speed of longitudinal waves in a medium is
\(
v=\sqrt{\frac{B}{\rho}}
\)
where, bulk modulus of the medium,
\(
B=-\frac{\Delta p}{\Delta V / V}
\)
where, \(\Delta p=\) change in pressure, \(\Delta V=\) change in volume and \(\quad V=\) initial volume. \(\rho=\text { density. }\)
The speed of a longitudinal wave (sound wave) in a medium depends on its elastic properties and inertial properties.
For a linear medium like a solid bar, the modulus of elasticity will be Young’s modulus \((Y)\).

Derivation: Speed \(v\) of a transverse wave depends upon bulk modulus \(B\) of the medium and density of the medium \(\rho\)
\(
\begin{aligned}
& v \propto B^a \rho^b \\
& \Rightarrow v=C B^a \rho^b \\
& \Rightarrow\left[L T^{-1}\right]=C\left[M L^{-1} T^{-2}\right]^a\left[M L^{-3}\right]^b \\
& \Rightarrow\left[L T^{-1}\right]=C\left[M^{a+b} L^{-a-3 b} T^{-2 a}\right] \\
& a+b=0,-a-3 b=1,
\end{aligned}
\)
\(
\begin{aligned}
& \therefore a=\frac{1}{2}, b=-\frac{1}{2} \\
& \therefore v=C \sqrt{\frac{B}{\rho}}
\end{aligned}
\)
As \(C=1\), \(v=\sqrt{\frac{B}{\rho}}\)

Note: The speed of sound is greater in solids and liquids than in gases even though they are denser than gases. This happens because they are much more difficult to compress than gases and hence, have much higher values of bulk modulus. This factor compensates for their higher densities than gases.

Example 3: The volumetric strain of water at a pressure of \(10^5 \mathrm{Nm}^{-2}\) is \(5 \times 10^{-5}\). Determine the speed of sound in water. Density of water is \(10^3 \mathrm{~kg} \mathrm{~m}^{-3}\).

Solution: Bulk modulus of water, \(B=\frac{\text { Normal stress }}{\text { Volumetric strain }}=\frac{10^5}{5 \times 10^{-5}}\)
\(
=2 \times 10^9 \mathrm{Nm}^{-2}
\)
\(\because\) Density of water, \(\rho=10^3 \mathrm{~kg} \mathrm{~m}^{-3} \quad \text { (Given) }\)
\(\therefore\) Speed of sound in water, \(v=\sqrt{\frac{B}{\rho}}\)
\(
\Rightarrow \quad v=\sqrt{\frac{2 \times 10^9}{10^3}}=1.41 \times 10^3 \mathrm{~ms}^{-1}
\)

Speed of longitudinal waves in a solid bar (thin rod or wire)

So, the speed of longitudinal waves in a solid bar is given by
\(
v=\sqrt{\frac{Y}{\rho}}
\)
where, \(Y\) is the Young’s modulus of the material of the bar and \(\rho\) is density. Also, the speed of a longitudinal wave through a solid of bulk modulus \(B\), modulus of rigidity \(\eta\) and density \(\rho\) is given by
\(
v=\sqrt{\frac{B+(4 / 3) \eta}{\rho}}
\)

Example 4: If the Young’s modulus of elasticity for a steel rod is \(2.9 \times 10^{11} \mathrm{Nm}^{-2}\) and density is \(8 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\). Determine the velocity of longitudinal waves in the steel rod.

Solution: Given, Young’s modulus, \(Y=2.9 \times 10^{11} \mathrm{Nm}^{-2}\)
Density, \(\rho=8 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\)
\(\therefore\) Velocity of longitudinal waves in steel,
\(
\begin{aligned}
v & =\sqrt{\frac{Y}{\rho}}=\sqrt{\frac{2.9 \times 10^{11}}{8 \times 10^3}} \\
& =6.02 \times 10^3 \mathrm{~ms}^{-1}
\end{aligned}
\)

Example 5: The bulk modulus and modulus of rigidity for aluminium are \(7.5 \times 10^{10} \mathrm{Nm}^{-2}\) and \(2.1 \times 10^{10} \mathrm{Nm}^{-2}\), respectively. Determine the velocity of the waves in the medium. Density of aluminium is \(2.7 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\).

Solution: Given, bulk modulus, \(B=7.5 \times 10^{10} \mathrm{Nm}^{-2}\)
Modulus of rigidity, \(\eta=2.1 \times 10^{10} \mathrm{Nm}^{-2}\)
Density, \(\rho=2.7 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\)
The velocity of longitudinal waves in aluminium,
\(
\begin{aligned}
& v=\sqrt{\frac{B+\frac{4}{3} \eta}{\rho}}=\sqrt{\frac{7.5 \times 10^{10}+\frac{4}{3} \times 2.1 \times 10^{10}}{2.7 \times 10^3}} \\
& v=6.18 \times 10^3 \mathrm{~ms}^{-1}
\end{aligned}
\)

Example 6: Determine the speed of sound waves in water and find the wavelength of a wave having a frequency of 242 Hz. (Take, \(B_{\text {water }}=2 \times 10^9 \mathrm{~Pa}\) )

Solution: Speed of sound wave in water,
\(
v=\sqrt{\frac{B_{\text {water }}}{\rho}}=\sqrt{\frac{\left(2 \times 10^9\right)}{10^3}}=1414 \mathrm{~ms}^{-1}
\)
Wavelength, \(\lambda=\frac{v}{f}=\frac{1414}{242}=5.84 \mathrm{~m}\)

Speed of longitudinal waves in gas

Speed of longitudinal waves in gases,
\(
v=\sqrt{\frac{p}{\rho}}
\)
where, \(p\) is the pressure exerted by gas.
For gases, the bulk modulus is equal to the pressure of the gas. This formula is known as Newton’s formula. Newton assumed that when sound waves are propagated through a gas, the temperature variations in the layers of compression and rarefaction are negligible (isothermal process).

Example 7: The speed of sound in air at NTP is \(332 \mathrm{~ms}^{-1}\). Calculate the percentage error in the speed of sound as calculated from Newton’s formula. Given that the density of air at 1 atm pressure is \(1.293 \mathrm{kgm}^{-3}\).

Solution: From Newton’s formula, \(v=\sqrt{\frac{p}{\rho}}\)
At NTP, \(p=1.01 \times 10^5 \mathrm{~Pa} \Rightarrow v=\sqrt{\frac{1.01 \times 10^5}{1.293}}=280 \mathrm{~ms}^{-1}\)
Difference in speed \(=332-280=52 \mathrm{~ms}^{-1}\)
\(
\% \text { error }=\frac{52}{332} \times 100=15.7 \%
\)

Laplace correction

The result obtained for the speed of sound in air at STP from the above formula is \(280 \mathrm{~ms}^{-1}\), which is about \(15 \%\) smaller as compared to the experimental value, i.e. \(331 \mathrm{~ms}^{-1}\).
The mistake in the formula was pointed out by Laplace and he told that the changes in pressure and volume of a gas, when sound waves are propagated through it, are not isothermal, but adiabatic.
He modified the formula as
\(
\text { Speed of sound in gases, } v=\sqrt{\frac{\gamma p}{\rho}}
\)
This modification in Newton’s formula is referred to as the Laplace correction.
For air, \(\gamma=7 / 5\).
By this formula, the speed of sound comes out to be \(331.3 \mathrm{~ms}^{-1}\), which agrees with the practical value.

Example 8: At normal temperature and pressure, the speed of sound in air is \(332 \mathrm{~m} / \mathrm{s}\). What will be the speed of sound in hydrogen at normal temperature and pressure? (Air is 16 times heavier than hydrogen)

Solution: The speed sound in a gas is given by
\(
\begin{aligned}
& v=\sqrt{\gamma p / \rho} \Rightarrow \frac{v_a}{v_H}=\sqrt{\frac{\rho_H}{\rho_a}} \\
& \frac{\rho_H}{\rho_a}=1 / 16 \Rightarrow \frac{v_a}{v_H}=\sqrt{\frac{1}{16}}=\frac{1}{4} \\
& v_H=4 v_a=4 \times 332=1328 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Example 9: Find speed of sound in hydrogen gas at \(27^{\circ} \mathrm{C}\), if \(C_p / C_V\) for \(\mathrm{H}_2\) is 1.4. Gas constant, \(R=8.31 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}\).

Solution: The speed sound in a gas, \(v=\sqrt{\gamma p / \rho}\)
\(
\therefore \quad v=\sqrt{\frac{\gamma R T}{M}} \quad\left(\because p V=R T \text { and } \rho=\frac{M}{V}\right)
\)
Here,
\(
\begin{aligned}
T & =27+273=300 \mathrm{~K}, \gamma=\frac{C_p}{C_V}=1.4 \\
R & =8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\
M & =2 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}^{-1}
\end{aligned}
\)
\(\therefore \quad v=\sqrt{\frac{1.4 \times 8.31 \times 300}{2 \times 10^{-3}}}=1321 \mathrm{~ms}^{-1}\)

Example 10: One end of 12.0 m long rubber tube with a total mass of 0.9 kg is fastened to a fixed support. A cord attached to the other end passes over a pulley and supports an object with a mass of 5.0 kg . The tube is struck a transverse blow at one end. Find the time required for the pulse to reach the other end. \(\left(g=9.8 \mathrm{~m} / \mathrm{s}^2\right)\)

Solution: Tension in the rubber tube \(A B, T=m g\)

\(
T=(5.0)(9.8)=49 \mathrm{~N}
\)
Mass per unit length of rubber tube, \(\mu=\frac{0.9}{12}=0.075 \mathrm{~kg} / \mathrm{m}\)
Speed of wave on the tube, \(v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{49}{0.075}}=25.56 \mathrm{~m} / \mathrm{s}\)
The required time is, \(t=\frac{A B}{v}=\frac{12}{25.56}=0.47 \mathrm{~s}\)

Factors affecting the speed of sound in air

The following factors affect the speed of sound in air (or gas)
1. Effect of temperature
As, \(v=\sqrt{\frac{\gamma p}{\rho}}\)
We can write
\(
v=\sqrt{\frac{\gamma R T}{M}} \quad(\because p V=R T)
\)
where, \(M=\) molecular mass of gas, \(T\) is absolute temperature and \(R\) is gas constant.
\(
\begin{array}{lc}
\Rightarrow & v \propto \sqrt{T} \\
\text { or } & \frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}
\end{array}
\)

Example 11: In given gaseous medium, at what temperature, the velocity of sound will be double that of velocity at \(27^{\circ} \mathrm{C}\).

Solution: Given, the velocity of sound, \(v_2=2 v_1\)
Temperature, \(T_1=27^{\circ} \mathrm{C}=(27+273)=300 \mathrm{~K}\)
As,\(v \propto \sqrt{T}\)
\(
\begin{array}{ll}
\Rightarrow & \frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}} \Rightarrow \frac{v_1}{2 v_1}=\sqrt{\frac{300}{T_2}} \\
\Rightarrow & T_2=1200 \mathrm{~K} \text { or } T_2=927^{\circ} \mathrm{C}
\end{array}
\)

Example 12: A wire of uniform cross-section is stretched between two points 100 cm apart. The wire is fixed at one end and a weight is hung over a pulley at the other end. A weight of 9 kg produces a fundamental frequency of 750 Hz .
(a) What is the velocity of the wave in the wire?
(b) If the weight is reduced to 4 kg, what is the velocity of the wave?

Solution: Fundamental frequency is given by
\(
f=\frac{v}{2 L}
\)
(a) \(L=100 \mathrm{~cm}, \quad \begin{aligned} f_1 & =750 \mathrm{~Hz} \\ v_1 & =2 L f_1=2 \times 100 \times 750 \\ & =150000 \mathrm{cms}^{-1}=1500 \mathrm{~ms}^{-1}\end{aligned}\)
(b) \(\quad v_1=\sqrt{\frac{T_1}{\mu}}\) and \(\quad v_2=\sqrt{\frac{T_2}{\mu}}\)
\(
\begin{aligned}
\frac{v_2}{v_1} & =\sqrt{\frac{T_2}{T_1}} \\
\frac{v_2}{1500} & =\sqrt{\frac{4}{9}} \\
v_2 & =1000 \mathrm{~ms}^{-1}
\end{aligned}
\)

2. Effect of pressure

The speed of sound in a gas, \(v=\sqrt{\frac{\gamma p}{\rho}}\). From this formula, it appears that \(v \propto \sqrt{p}\). But actually, it is not so.
Because \(\frac{p}{\rho}=\frac{R T}{M}=\) constant at constant temperature.
Hence, if the temperature of a gas remains constant, then there is no effect of pressure change on the speed of sound in the gas.

Example 13: A sample of oxygen at NTP has volume \(V\) and a sample of hydrogen at NTP has volume \(4 V\). Both the gases are mixed and the mixture is maintained at NTP. If the speed of sound in hydrogen at NTP is \(1270 \mathrm{~ms}^{-1}\), then calculate the speed of sound in the mixture.

Solution: \(\quad p_{\text {mix }}=\frac{p_{\mathrm{O}_2} V_{\mathrm{O}_2}+p_{\mathrm{H}_2} V_{\mathrm{H}_2}}{V_{\mathrm{O}_2}+V_{\mathrm{H}_2}}=\frac{16 \times V+1 \times 4 V}{V+4 V}=4\)
As the temperature remains the same, therefore pressure will remain constant.
\(
\begin{array}{ll}
\therefore & \frac{v_{\text {mix }}}{v_{\mathrm{H}_2}}=\sqrt{\frac{p_{\mathrm{H}_2}}{p_{\text {mix }}}}=\sqrt{\frac{1}{4}}=\frac{1}{2} \\
\Rightarrow & v_{\text {mix }}=\frac{v_{\mathrm{H}_2}}{2}=\frac{1270}{2}=635 \mathrm{~ms}^{-1}
\end{array}
\)

3. Effect of humidity

The density of moist air (i.e. air mixed with water vapour) is less than the density of dry air. It is clear from the formula, \(v=\sqrt{\gamma \rho / \rho}\) that the speed of sound in moist air is slightly greater than that in dry air. So, speed of sound increases with an increase in humidity.

4. Effect of frequency

The speed of sound in air is independent of its frequency. Sound waves of different frequencies travel with the same speed in air but their wavelengths in air are different.

Example 14: A sound wave propagating in air has a frequency of 4000 Hz. Calculate the percentage change in wavelength when the wavefront initially in a region, where \(T=27^{\circ} \mathrm{C}\), enters a region, where the temperature decreases to \(10^{\circ} \mathrm{C}\).

Solution: \(\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{273+10}{273+27}}=\sqrt{\frac{283}{300}}=0.97\)
As, frequency remains unchanged, so
\(
\frac{v_2}{v_1}=\frac{v \lambda_2}{v \lambda_1}=\frac{\lambda_2}{\lambda_1}=0.97 \quad(\because v=v \lambda)
\)
Percentage change in wavelength,
\(
\frac{\lambda_1-\lambda_2}{\lambda_1} \times 100=\left(1-\frac{\lambda_2}{\lambda_1}\right) \times 100=(1-0.97) \times 100=3 \%
\)

Relation between speed of sound in a gas and root mean square speed of molecules

If sound travels in a gaseous medium, then the velocity of sound in gas will be \(v_{\text {sound }}=\sqrt{\frac{\gamma R T}{M}}=\sqrt{\frac{\gamma p}{\rho}}\)
where, \(M\) is the molecular mass of the gas.
rms speed of gas molecule is given by, \(v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}\) or \(\sqrt{\frac{3 p}{\rho}}\)
So, \(\frac{v_{\text {rms }}}{v_{\text {sound }}}=\sqrt{\frac{3}{\gamma}} \text { or } v_{\text {sound }}=[\gamma / 3]^{1 / 2} v_{\text {rms }}\)
\(\because \quad \sqrt{\frac{\gamma}{3}}<1 \quad \Rightarrow \quad v_{\text {sound }}<v_{\text {rms }}\)
Hence, the speed of sound in a gas is less than its rms speed of molecules.

Example 15: Compute the ratio of the speed of sound in hydrogen gas to the rms speed of hydrogen molecules.

Solution: As hydrogen is a diatomic gas, \(\gamma=\frac{7}{5}\)
Speed of sound in hydrogen, \(\left(v_{\text {sound }}\right)_{\mathrm{H}}=\sqrt{\frac{\gamma_p}{\rho}}\)
rms speed of hydrogen molecules, \(\left(v_{\mathrm{rms}}\right)_{\mathrm{H}}=\sqrt{\frac{3 p}{\rho}}\)
\(
\therefore \quad \frac{\left(v_{\text {sound }}\right)_H}{\left(v_{\mathrm{rms}}\right)_{\mathrm{H}}}=\sqrt{\frac{\gamma}{3}}=\sqrt{\frac{7 / 5}{3}}=\sqrt{\frac{7}{15}}
\)

Example 16: Equation of a transverse wave travelling in a rope is given by \(y=5 \sin (4.0 t-0.02 x)\), where \(y\) and \(x\) are expressed in cm and time in seconds. Calculate
(i) the amplitude, frequency, velocity and wavelength of the wave.
(ii) the maximum transverse velocity speed and acceleration of a particle in the rope.

Solution: (i) Comparing the given equation with the standard equation of wave motion, \(y=A \sin \left(2 \pi f t-\frac{2 \pi}{\lambda} x\right)\)
where, \(A, f\) and \(\lambda\) are amplitude, frequency and wavelength, respectively.
Thus, amplitude, \(A=5 \mathrm{~cm}, 2 \pi f=4\)
\(\Rightarrow\) Frequency, \(f=\frac{4}{2 \pi}=0.637 \mathrm{~Hz}\)
Again, \(\frac{2 \pi}{\lambda}=0.02\)
\(\Rightarrow\) Wavelength, \(\lambda=\frac{2 \pi}{0.02}=100 \pi \mathrm{~cm}\)
Velocity of the wave, \(v=f \lambda=\frac{4}{2 \pi} \frac{2 \pi}{0.02}=200 \mathrm{~cm} \mathrm{~s}^{-1}\)
(ii) Transverse velocity of the particle,
\(
\begin{aligned}
u & =\frac{d y}{d t}=5 \times 4 \cos (4.0 t-0.02 x) \\
& =20 \cos (4.0 t-0.02 x)
\end{aligned}
\)
Maximum velocity of the particle \(=20 \mathrm{~cm} \mathrm{~s}^{-1}\)
Particle acceleration,
\(
a=\frac{d^2 y}{d t^2}=-20 \times 4 \sin (4.0 t-0.02 x)
\)
Maximum particle acceleration \(=80 \mathrm{~cm} \mathrm{~s}^{-2}\).

Example 17: Figure shows a snapshot of a sinusoidal travelling wave taken at \(t=0.3 \mathrm{~s}\). The wavelength is 7.5 cm and the amplitude is 2 cm. If the crest \(P\) was at \(x=0\) at \(t=0\), write the equation of travelling wave.

Solution: Given, \(A=2 \mathrm{~cm}, \lambda=7.5 \mathrm{~cm}\)
\(
\therefore \quad k=\frac{2 \pi}{\lambda}=0.84 \mathrm{~cm}^{-1}
\)
The wave has travelled a distance of 1.2 cm in 0.3 s. Hence, the speed of the wave,
\(
v=\frac{1.2}{0.3}=4 \mathrm{~cm} \mathrm{~s}^{-1}
\)
\(\therefore\) Angular frequency, \(\omega=(v)(k)=3.36 \mathrm{rad} \mathrm{s}^{-1}\)
Since the wave is travelling along positive \(x\)-direction and crest (maximum displacement) is at \(x=0\) at \(t=0\), we can write the wave equation as
\(
\begin{aligned}
& y(x, t)=A \cos (k x-\omega t) \\
& y(x, t)=A \cos (\omega t-k x) \quad[\because \cos (-\theta)=\cos \theta]
\end{aligned}
\)
Therefore, the desired equation is
\(
y(x, t)=2 \cos (0.84 x-3.36 t) \mathrm{cm}
\)

Example 18: The displacement of a standing wave on a string is given by
\(
y(x, t)=0.4 \sin (0.5 x) \cos (30 t)
\)
where, \(x\) and \(y\) are in centimetres.
(i) Find the frequency, amplitude and wave speed of the component waves.
(ii) What is the particle velocity at \(x=2.4 \mathrm{~cm}\) and \(t=0.8 \mathrm{~s}\) ?

Solution: (i) The given wave can be written as the sum of two component waves as
\(
y(x, t)=0.2 \sin (0.5 x-30 t)+0.2 \sin (0.5 x+30 t)
\)
The two-component waves are
\(
y_1(x, t)=0.2 \sin (0.5 x-30 t)
\)
(Travelling in positive \(x\)-direction)
and \(y_2(x, t)=0.2 \sin (0.5 x+30 t)\)
(Travelling in negative \(x\)-direction)
Now, \(\omega=30 \mathrm{rad} / \mathrm{s} \text { and } k=0.5 \mathrm{~cm}^{-1}\)
\(\therefore\) Frequency, \(f=\frac{\omega}{2 \pi}=\frac{15}{\pi} \mathrm{~Hz}\)
Amplitude, \(A=0.2 \mathrm{~cm}\)
and wave speed, \(v=\frac{\omega}{k}=\frac{30}{0.5}=60 \mathrm{~cm} \mathrm{~s}^{-1}\)
(ii) Particle velocity,
\(
\begin{aligned}
v_P(x, t) & =\frac{d y}{d t}=-12 \sin (0.5 x) \sin (30 t) \\
\therefore \quad v_p(x & =2.4 \mathrm{~cm}, t=0.8 \mathrm{~s}) \\
& =-12 \sin (1.2) \sin (24) \\
& =10.12 \mathrm{~cm} \mathrm{~s}^{-1}
\end{aligned}
\)

Relation between the phase difference and path difference of a plane progressive wave

At any instant \(t\), if \(\phi_1\) and \(\phi_2\) are the phases of two particles whose distances from the origin are \(x_1\) and \(x_2\) respectively, then
\(
\begin{aligned}
\phi_1 & =\left(\omega t-k x_1\right) \text { and } \phi_2=\left(\omega t-k x_2\right) \\
\Rightarrow \quad \phi_1-\phi_2 & =k\left(x_2-x_1\right)
\end{aligned}
\)
Phase difference, \(\Delta \phi=\frac{2 \pi}{\lambda}\) (Path difference, \(\Delta x\) )

Example 19: If the phase difference between two waves is \(60^{\circ}\), then find the value of the path difference between them.

Solution: Given, \(\Delta \phi=60^{\circ}=\frac{\pi}{3}\)
As,\(\Delta \phi=\frac{2 \pi}{\lambda} \times \Delta x\)
\(
\Rightarrow \quad \Delta x=\frac{\Delta \phi}{2 \pi} \times \lambda=\frac{\lambda \times \pi}{3 \times 2 \pi} \Rightarrow \Delta x=\frac{\lambda}{6}
\)

Relation between phase difference and time difference of a plane progressive wave

If the phase of a particle distance \(x\) from the origin is \(\phi_1\) at time \(t_1\) and \(\phi_2\) at time \(t_2\), then \(\phi_1=\left(\omega t_1-k x\right)\) and
\(
\phi_2=\left(\omega t_2-k x\right) \Rightarrow \phi_1-\phi_2=\omega\left(t_1-t_2\right)
\)
Phase difference, \(\Delta \phi=\frac{2 \pi}{T}\) (time difference, \(\Delta t\) )

Example 20: The frequency of a progressive wave travelling in a medium is 40 Hz. Calculate the change in phase at a given place at 0.02 s.

Solution: Given, \(\Delta t=0.02 \mathrm{~s}, {f}=40 \mathrm{~Hz}\) and \(\Delta t=0.02 \mathrm{~s}\)
Time period, \(\quad T=\frac{1}{f}=\frac{1}{40} \mathrm{~s}\)
Phase difference, \(\Delta \phi=\frac{2 \pi}{T} \times \Delta t=\frac{2 \pi}{\left(\frac{1}{40}\right)} \times 0.02\)
\(
=2 \pi \times 40 \times \frac{2}{100}=\frac{8 \pi}{5} \mathrm{rad}
\)

Energy in Wave Motion

In a wave, many particles oscillate. In a sinusoidal wave, these oscillations are simple harmonic in nature. Each particle has some energy of oscillation. At the same time, energy transfer also takes place. Related to the energy of oscillation and energy transfer, there are three terms, namely, energy density \((u)\), power \((P)\) and intensity \((I)\).

Energy density \((u)\)

The energy of oscillation per unit volume is called energy density \((u)\). Its SI unit is \(\mathrm{J} / \mathrm{m}^3\). In the case of SHM, the energy of oscillation (of a single particle) is \(E=\frac{1}{2} m \omega^2 A^2\). In a sinusoidal wave, each particle of the string oscillates simple harmonically.
\(
\begin{aligned}
&\text { Therefore, }\\
&\text { Energy density }=\frac{\text { energy of oscillation }}{\text { volume }} \text { or } u=\frac{E}{V}=\frac{\frac{1}{2} m \omega^2 A^2}{V}
\end{aligned}
\)

Power \((P)\)

The instantaneous rate at which energy is transferred along the string, if we consider a transverse wave on a string, is called power.
In unit time, the wave will travel a distance \(v\). If \(S\) be the area of cross-section of the string, then the volume of this length would be \(S V\). Thus,
Power, \(\quad P=\) Energy density \(\times\) Volume \(=\frac{1}{2} \rho \omega^2 A^2 \times S V\)
\(
P=\frac{1}{2} \rho \omega^2 A^2 S V
\)

Intensity (I)

Energy transferred per unit cross-sectional area per unit time is called intensity. Thus,
\(
I=\frac{\text { Energy transferred }}{(\text { time })(\text { cross – sectional area })}=\frac{\text { Power }}{\text { cross – sectional area }}=\frac{P}{S}
\)
\(
I=\frac{1}{2} \rho \omega^2 A^2 v
\)

Note: \(\text { The intensity of sound waves is given by } I=\frac{p_{\max }^2}{2 \rho v}\)
where \(p_{\max }\) is the maximum change of pressure in the medium.
The intensity of waves emitting in all directions due to a point source varies inversely as the square of the distance \((r)\).
i.e. \(\quad I=\frac{P}{4 \pi r^2} \quad\) or \(\quad l \propto \frac{1}{r^2}\)
The intensity of waves from a linear source varies inversely as the distance \((r)\) perpendicular to the source, i.e. \(l \propto \frac{1}{r}\)

Example 21: The faintest sound that the human ear can detect at a frequency of 1 kHz corresponds to an intensity of about \(10^{-12} \mathrm{Wm}^{-2}\). Determine the pressure amplitude and the maximum displacement associated with this sound, assuming the density of the air \(=1.3 \mathrm{~kg} \mathrm{~m}^{-3}\) and velocity of sound in the air \(=332 \mathrm{~ms}^{-1}\).

Solution: Intensity of sound wave, \(I=\frac{p^2}{2 \rho v}\)
\(
\begin{aligned}
\Rightarrow \quad p & =\sqrt{I \times 2 \rho v}=\sqrt{10^{-12} \times 2 \times 1.3 \times 332} \\
& =2.94 \times 10^{-5} \mathrm{Nm}^{-2}
\end{aligned}
\)
Now, \(\quad p=\rho v \omega A\)
\(
\begin{aligned}
\Rightarrow \quad A & =\frac{p}{\rho v \omega}=\frac{2.94 \times 10^{-5}}{1.3 \times 332 \times 2 \pi \times 10^3} \\
& =1.1 \times 10^{-11} \mathrm{~m}
\end{aligned}
\)

Example 22: For a person with normal hearing, the faintest sound that can be heard at a frequency of 400 Hz has a pressure amplitude of about \(6.0 \times 10^{-5} \mathrm{~Pa}\). Calculate the corresponding intensity (in \(\mathrm{Wm}^{-2}\) ). (Take, speed of sound in air is \(344 \mathrm{~m} / \mathrm{s}\) and density of air is \(1.2 \mathrm{kgm}^{-3}\) )

Solution: Intensity, \(I=\frac{p_m^2}{2 \rho v}\) where, \(p_m\) is the pressure amplitude.
\(
I=\frac{\left(6.0 \times 10^{-5}\right)^2}{2 \times 1.2 \times 344}=4.4 \times 10^{-12} \mathrm{Wm}^{-2}
\)

Example 23: A stretched string is forced to transmit transverse waves by means of an oscillator coupled to one end. The string has a diameter of 4 mm . The amplitude of the oscillation is \(10^{-4} \mathrm{~m}\) and the frequency is 10 Hz . Tension in the string is 100 N and the mass density of wire is \(4.2 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\). Find
(i) the equation of the waves along the string,
(ii) the energy per unit volume of the wave,
(iii) the average energy flow per unit time across any section of the string, and
(iv) power required to drive the oscillator.

Solution: (i) Speed of transverse waves in the string, \(v=\sqrt{\frac{T}{\rho S}}\)
\(
(\mathrm{As}, \mu=\rho S)
\)
Substituting the given values in above equation, we get
\(
\begin{aligned}
v & =\sqrt{\frac{100}{\left(4.2 \times 10^3\right)(\pi / 4)\left(4.0 \times 10^{-3}\right)^2}}=43.53 \mathrm{~m} / \mathrm{s} \\
\omega & =2 \pi f=20 \pi \mathrm{rad} / \mathrm{s}=62.83 \mathrm{rad} / \mathrm{s} \quad(\because f=10 \mathrm{~Hz}) \\
k & =\omega / v=1.44 \mathrm{~m}^{-1}
\end{aligned}
\)
\(\therefore\) Equation of the waves along the string,
\(
y(x, t)=A \sin (k x-\omega t)=10^{-4} \sin (1.44 x-62.83 t) \mathrm{m}
\)

(ii) Energy per unit volume of the string,
\(
u=\text { energy density }=\frac{1}{2} \rho \omega^2 A^2
\)
Substituting all the values in above equation, we get
\(
\begin{aligned}
u & =\left(\frac{1}{2}\right)\left(4.2 \times 10^3\right)(62.83)^2\left(10^{-4}\right)^2 \\
& =8.29 \times 10^{-2} \mathrm{Jm}^{-3}
\end{aligned}
\)

(iii) Average energy flow per unit time,
\(
P=\text { Power }=\left(\frac{1}{2} \rho \omega^2 A^2\right)(S v)=(u)(S v)
\)
Substituting the values, we get
\(
\begin{aligned}
P & =\left(8.29 \times 10^{-2}\right)\left(\frac{\pi}{4}\right)\left(4.0 \times 10^{-3}\right)^2 \\
& =4.53 \times 10^{-5} \mathrm{Js}^{-1}
\end{aligned}
\)

(iv) The power required to drive the oscillator is obviously \(4.53 \times 10^{-5} \mathrm{~W}\).

Principle of superposition of waves

Resultant Amplitude and Intensity due to Coherent Sources

Two or more waves can travel simultaneously in a medium without affecting the motion of one another. The resultant displacement of each particle of the medium at any instant is equal to the vector sum of the individual displacements produced by the two waves. This principle is called principle of superposition.
If \(\mathbf{y}_1, \mathbf{y}_2, \mathbf{y}_3, \ldots\) are the displacements of particles at a particular time due to individual waves, then the resultant displacement is given by
\(
\mathbf{y}=\mathbf{y}_1+\mathbf{y}_2+\ldots
\)

Resultant Amplitude

Consider the superposition of two sinusoidal waves of the same frequency at a point. Let us assume that the two waves are travelling in the same direction with same velocity. The equation of the two waves reaching at a point can be written as
\(
\begin{array}{ll}
& y_1=A_1 \sin (k x-\omega t) \\
\text { and } & y_2=A_2 \sin (k x-\omega t+\phi)
\end{array}
\)

Assume a vector \({A}_1\) of length \(A_1\) to represent the amplitude of first wave. Another vector \(\mathbf{A}_2\) of length \(A_2\), making an angle \(\phi\) with \({A}_1\) represent the amplitude of second wave. The resultant of \({A}_1\) and \({A}_2\) represent the amplitude of resulting function \(y\). The angle \(\theta\) represents the phase difference between the resulting function and the first wave. The resultant displacement of the point, where the waves meet is

\(
\begin{aligned}
y= & y_1+y_2=A_1 \sin (k x-\omega t)+A_2 \sin (k x-\omega t+\phi) \\
= & A_1 \sin (k x-\omega t)+A_2 \sin (k x-\omega t) \cos \phi \\
& \quad+A_2 \cos (k x-\omega t) \sin \phi \\
= & \left(A_1+A_2 \cos \phi\right) \sin (k x-\omega t)+A_2 \sin \phi \cos (k x-\omega t)
\end{aligned}
\)
Substituting \(A_1+A_2 \cos \phi=A \cos \theta\) and \(A_2 \sin \phi=A \sin \theta\) in above equation, we get
\(
\begin{aligned}
& y=A \cos \theta \sin (k x-\omega t)+A \sin \theta \cos (k x-\omega t) \\
& y=A \sin (k x-\omega t+\theta)
\end{aligned}
\)
\(
\begin{aligned}
& A^2=\left(A_1+A_2 \cos \phi\right)^2+\left(A_2 \sin \phi\right)^2 \\
& A=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \phi} \dots(i)
\end{aligned}
\)
and \(\tan \theta=\frac{A \sin \theta}{A \cos \theta}=\frac{A_2 \sin \phi}{A_1+A_2 \cos \phi}\)

Example 24: Two harmonic waves are represented in SI units by
\(
y_1(x, t)=0.2 \sin (x-3.0 t)
\)
and \(y_2(x, t)=0.2 \sin (x-3.0 t+\phi)\)
(i) Write an expression for the sum \(y=y_1+y_2\) for \(\phi=\pi / 2 \mathrm{rad}\).
(ii) Suppose the phase difference \(\phi\) between the waves is unknown and the amplitude of their sum is 0.32 m, what is \(\phi\)?

Solution: (i) \(y=y_1+y_2=0.2 \sin (x-3.0 t)+0.2 \sin \left(x-3.0 t+\frac{\pi}{2}\right)\)
\(
=A \sin (x-3.0 t+\theta)
\)

Here, \(\quad A=\sqrt{(0.2)^2+(0.2)^2}=0.28 \mathrm{~m}\) and \(\theta=\frac{\pi}{4}\)
\(
\therefore \quad y=0.28 \sin \left(x-3.0 t+\frac{\pi}{4}\right)
\)
(ii) Since, the amplitude of the resulting wave is 0.32 m and
\(
\begin{aligned}
& A=0.2 \mathrm{~m}, \text { we get } \\
& 0.32=\sqrt{(0.2)^2+(0.2)^2+(2)(0.2)(0.2) \cos \phi}
\end{aligned}
\)
Solving this, we get \(\phi= \pm 1.29 \mathrm{rad}\)

Resultant Intensity

we have read that the intensity of a wave is given by
\(
I=\frac{1}{2} \rho \omega^2 A^2 v \quad \text { or } \quad I \propto A^2
\)
So, if \(\rho, \omega\) and \(v\) are same for the both interfering waves, then Eq. (i) can also be written as
\(
I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi \dots(ii)
\)
Here, the proportionality constant \(\left(I \propto A^2\right)\) cancels out on right-hand side and left-hand side.

Note: The special case of the above two equations is when the individual amplitudes (or intensities) are equal. or
\(
A_1=A_2=A_0 \text { (say) } \Rightarrow I_1=I_2=I_0 \text { (say) }
\)
In this case, Eqs. (i) and (ii) become
\(
\begin{aligned}
& A=2 A_0 \cos \frac{\phi}{2} \dots(iii) \\
& I=4 I_0 \cos ^2 \frac{\phi}{2} \dots(iv)
\end{aligned}
\)

Interference of waves

When two coherent waves of same frequency propagate in same direction and superimpose on each other, then the intensity of resultant wave becomes maximum at some points and minimum at some other points, this phenomenon of intensity variation is called interference.

If two sinusoidal waves \(S_1\) and \(S_2\) of same angular frequency \(\omega\) meet at a point \(B\), where the phase difference between them is \(\phi\) and path difference is \(\Delta x\) as shown in the figure, then according to the principle of superposition, the resulting amplitude is given by

\(
A^2=A_1^2+A_2^2+2 A_1 A_2 \cos \phi \dots(i)
\)
As the intensity of a wave is given by
\(
I=\frac{1}{2} \rho A^2 \omega^2 v, \text { i.e. } I \propto A^2
\)
So, if \(\rho, \omega\) and \(v\) are the same for both interfering waves, Eq. (i) can also be written as
\(
I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi \dots(ii)
\)

Constructive interference

From Eqs. (i) and (ii), we see that the resulting amplitude \(A\) and intensity \(I\) depends on the phase difference \(\phi\) between the interfering waves. If the resultant wave has maximum amplitude, then the interference is called constructive interference (these are the points where the resultant amplitude or intensity is maximum).

For maximum amplitude, \(\cos \phi=+1\)
\(
\phi=0,2 \pi, \ldots, 2 n \pi
\)
\(
\Delta x=0, \lambda, 2 \lambda, \ldots, n \lambda \quad\left[\text { as } \Delta x=\phi\left(\frac{\lambda}{2 \pi}\right)\right]
\)
Resultant amplitude, \(A=A_{\text {max }}=A_1+A_2\)
\(\therefore\) Resultant intensity, \(I=I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2\)
\(A_{\max }= \pm 2 a\)
\(I_{\max }=4 I_0\)

Destructive interference

If the resultant wave has minimum amplitude, then the interference is called destructive interference.
For minimum amplitude, \(\cos \phi=-1\)
\(
\phi=\pi, 3 \pi, \ldots, (2n-1) \pi
\)
\(
\Delta x=\frac{\lambda}{2}, \frac{3 \lambda}{2} \ldots(2 n-1) \frac{\lambda}{2} \quad\left[\operatorname{as} \Delta x=\phi\left(\frac{\lambda}{2 \pi}\right)\right]
\)
Resultant amplitude, \(A=A_{\min }=A_1-A_2\)
\(\therefore\) Resultant intensity, \(I=I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\)
\(A_{\min }=0\)
\(I_{\min }=0\)

The graph shown below represents the variation of resultant intensity, \(I\) with \(\left(I_1+I_2\right)\), i.e. sum of individual intensities.

Points to remember: 

In interference
\(
\begin{aligned}
\frac{I_{\max }}{I_{\min }} & =\left(\frac{\sqrt{l_1}+\sqrt{I_2}}{\sqrt{l_1}-\sqrt{I_2}}\right)^2=\left(\frac{\sqrt{l_1 / I_2}+1}{\sqrt{I_1 / I_2}-1}\right)^2 \\
& =\left(\frac{A_1 / A_2+1}{A_1 / A_2-1}\right)^2=\left(\frac{A_1+A_2}{A_1-A_2}\right)^2=\left(\frac{A_{\max }}{A_{\min }}\right)^2
\end{aligned}
\)

Example 25: Two waves of equal frequencies have their amplitudes in the ratio of \(3: 5\). They are superimposed on each other. Calculate the ratio of maximum and minimum intensities of the resultant wave.

Solution: Given, \(\frac{A_1}{A_2}=\frac{3}{5}\)
\(
\therefore \quad \sqrt{\frac{I_1}{I_2}}=\frac{3}{5} \quad\left(\text { As, } I \propto A^2\right)
\)
Maximum intensity is obtained, where
\(
\cos \phi=1 \text { and } I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2
\)
Minimum intensity is obtained, where
\(
\cos \phi=-1 \text { and } I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2
\)
Hence, \(\quad \frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2=\left(\frac{\sqrt{I_1 / I_2}+1}{\sqrt{I_1 / I_2}-1}\right)^2\)
\(
=\left(\frac{3 / 5+1}{3 / 5-1}\right)^2=\frac{64}{4}=\frac{16}{1}
\)

Example 26: Two coherent sound sources are at distances \(x_1=0.2 \mathrm{~m}\) and \(x_2=0.48 \mathrm{~m}\) from a point. Calculate the intensity of the resultant wave at that point, if the frequency of each wave is \(f=400 \mathrm{~Hz}\) and the velocity of the wave in the medium is \(v=448 \mathrm{~ms}^{-1}\). The intensity of each wave is \(I_0=60 \mathrm{Wm}^{-2}\).

Solution: Path difference, \(\Delta x=x_2-x_1=0.48-0.2=0.28 \mathrm{~m}\)
Now, phase difference, \(\phi=\frac{2 \pi}{\lambda} \Delta x=\left(\frac{2 \pi f}{v}\right) \Delta x\)
\(
=\frac{2 \pi(400)(0.28)}{448}=\frac{\pi}{2}
\)
Now, the intensity of the resultant wave,
\(
I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi
\)
\(
\begin{aligned}
I & =I_0+I_0+2 I_0 \cos \left(\frac{\pi}{2}\right) \\
& =2 I_0=2(60)=120 \mathrm{Wm}^{-2}
\end{aligned}
\)

Example 27: The figure shows a tube structure in which a sound signal is sent from one end and received at the other end. The semicircular part has a radius of 20.0 cm. The frequency of the sound source can be varied electronically between 1000 and 4000 Hz. Find the frequencies at which maxima of intensity are detected. The speed of sound in air \(=340 \mathrm{~m} / \mathrm{s}\).

Solution: According to the question,

Path difference, \(\Delta x=\pi r-2 r=(\pi-2) r\) For maximum intensity,
\(
\Delta x=n \lambda
\)
\(
\Rightarrow \quad(\pi-2) r=n \lambda=\frac{n v}{f}
\)
\(
\begin{aligned}
\Rightarrow \quad f & =\frac{n v}{(\pi-2) r}=\frac{n \times 340}{(3.14-2) \times 0.2} \\
& =1491 n \mathrm{~Hz}
\end{aligned}
\)
\(
\text { If } n=1 \text {, then } f=1491 \mathrm{~Hz}
\)
If \(n=2\), then \(f=2982 \mathrm{~Hz}\)
If \(n=3\), then \(f=4473 \mathrm{~Hz}\)
The frequencies in the given range are 1491 Hz and 2982 Hz.

Example 28: Two sources are placed from a person \(P\) as shown in figure. The speed of sound in air is \(320 \mathrm{~m} / \mathrm{s}\). If sound signal is continuously varied from 500 Hz to 2500 Hz, for which frequency listener will hear minimum sound intensity?

Solution: Path difference, \(\Delta x=S_2 P-S_1 P=6.4-6.0=0.4 \mathrm{~m}\)
For minimum sound intensity,
\(
\Delta x=(2 n+1) \frac{\lambda}{2}=\frac{(2 n+1)}{2} \frac{v}{f}
\)
\(
\Rightarrow \quad f=\frac{(2 n+1) v}{2 \Delta x}=\frac{(2 n+1) 320}{2(0.4)}=400(2 n+1) \mathrm{Hz}
\)
where, \(n=0,1,2, \ldots\)
If \(\quad n=0\), then \(f=400 \mathrm{~Hz}\)
If \(\quad n=1\), then \(f=1200 \mathrm{~Hz}\)
If \(\quad n=2\), then \(f=2000 \mathrm{~Hz}\)
If \(n=3\), then \(f=2800 \mathrm{~Hz}\)
The frequencies in the given range are 1200 Hz and 2000 Hz.

Example 29: In interference, two individual amplitudes are \(A_0\) each and the intensity is \(I_0\) each. Find the resultant amplitude and intensity at a point, where
(a) the phase difference between two waves is \(60^{\circ}\)
(b) path difference between two waves is \(\frac{\lambda}{3}\).

Solution: (a) Substituting \(\phi=60^{\circ}\) in the equations,
\(
A=2 A_0 \cos \frac{\phi}{2}
\)
and \(I=4 I_0 \cos ^2 \frac{\phi}{2}\)
We get, \(A=\sqrt{3} A_0 \text { and } I=3 I_0\)

(b) Given, \(\Delta x=\frac{\lambda}{3}\)
\(\phi\) or \(\Delta \phi=\left(\frac{2 \pi}{\lambda}\right) \cdot \Delta x\)
\(
=\left(\frac{2 \pi}{\lambda}\right)\left(\frac{\lambda}{3}\right)=\frac{2 \pi}{3} \text { or } 120^{\circ}
\)
\(
\begin{aligned}
&\text { Now, substituting } \phi=120^{\circ} \text { in the above two equations we get }\\
&A=A_0 \text { and } I=I_0
\end{aligned}
\)

Example 30: In interference, \(\frac{I_{\max }}{I_{\min }}=\alpha\), find
(a) \(\frac{A_{\max }}{A_{\min }}\)
(b) \(\frac{A_1}{A_2}\)
(c) \(\frac{I_1}{I_2}\)

Solution: (a) \(\frac{A_{\max }}{A_{\min }}=\sqrt{\frac{I_{\max }}{I_{\min }}}=\sqrt{\alpha}\)
(b) \(\frac{A_{\max }}{A_{\min }}=\sqrt{\alpha}=\frac{A_1+A_2}{A_1-A_2}=\frac{A_1 / A_2+1}{A_1 / A_2-1}\)
Solving this equation, we get
\(
\frac{A_1}{A_2}=\frac{\sqrt{\alpha}+1}{\sqrt{\alpha}-1}
\)
(c) \(\frac{I_1}{I_2}=\left(\frac{A_1}{A_2}\right)^2\left(\frac{\sqrt{\alpha}+1}{\sqrt{\alpha}-1}\right)^2\)

Reflection Of Waves

When a progressive wave travelling through a medium reaches a rigid boundary, it gets reflected and this phenomenon is called reflection of waves. If the wave is reflected from a free end, then the reflected wave has the same phase as that of the incident wave and in the case of a hard boundary, the reflected wave is out of phase (\(\text { phase change of } \pi\)) with respect to the incident wave. If the equation of incident travelling wave is
\(
y_i(x, t)=a \sin (k x-\omega t)
\)
At a rigid boundary, the reflected wave is given by
\(
\begin{aligned}
y_r(x, t) & =a \sin (k x-\omega t+\pi) . \\
& =-a \sin (k x-\omega t)
\end{aligned}
\)
At an open boundary, the reflected wave is given by
\(
\begin{aligned}
y_r(x, t) & =a \sin (k x-\omega t+0) . \\
& =a \sin (k x-\omega t)
\end{aligned}
\)
Clearly, at the rigid boundary, \(y=y_i+y_r=0\) at all times. The reflection of waves travelling along a stretched string and reflected by a rigid boundary is shown below.

Echo

Multiple reflection of sound is called an echo. e.g. Reflection of sound from hill or cliff.

If the distance of the reflector from the source is \(d\), then
\(
2 d=v t
\)
Hence, \(v=\) speed of sound and \(t=\) the time of echo.
\(
\therefore \quad d=\frac{v t}{2}
\)

Example 31: The echo of a gunshot is heard \(5 s\) after it is fired. Calculate the distance of the surface which reflects the sound. (Take, velocity of sound is \(332 \mathrm{~ms}^{-1}\) )

Solution: Distance of the surface which reflects the sound,
\(
d=\frac{v \times t}{2}=\frac{332 \times 5}{2}=166 \times 5=830 \mathrm{~m}
\)

Example 32: (i) An engine approaches a hill with a constant speed. When it is at a distance of 0.8 km, it blows a whistle whose echo is heard by the driver after 4 s.
If the speed of the engine in the air is \(330 \mathrm{~m} / \mathrm{s}\). Calculate the speed of the engine. (ii) A person standing between two parallel hills fires a gun. He hears the first echo after 1.5 s and the second after 2.5 s. If the speed of sound is \(332 \mathrm{~m} / \mathrm{s}\), calculate the distance between the hills. When will he hear the third echo?

Solution: (i) The given situation is shown in the following figure.

Distance travelled by sound when it again meets the person
\(
\begin{aligned}
& =800+(800-x) \\
& =1600-u t=1600-u \times 4
\end{aligned}
\)
Now, \(\frac{1600-4 u}{v}=4\)
\(
\begin{aligned}
\Rightarrow \quad 1600-4 u & =4 v \Rightarrow 1600-4 u=4 \times 330 \\
4 u & =1600-1320 \\
4 u & =280 \Rightarrow u=70 \mathrm{~ms}^{-1}
\end{aligned}
\)
(ii) The given situation is shown in the following figure.

where \(d\) is the distance between two parallel hills and \(x\) is the distance between the man and the hill in front of it (ie. hill 1).
\(\because\) The person hears the first echo after 1.5 s,
\(
1.5=\frac{2 x}{v} \dots(i)
\)
For second echo, \(\quad 2.5=\frac{2(d-x)}{v} \dots(ii)\)
From Eqs. (i) and (ii), we get
\(
4=\frac{2 d}{v} \Rightarrow d=2 v=2 \times 332=664 \mathrm{~m}
\)

For the third echo, the sound will be reflected by one hill and then by another hill, and then to person
\(
=\frac{2 d}{v}=\frac{2 \times 664}{332}=4 \mathrm{~s}
\)

Standing/stationary waves

When two harmonic waves of equal frequency and amplitude travelling through a medium (say, string) in opposite directions superimpose each other, we get stationary waves. Suppose the two waves are (assume \(\phi=0\))
\(
\begin{aligned}
& y_1(x, t)=A \sin (k x-\omega t) \\
& y_2(x, t)=A \sin (k x+\omega t)
\end{aligned}
\)
By the principle of superposition, the resultant wave will be
\(
y(x, t)=y_1(x, t)+y_2(x, t)
\)
\(
=A[\sin (k x-\omega t)+\sin (k x+\omega t)]
\)
Using the familiar trigonometric identity \(\operatorname{Sin}(A+B)+\operatorname{Sin}(A-B)=2 \sin A \cos B\) we get,
\(
y(x, t)=2 A \sin k x \cos \omega t \dots(i)
\)
Or \(y(x, t)=2 A \sin \left(\frac{2 \pi x}{\lambda}\right) \cos \left(\frac{2 \pi t}{T}\right) \dots(ii)\)

From Eqs. (i) and (ii), it is clear that the amplitude of the resultant wave depends on the position \(x\) (resultant amplitude \(=2 A \sin k x\) ). The points at which the amplitude is zero are called nodes. At nodes, there is no motion. The points at which the amplitude is maximum are called antinodes. Also, the distance between successive nodes or antinodes is \(\frac{\lambda}{2}\), whereas the distance between adjacent node and antinode is \(\frac{\lambda}{4}\).

Position of nodes:

Nodes are the points on the string, where the amplitude of the oscillation of constituents is zero.
i.e. \(\sin k x=0\)
\(
k x=n \pi \quad(\text { where }, n=0,1, \ldots)
\)
\(
\frac{2 \pi}{\lambda} x=n \pi \quad x=\frac{n \lambda}{2}
\)

Position of antinodes

Antinodes are the points on the string, where the amplitude of oscillation of the constituents is maximum.
For maximum amplitude,
\(
\sin k x= \pm 1 \Rightarrow k x=(2 n+1) \frac{\pi}{2}
\)
where, \(n=0,1,2, \ldots\)
\(
\begin{array}{r}
\frac{2 \pi}{\lambda} x=(2 n+1) \frac{\pi}{2} \\
x=(2 n+1) \frac{\lambda}{4}
\end{array}
\)

Note:
(i) The following equations also represent stationary waves
\(
\begin{aligned}
& y=2 a \sin (k x) \sin (\omega t) \quad \text { – } y=2 a \cos (k x) \sin (\omega t) \\
& y=2 a \cos (k x) \cos (\omega t)
\end{aligned}
\)

(ii) In stationary waves, the system cannot oscillate with any arbitrary frequency but frequencies are characterised by a set of natural frequencies called normal modes of oscillation.

(iii) The standing wave ratio (SWR) is defined as
\(
\frac{A_{\max }}{A_{\min }}=\frac{A_i+A_r}{A_i-A_r}
\)
where, \(A_i\) and \(A_r\) are the amplitudes of incident and reflected rays, respectively.
For \(100 \%\) reflection, \(\mathrm{SWR}=\infty\) and for no reflection, \(S W R=1\).

(iv) If the amplitude of the incident wave in medium- 1 is \(A_i\), it is partly reflected and partly transmitted at the boundary of two media-1 and 2. Wave speeds in two media are \(v_1\) and \(v_2\). If amplitudes of reflected and transmitted waves are \(A_r\) and \(A_t\), then
\(
A_r=\left(\frac{v_2-v_1}{v_2+v_1}\right) A_i \text { and } A_t=\left(\frac{2 v_2}{v_1+v_2}\right) A_i
\)
From the above two expressions, we can make the following conclusions :

Conclusion 1: If \(v_1=v_2\), then \(A_r=0\) and \(A_t=A_i\)
Basically \(v_1=v_2\) means both media are same from wave point of view. So, in this case there is no reflection \(\left(A_r=0\right)\), only transmission \(\left(A_t=A_i\right)\) is there.

Conclusion 2: If \(v_2<v_1\), then \(A_r\) comes out to be negative. Now, \(v_2<v_1\) means the second medium is denser. \(A_r\) in this case is negative means, there is a phase change of \(\pi\).

Conclusion 3: If \(v_2>v_1\), then \(A_t>A_i\). This implies that amplitude always increases as the wave travels from a denser medium to rarer medium (as \(v_2>v_1\), so second medium is rarer).

(v) Power At the boundary of two media, energy incident per second = energy reflected per second + energy transmitted per second.
or Power incident = power reflected + power transmitted or \(P_i=P_r+P_t\)

(vi) The intensity of a travelling wave is given by
\(
I=\frac{1}{2} \rho A^2 \omega^3 v
\)
i.e.\(I \propto A^2\)
So, we can write, \(\frac{I_1}{I_2}=\left(\frac{A_1}{A_2}\right)^2\), if \(\rho, \omega\) and \(v\) are same for two waves.
e.g. When an incident travelling wave is partly reflected and partly transmitted from a boundary, we can write
\(
\frac{I_i}{I_r}=\left(\frac{A_i}{A_r}\right)^2
\)
The incident and reflected waves are in the same medium, hence they have same values of \(\rho\) and \(v\).

Example 33: The equation given below represents \(a\) stationary wave set up in a medium
\(
y=12 \sin (4 \pi x) \sin (40 \pi t)
\)
where, \(y\) and \(x\) are in cm and \(t\) is in second. Calculate the amplitude, wavelength and velocity of the component waves.

Solution: Comparing the given equation with the stationary wave equation, \(y=2 A \sin k x \sin \omega t\), we get
\(
2 A=12 \mathrm{~cm}, k=4 \pi \mathrm{~cm}^{-1}
\)
and \(\omega=40 \pi \mathrm{rad} \mathrm{~s}^{-1}\)
Therefore, amplitude of the component wave \(=6 \mathrm{~cm}\)
As, \(k=4 \pi \Rightarrow \frac{2 \pi}{\lambda}=4 \pi\)
\(
\Rightarrow \quad \lambda=\frac{1}{2}=0.5 \mathrm{~cm}
\)
Further, the velocity of the wave,
\(
v=\frac{\omega}{k}=\frac{40 \pi}{4 \pi}=10 \mathrm{~cm} \mathrm{~s}^{-1}
\)

Example 34: The vibrations of a string fixed at both ends are described by the equation
\(
y=(5.00 \mathrm{~mm}) \sin \left[\left(1.57 \mathrm{~cm}^{-1}\right) x\right] \sin \left[\left(314 \mathrm{~s}^{-1}\right) t\right]
\)
If the length of the string is 10.0 cm, locate the nodes and the antinodes. How many loops are formed in the vibration?

Solution: The given equation can be written as
\(
\begin{aligned}
y & =5 \sin (1.57 x) \sin (314 t) \\
& =5 \sin \left(\frac{\pi x}{2}\right) \sin (100 \pi t) \dots(i)
\end{aligned}
\)
General equation of standing wave,
\(
y=2 A \sin (k x) \sin (\omega t) \dots(ii)
\)
Comparing both the equations, we get
\(
k=\frac{\pi}{2} \mathrm{~cm}^{-1}, \omega=100 \pi \mathrm{rad} \mathrm{~s}^{-1}
\)
\(
\begin{aligned}
\therefore \quad k & =\frac{2 \pi}{\lambda}=\frac{\pi}{2} \Rightarrow \lambda=4 \mathrm{~cm} \\
v & =\frac{\omega}{k}=\frac{100 \pi}{\pi / 2}=200 \mathrm{~cm} \mathrm{~s}^{-1}=2 \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}
\)
\(\therefore\) Length, \(L=n \frac{\lambda}{2} \Rightarrow 10=n \times \frac{4}{2} \Rightarrow n=5\)
Hence, the string is vibrating in 5 loops.

Nodes are formed at \(L=\frac{n \lambda}{2}\), whereas antinodes are formed at \((2 n+1) \frac{\lambda}{2}\).
\(\therefore\) Nodes, \(x=0,2,4,6,8,10 \mathrm{~cm}\)
and antinodes, \(x=1,3,5,7,9 \mathrm{~cm}\)

Terms related to standing waves
Few important terms related to standing waves are given below: Fundamental tones, overtones, and harmonics of a sound source

When a sound source produces sound waves, it consists of mixture of many frequencies. This mixture of sound is called note. If produced sound contains only one frequency, then it is called tone. Tone of minimum frequency is called fundamental tone.

Overtones: The tones which are having frequencies greater than fundamental tones are called overtones.

Harmonics: Sound produced from a source consists of frequencies which contain fundamental frequency as well as frequencies which are multiples of the fundamental frequency, called harmonics.
Odd multiples of the fundamental frequency are called odd harmonics. Even multiples of the fundamental frequency are called even harmonics.

VIBRATIONS OF AIR COLUMNS IN ORGAN PIPES

Organ pipes are musical instruments which are used to produce musical sound by blowing air into the pipe. This pipe is a cylindrical tube which may be open at one end called closed organ pipe [Fig(a)] or open at both ends called open organ pipe [Fig(b)].

If the air in the pipe at its open end is made to vibrate, the longitudinal wave is produced. This wave travels along the pipe towards its far end and is reflected back. Thus, due to the superposition of incident and reflected waves, stationary waves are formed in pipe.

Standing waves in a closed organ pipe

Closed organ pipes are those which are closed at one end and open at the other. A glass tube partially filled with water is an example of such a system. A node formed is at closed end and an antinode is formed at open end. If the length of air column is \(L\), then the closed end can be denoted by \(x=0\), while the open end can be taken as \(x=L\).

Position of Nodes

We know standing wave equation \(y=2 A \sin (k x) \cos (\omega t)\)
Nodes are the points of zero oscillation.
\(
\begin{aligned}
\sin k x & =0 \\
k x & =n \pi \\
\frac{2 \pi}{\lambda} x & =n \pi \\
x & =\frac{n \lambda}{2}
\end{aligned}
\)
where,\(n=0,1,2, \ldots\)

Position of antinodes

Antinodes are the points of maximum displacement.
\(
\therefore \quad \sin k x= \pm 1 \Rightarrow k x=(2 n-1) \frac{\pi}{2}
\)
\(
x=(2 n-1) \frac{\lambda}{4} \quad\left(\because k=2 \frac{\pi}{\lambda}\right)
\)
where, \(\quad n=1,2, \ldots\)

Modes of vibration of air columns in a closed organ pipe

At the open end \(x=L\), an antinode is formed
\(
\begin{array}{ll}
\therefore & L=(2 n-1) \frac{\lambda}{4} \Rightarrow \lambda=\frac{4 L}{(2 n-1)} \\
\text { where, } & n=1,2, \ldots
\end{array}
\)
The corresponding frequencies are given by
\(
f=\frac{v}{\lambda} \Rightarrow f=\frac{(2 n-1) v}{4 L}
\)

(i) First mode
For \(\quad n=1, \quad \lambda=4 L\)
or \(\quad L=\frac{\lambda}{4}\) and \(f_1=\frac{v}{4 L}\)
where, \(f_1=\) fundamental frequency,
\(v=\) speed of sound in air and \(\quad L=\) length of pipe.

This mode is called fundamental mode or 1st harmonic.

(ii) Second mode
For \(n=2, L=\frac{3 \lambda}{4} \Rightarrow f_2=\frac{3 v}{4 L}\) or \(f_2=3 f_1\)

This mode is called 3rd harmonic or 1st overtone.

(iii) Third mode
For \(n=3, L=\frac{5 \lambda}{4}, f_3=\frac{5 v}{4 L} \Rightarrow f_3=5 f_1\)

This mode is called 5th harmonic or 2nd overtone.
From the above discussion, we can say that in a closed organ pipe, only odd harmonics are present.
Ratio of harmonics is \(f_1: f_2: f_3 \ldots=1: 3: 5 \ldots\)
Ratio of overtones \(=3: 5: 7\)…….
Also, \(f_n=(2 n-1) f_1\) and frequency of \(n\)th overtone,
\(
f_n=(2 n+1) f_1
\)

Example 35: Calculate the fundamental frequency of a closed organ pipe of length 66.4 cm at \(0^{\circ} \mathrm{C}\), if the velocity of sound at \(0^{\circ} \mathrm{C}\) is \(332 \mathrm{~ms}^{-1}\).

Solution: Fundamental frequency of a closed organ pipe,
\(
f_1=\frac{v}{4 L}=\frac{332}{4 \times 66.4} \times 100=125 \mathrm{~Hz}
\)

Example 36: For a certain organ pipe, three successive resonance frequencies are observed at 425, 595 and 765 Hz , respectively. Taking the speed of sound in air to be \(340 \mathrm{~ms}^{-1}\).
(i) Explain whether the pipe is closed at one end or open at both ends.
(ii) Determine the fundamental frequency and length of the pipe.

Solution: (i) The given frequencies are in the ratio \(5: 7: 9\). As the frequencies are odd multiples of 85 Hz , the pipe must be closed at one end.
(ii) Now, the fundamental frequency is the lowest one, i.e. 85 Hz.
\(
\therefore \quad 85=\frac{v}{4 l} \Rightarrow l=\frac{340}{4 \times 85}=1 \mathrm{~m}
\)

Standing waves in an open organ pipe

Open organ pipes are those which are open at both ends. Therefore, a displacement antinode or pressure node is formed at each end. The standing wave in an open organ pipe is given by
\(
y(x, t)=2 A \cos k x \cos \omega t
\)
If the length of air column is \(L\), then \(y=\) maximum at \(x=0\) and at \(x=L\).

Position of antinodes

For position of antinodes,
\(
\begin{gathered}
\cos k x= \pm 1 \Rightarrow k x=n \pi, \text { where } n=0,1,2,3, \ldots \\
\frac{2 \pi}{\lambda} x=n \pi \Rightarrow x=\frac{n \lambda}{2}
\end{gathered}
\)
Further at \(x=L\), an antinode is formed
\(
\Rightarrow \quad L=\frac{n \lambda}{2} \Rightarrow \lambda=\frac{2 L}{n}
\)

Modes of vibrations of air column in an open organ pipe

The corresponding frequencies are given by
\(
f=\frac{v}{\lambda} \Rightarrow f=\frac{n v}{2 L}
\)
where, \(n=1,2,3\),
(i) First mode
For \(n=1, \lambda_1=2 L \Rightarrow L=\frac{\lambda_1}{2} \Rightarrow f_1=\frac{v}{2 L}\)

This mode is called fundamental mode or 1st harmonic. 

(ii) Second mode
For \(n=2, \quad \lambda_2=\frac{2 L}{2} \Rightarrow L=\frac{2 \lambda_2}{2}\)
\(\Rightarrow f_2=\frac{2 v}{2 L} \quad\) or \(\quad f_2=2 f_1\)

This is called 2nd harmonic or 1st overtone.

(iii) Third mode
For \(n=3, \quad \lambda_3=\frac{2 L}{3} \Rightarrow L=\frac{3 \lambda_3}{2}\)
\(\Rightarrow \quad f_3=\frac{3 v}{2 L}\) or \(f_3=3 f_1\)

This is called 3rd harmonic or 2nd overtone.
On generalising, \(f_n=n f_1\)
From the above discussion, we can say that in open organ pipe, all (even and odd) harmonics are present.
Ratio of harmonics is \(f_1: f_2: f_3 \ldots=1: 2: 3 \ldots\) and ratio of overtones \(=2: 3: 4: 5 \ldots \ldots\).
Frequency of \(n\)th over tone is \(f_n=(n+1) f_1\).

Example 37: Calculate the frequency of 2nd harmonic in an open organ pipe of length 34 cm , if the velocity of sound is \(340 \mathrm{~ms}^{-1}\).

Solution: In an open organ pipe, the frequency of the second harmonic, \(f_2=2 \times \frac{v}{2 L}=\frac{2 \times 340}{2 \times 0.34}=1000 \mathrm{~Hz}\)

Example 38: Third overtone of a closed organ pipe is in unison with fourth harmonic of an open organ pipe. Find the ratio of the lengths of the pipes.

Solution: The third overtone of the closed organ pipe means the seventh harmonic.
Given,
\(
\begin{aligned}
& \left(f_7\right)_{\text {closed }}=\left(f_4\right)_{\text {open }} \\
& 7\left(\frac{v}{4 l_c}\right)=4\left(\frac{v}{2 l_o}\right) \quad \therefore \quad \frac{l_c}{l_o}=\frac{7}{8}
\end{aligned}
\)

Example 39: Consider the situation shown in the figure. The wire which has a mass of 5 g oscillates in its second harmonic and sets the air column in the tube into vibrations in its fundamental mode. Assuming that the speed of sound in air is 340 \(\mathrm{ms}^{-1}\), find the tension in the wire.

Solution: Frequency of wire, \(f_2=\frac{2}{2 l} \sqrt{\frac{T}{\mu}}=\frac{1}{l} \sqrt{\frac{T}{M / l}}\)
\(
\begin{aligned}
& =\frac{1}{0.5} \sqrt{\frac{T}{5 \times 10^{-3} / 0.5}} \\
& =2 \sqrt{100 T}=20 \sqrt{T}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Fundamental frequency of closed pipe, }\\
&\begin{array}{ll}
& f=\frac{v}{4 L}=\frac{340}{4 \times 1}=85 \\
\because & f_2=f_1 \Rightarrow 20 \sqrt{T}=85 \\
\text { Therefore, } & T=(4.25)^2=18.06 \mathrm{~N}
\end{array}
\end{aligned}
\)

Vibration of Stretched String (Normal Modes of a String)

In an unbounded continuous medium, there is no restriction on the frequencies or wavelengths of the standing waves. However, if the waves are confined in space, (e.g. when a string is tied at both ends) standing waves can be set up for a discrete set of frequencies or wavelengths. The system cannot oscillate with any arbitrary frequency (contrast this with a harmonic travelling wave) but is characterised by a set of natural frequencies or normal modes of oscillation.
Consider a string of definite length \(l\), rigidly held at both ends. When we set up a sinusoidal wave on such a string, it gets reflected from the fixed ends. By the superposition of two identical waves travelling in opposite directions, standing waves are established on the string. Here, the only requirement we have to satisfy is that the endpoints should be nodes, as these points cannot oscillate.

The nodes are permanently at rest. There may be any number of nodes in between or none at all, so that the wavelength associated with the standing waves can take many different values. The distance between adjacent nodes is \(\lambda / 2\), so that in a string of length \(l\) there must be exactly an integral number \(n\) of half wavelengths, \(\lambda / 2\). That is,
\(
\frac{n \lambda}{2}=l \text { or } \lambda=\frac{2 l}{n} \quad(\text { where }, n=1,2,3, \ldots)
\)

But \(\lambda=\frac{v}{f}\) and \(v=\sqrt{\frac{T}{\mu}}\), so that the natural frequencies of oscillation of the system are
\(
f=n\left(\frac{v}{2 l}\right)=\frac{n}{2 l} \sqrt{\frac{T}{\mu}}
\)

The smallest frequency \(f_1\) corresponds to the largest wavelength \((n=1), \lambda_1=2 l\).
\(
f_1=\frac{v}{2 l}
\)
This is called the fundamental frequency. The other standing wave frequencies are
\(
\begin{aligned}
& f_2=\frac{2 v}{2 l}=2 f_1 \\
& f_3=\frac{3 v}{2 l}=3 f_1 \text { and so on }
\end{aligned}
\)
These frequencies are called harmonics. Musicians sometimes call them overtones. Students are advised to remember these frequencies by name.
For example,
\(
\begin{aligned}
& f_1=\text { fundamental tone or first harmonic } \\
& f_2=2 f_1=\text { first overtone or second harmonic } \\
& f_3=3 f_1=\text { second overtone or third harmonic and so on. }
\end{aligned}
\)

Laws of transverse vibration of a stretched string

(i) Law of length:

For a given string under a tension \((T)\), the frequency of the vibrating string is inversely proportional to the vibrating length \((L)\) of the string.
i.e. \(f \propto \frac{1}{L} \Rightarrow f L=\) constant \(\Rightarrow f_1 L_1=f_2 L_2\)

(ii) Law of tension

The frequency of the uniform string of a given length is proportional to the square root of the tension \((T)\) in the string.
i.e. \(f \propto \sqrt{T} \Rightarrow \frac{f}{\sqrt{T}}=\) constant \(\Rightarrow \frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}}\)

(iii) Law of mass

For a given vibrating length and tension of the string, the frequency of the vibrating string is inversely proportional to the square root of the mass per unit length of the string.
i.e.
\(
f \propto \frac{1}{\sqrt{\mu}}
\)
Therefore, with an increase in mass of the string frequency decreases.
As, Mass per unit length of the string \(=\) Volume of unit length \(\times\) Density or
\(
\mu=\pi r^2 \rho
\)
where, \(\rho=\) density of the wire, \(r\) = radius of the wire
\(
\therefore \quad f \propto \frac{1}{\sqrt{\pi r^2 \rho}}
\)
Now, the law of mass can be categorised in two parts as follows
(a) Law of radius: If length \((L)\), density \((\rho)\), and tension \((T)\) of a vibrating wire are fixed, thenthe  frequency of vibration root of the wire is inversely proportional to its radius.
i.e.
\(
f \propto \frac{1}{r}
\)
(b) Law of density: If length \((L)\), radius \((r)\), and tension \((T)\) of a vibrating wire are fixed, then the frequency of vibration of the wire is inversely proportional to the square root of its density.
i.e.
\(
f \propto \frac{1}{\sqrt{\rho}}
\)

Example 40: A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz . The mass of the wire is \(3.5 \times 10^{-2} \mathrm{~kg}\) and its linear mass density is \(4 \times 10^{-2} \mathrm{kgm}^{-1}\). What is the speed of transverse wave on the string and tension in the wire?

Solution: Length of wire \(=\frac{\text { Mass of wire }}{\text { Linear density }}=\frac{3.5 \times 10^{-2}}{4 \times 10^{-2}}=0.875 \mathrm{~m}\)
Now, \(f=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}\)
\(\therefore\) Tension in the wire, \(T=4 f^2 l^2 \mu\)
\(
=4(45)^2(0.875)^2\left(4 \times 10^{-2}\right)=248.1 \mathrm{~N}
\)
Speed of transverse wave, \(v=f \lambda=f(2 l)\)
\(
=45 \times 2 \times 0.875=78.75 \mathrm{~ms}^{-1}
\)

Example 41: A string on a musical instrument is 50 cm long and its fundamental frequency is 270 Hz. Calculate the length of the string, if the desired frequency of 1000 Hz is to be produced.

Solution: As, \(\Rightarrow \quad l_2=l_1\left(\frac{f_1}{f_2}\right)\)
\(
\begin{aligned}
f & \propto \frac{1}{l} \Rightarrow \frac{l_2}{l_1}=\frac{f_1}{f_2} \\
l_2 & =l_1\left(\frac{f_1}{f_2}\right) \\
& =50 \times\left(\frac{270}{1000}\right)=13.5 \mathrm{~cm}
\end{aligned}
\)

Example 42: The tension in a piano wire is 10 N. What should be the tension in the wire to produce a note of double the frequency.

Solution: 

\(
\begin{array}{rlrl}
\text { As, } & & f \propto \sqrt{T} \\
\therefore & \frac{f_1}{f_2}=\frac{\sqrt{T_1}}{\sqrt{T_2}}
\end{array}
\)
\(
\begin{array}{ll}
\Rightarrow & \frac{f}{2 f}=\sqrt{\frac{10}{T_2}} \\
\Rightarrow & T_2=40 \mathrm{~N}
\end{array}
\)

Standing waves in a string fixed at both ends

Consider a stretched string having length \(L\) fixed at both ends. Let the one end be fixed at \(x=0\) while the other one at \(x=L\), and at \(x=0\) and \(x=L\), nodes will be obtained as these ends are fixed.
As,
\(
L=\frac{n \lambda}{2}, n=1,2,3, \ldots
\)
\(\Rightarrow\) Possible wavelengths of stationary waves,
\(
\lambda=\frac{2 L}{n}, n=1,2,3, \ldots
\)
and corresponding frequencies are
\(
f=\frac{n v}{2 L}, n=1,2,3, \ldots
\)
Here, \(v\) is the speed of the wave in the givens medium. This relation gives the expression for the natural frequency of vibration.

Modes of vibration

The manner in which the string vibrates and gives rise to a standing wave is known as the mode of vibration of the string.

First mode of vibration

The mode of vibration in which string will vibrate in one segment or the lowest possible natural frequency [as shown in Fig. (a)] is called fundamental mode or first harmonic.
As, \(\quad L=\frac{n \lambda}{2}\)
For first harmonic, put \(n=1\)
\(
\therefore \quad L=\frac{\lambda_1}{2} \text { and } f_1=\frac{v}{2 L}
\)

Second mode of vibration

The mode of vibration in which string will vibrate in two segments [as shown in Fig. (b)] is called second harmonic or first overtone.
For the second harmonic, put
\(
\begin{array}{ll}
& n=2 \text { in } L=\frac{n \lambda}{2} . \\
\therefore & L=\lambda_2 \\
\text { and } & f_2=\frac{2 v}{2 L} \\
\text { or } & f_2=2 f_1 \\
\text { As, } & f=\frac{n v}{2 L}
\end{array}
\)
If \(n=3\), the frequency, \(f=\frac{3 v}{2 L}\) or \(f=3 f_1\) is called third harmonic and so on.
The figure below shows the first four harmonics of a stretched string fixed at both ends.

The vibration of a string will be a superposition of different modes. In this case, \(f_1: f_2: f_3=1: 2: 3 \ldots\), i.e. all harmonics (odd as well as even) can be produced.
Note: Musical instruments like sitar or violin are based on this principle.

Example 43: A string fixed at both ends has consecutive standing wave modes for which the distances between adjacent nodes are 18 cm and 16 cm respectively.
(i) What is the minimum possible length of the string?
(ii) If the tension is 10 N and the linear mass density is \(4 \mathrm{gm}^{-1}\), what is the fundamental frequency?

Solution:

Let \(l\) be the length of the string. Then,
\(
\begin{array}{r}
18 n=l \dots(i) \\
16(n+1)=l \dots(ii)
\end{array}
\)
From Eqs. (i) and (ii), we get
\(
n=8 \text { and } l=144 \mathrm{~cm}
\)
Therefore, the minimum possible length of the string can be 144 cm
(ii) For fundamental frequency, \(l=\lambda / 2\) or
\(
\lambda=2 l=288 \mathrm{~cm}=2.88 \mathrm{~m}
\)

Vibration of a string fixed at one end

Standing waves can also be produced on a string or rod which is fixed at one end and whose other end is free to move in a transverse direction. Here, in this case, we will have antinode at the free end and node at the fixed end.
Now, consider the equation of standing wave,
\(
y(x, t)=2 a \sin k x \cos \omega t
\)
As, we are having antinode at the end \(x=L\)
\(
\sin k L= \pm 1 \Rightarrow k L=(2 n+1)(\pi / 2)
\)
where, \(n=0,1,2, \ldots\)
\(
\begin{array}{rlrl}
& \frac{2 \pi}{\lambda} \times L  =(2 n+1) \frac{\pi}{2} \\
\Rightarrow & \frac{2 L}{\lambda}  =(2 n+1) \times \frac{1}{2} \\
& \text { or }  \frac{2 L f}{v}  =(2 n+1) \frac{1}{2}
\end{array}
\)
\(\therefore\) Frequency of the vibration,
\(
f=\frac{(2 n+1)}{4 L} v
\)
where, \(f\) is frequency and \(v\) is the speed of the wave.
The above frequencies are the normal frequencies of vibration. The fundamental frequency is obtained when \(n=0\).
\(
\therefore \quad f_0=\frac{v}{4 L} \quad \text { (first harmonic) }
\)
\(
\begin{array}{rlr}
f_1  =\frac{3 v}{4 L}=3 f_0 & \text { (third harmonic) } \\
f_2  =\frac{5 v}{4 L}=5 f_0 & \text { (fifth harmonic) } \\
f_3  =\frac{7 v}{4 L}=7 f_0 & \text { (seventh harmonic) } \\
f_0: f_1: f_2: f_3 \ldots=1: 3: 5: 7: \ldots &
\end{array}
\)
Here, we see that only odd harmonics are present (i.e. contains odd multiples of the fundamental frequency). The figure below shows shapes of the string.

Vibration of composite strings

Suppose two strings of different materials and lengths are joined end-to-end and tied between clamps as shown in the figure.

Now, after plucking, stationary waves are established only at those frequencies which matches with any one harmonic of both the independent strings \(S_1\) and \(S_2\). As the frequency of the wave in both strings must be same, so
\(
\frac{x}{2 l_1} \sqrt{\frac{T}{\mu_1}}=\frac{y}{2 l_2} \sqrt{\frac{T}{\mu_2}}
\)
\(
\frac{x}{y}=\frac{l_1}{l_2} \sqrt{\frac{\mu_1}{\mu_2}}=\frac{l_1}{l_2} \sqrt{\frac{\rho_1 A_1}{\rho_2 A_2}}
\)

Example 44: Figure shows an aluminium wire of length 60 cm joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40 N. The cross-sectional area of the steel wire is \(1.0 \mathrm{~mm}^2\) and that of the aluminium wire is \(3.0 \mathrm{~mm}^2\). What could be the minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node? The density of aluminium is \(2.6 \mathrm{~g} / \mathrm{cm}^3\) and that of steel is \(7.8 \mathrm{~g} / \mathrm{cm}^3\).

Solution: Let \(p\) th harmonic of steel wire coincide with \(q\) th harmonic of aluminium wire.
\(
\begin{aligned}
& n=\frac{p}{2 L_s} \sqrt{\frac{T}{\rho_s A_s}}=\frac{q}{2 L_a} \sqrt{\frac{T}{\rho_a A_a}} \\
& \frac{p}{q}=\frac{L_s}{L_a} \sqrt{\frac{\rho_s A_s}{\rho_a A_a}}=\frac{80}{60} \sqrt{\frac{7.8 \times 1}{2.6 \times 3}}=\frac{4}{3}
\end{aligned}
\)
4th harmonic of steel wire coincides with 3rd harmonic of aluminium wire, the minimum frequency of tuning fork can be determined by taking \(p=4\) or \(q=3\)
\(
\begin{aligned}
n_{\min } & =\frac{p}{2 L_s} \sqrt{\frac{T}{\rho_s A_s}} \\
& =\frac{4}{2 \times 0.8} \sqrt{\frac{40}{7.8 \times 10^3 \times 1 \times 10^{-6}}} \\
& =2.5 \times 71.6=179 \mathrm{~Hz}
\end{aligned}
\)

Beats

When two sound waves of the same amplitude travelling in the same direction with slightly different frequencies superimpose, the intensity varies periodically with time, this phenomenon is called beats.

A complete cycle of maximum intensity and minimum intensity is called one beat and the number of beats heard per second is called beat frequency. Beat frequency is equal to the difference of frequencies of two interfering waves.

Let us consider two harmonic waves of nearly equal angular frequencies \(\omega_1\) and \(\omega_2\). Further, let the waves have equal amplitude and zero phase difference. Let the initial phase angle for each wave be \(\phi=\frac{\pi}{2}\).
Then, at \(x=0\), the two waves can be given by
\(
S_1=a \cos \omega_1 t \text { and } S_2=a \cos \omega_2 t
\)
Symbol \(S\) refers to longitudinal displacement and not the transverse one.
Now, according to the principle of superposition, the resultant displacement is \(S=S_1+S_2\).
\(
S=a \cos \omega_1 t+a \cos \omega_2 t
\)

Using the identity
\(
\cos A+\cos B=2 \cos \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)
\)
\(
S=2 a \cos \left(\frac{\omega_1-\omega_2}{2}\right) t \cos \left(\frac{\omega_1+\omega_2}{2}\right) t
\)
If we write \(\omega_b=\left(\frac{\omega_1+\omega_2}{2}\right)\) and \(\omega_a=\left(\frac{\omega_1-\omega_2}{2}\right)\)
\(\therefore \quad S=\left[2 a \cos \omega_a t\right] \cos \omega_b t \dots(i)\)
Here, the resultant wave is a cosine function whose amplitude is \(\left(2 a \cos \omega_a t\right)\). The amplitude is maximum when \(\cos \omega_a t\) has the value +1 or -1, which happens twice in each repetition of the cosine function.
Therefore, the angular frequency of the beat,
\(
\omega_{\text {beat }}=2 \omega_a=2\left(\frac{\omega_1-\omega_2}{2}\right)
\)
\(
\begin{array}{lrl}
\Rightarrow & \omega_{\text {beat }} & =\omega_1-\omega_2 \\
\text { Also, } & \omega & =2 \pi f
\end{array}
\)
\(
\therefore \quad f_{\text {beat }}=f_1-f_2
\)
\(
\begin{aligned}
&\text { The reciprocal of beat frequency is beat period given by }\\
&T=\frac{1}{f}=\frac{1}{f_1-f_2}
\end{aligned}
\)

Essential condition for hearing beats

For beats to be audible, the difference in the frequency of the two sound waves should not exceed 10. If the difference is more than 10, we shall hear more than 10 beats per second. But due to the persistence of hearing, our ear is not able to distinguish between two sounds as separate, if the time interval between them is less than \((1 / 10)\) th of a second.
Hence, beats heard will not be distinct, if the number of beats produced per second is more than 10.

Example 45: Consider the two identical piano strings, each tuned exactly to the 420 Hz. The tension in any one of the strings is increased by \(2.0 \%\). If they are now struck, what is the beat frequency between the fundamentals of the two strings? (Take, length of the strings \(=65 \mathrm{~cm})\)

Solution: If \(f_1, v_1, T_1\) and \(f_2, v_2\) and \(T_2\) are the frequencies, velocities, and tensions in the first and second strings respectively, then
\(
\frac{v_2}{v_1}=\frac{v_2 / 2 L}{v_1 / 2 L}=\frac{f_2}{f_1}
\)
\(
\frac{f_2}{f_1}=\frac{\frac{1}{2 L} \sqrt{T_2 / \mu}}{\frac{1}{2 L} \sqrt{T_1 / \mu}}=\sqrt{\frac{T_2}{T_1}}
\)
Since, it is given that the tension in one string is \(2 \%\) larger than the other.
\(
\begin{aligned}
& T_2=T_1+\frac{2 T_1}{100}=1.02 T_1 \\
& \frac{f_2}{f_1}=\sqrt{\frac{1.02 T_1}{T_1}}=1.01
\end{aligned}
\)
Now, the frequency of the tightened string,
\(
\begin{aligned}
f_2 & =f_1(1.01)=1.01 \times 420 \\
& =424.2 \mathrm{~Hz}
\end{aligned}
\)
\(\therefore\) Beat frequency,
\(
\begin{aligned}
f_{\text {beat }} & =f_2-f_1=424.2-420 \\
& =4.2 \mathrm{~Hz}
\end{aligned}
\)

Example 46: Two strings 1 and 2 are taut between two fixed supports (as shown in the figure) such that the tension in both strings is the same. Mass per unit length of 2 is more than that of 1 . Explain which string is denser for a transverse travelling wave.

Solution: Speed of a transverse wave on a string, \(v=\sqrt{\frac{T}{\mu}}\)
\(
\begin{aligned}
v & \propto \frac{1}{\sqrt{\mu}} \\
\mu_2 & >\mu_1 \\
v_2 & <v_1
\end{aligned}
\)
i.e. Medium 2 is denser and medium 1 is rarer.

Example 47: One end of 12 m long rubber tube with a total mass of 0.9 kg is fastened to a fixed support. A cord attached to the other end passes over a pulley and supports an object with a mass of 5 kg. The tube is struck to a transverse blow at one end. Find the time required for the pulse to reach the other end. (Take, \(g=9.8 \mathrm{~ms}^{-2}\) )

Solution: Tension in the rubber tube \(A B\),

\(
T=m g \text { or } T=(5.0)(9.8)=49 \mathrm{~N}
\)
Mass per unit length of rubber tube,
\(
\mu=\frac{0.9}{12}=0.075 \mathrm{~kg} \mathrm{~m}^{-1}
\)
\(\therefore\) Speed of wave on the tube,
\(
v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{49}{0.075}}=25.56 \mathrm{~ms}^{-1}
\)
\(\therefore\) The required time,
\(
t=\frac{A B}{v}=\frac{12}{25.56}=0.47 \mathrm{~s}
\)

Example 48: A wave moves with speed \(300 \mathrm{~ms}^{-1}\) on a wire, which is under a tension of 500 N . Find how much the tension must be changed to increase the speed to \(312 \mathrm{~ms}^{-1}\).

Solution: Speed of a transverse wave on the wire, \(v=\sqrt{\frac{T}{\mu}} \dots(i)\)
Differentiating with respect to tension, we get
\(
\frac{d v}{d T}=\frac{1}{2 \sqrt{\mu T}} \dots(ii)
\)
On dividing Eq. (ii) by Eq. (i), we get
\(
\frac{d v}{v}=\frac{1}{2} \frac{d T}{T} \text { or } d T=(2 T) \frac{d v}{v}
\)
Substituting the values in the above equation, we get
\(
d T=\frac{(2)(500)(312-300)}{300}=40 \mathrm{~N}
\)
i.e. Tension should be increased by 40 N.

Example 49: A wire of uniform cross-section is stretched between two points 100 cm apart. The wire is fixed at one end and a weight is hung over a pulley at the other end. A weight of 9 kg produces a fundamental frequency of 750 Hz.
(i) What is the velocity of the wave in wire?
(ii) If the weight is reduced to 4 kg, what is the velocity of wave? What is the wavelength and frequency?

Solution: (i) Here, \(L=100 \mathrm{~cm}\) and \(f_1=750 \mathrm{~Hz}\)
\(
\begin{array}{ll}
\therefore v_1=2 L f_1=2 \times 100 \times 750 \\
=150000 \mathrm{cms}^{-1}=1500 \mathrm{~ms}^{-1}
\end{array}
\)
(ii) \(\because v_1=\sqrt{\frac{T_1}{\mu}}\) and \(v_2=\sqrt{\frac{T_2}{\mu}}\)
\(
\begin{array}{ll}
\therefore & \frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}} \Rightarrow \frac{v_2}{1500}=\sqrt{\frac{4}{9}} \\
\therefore & v_2=1000 \mathrm{~ms}^{-1}
\end{array}
\)
Wavelength, \(\lambda_2=2 L=200 \mathrm{~cm}=2 \mathrm{~m}\)
Frequency, \(f_2=\frac{v_2}{\lambda_2}=\frac{1000}{2}=500 \mathrm{~Hz}\)

Example 50: Length of a stretched wire is \(2 m\). It is oscillating in its fourth overtone mode. Maximum amplitude of oscillations is 2 mm . Find the amplitude of oscillation at a distance of 0.2 m from one fixed end.

Solution: Fourth overtone mode means five loops.

\(
\begin{aligned}
l & =5\left(\frac{\lambda}{2}\right) \\
\lambda & =\frac{2 l}{5} \\
& =\frac{2 \times 2}{5}=0.8 \mathrm{~m} \\
k & =\frac{2 \pi}{\lambda}=\frac{2 \pi}{0.8} \\
& =(2.5 \pi) \mathrm{m}^{-1}
\end{aligned}
\)
Now, \(P\) is a node. So, take \(x=0\) at \(P\).
Then, at a distance \(x\) from \(P\),
\(
\begin{aligned}
A_x & =A_{\max } \sin k x \\
& =(2 \mathrm{~mm}) \sin (2.5 \pi)(0.2) \\
& =(2 \mathrm{~mm}) \sin \left(\frac{\pi}{2}\right) \\
& =2 \mathrm{~mm}
\end{aligned}
\)

Example 51: A standing wave is formed by two harmonic waves, \(y_1=A \sin (k x-\omega t)\) and \(y_2=A \sin (k x+\omega t)\) travelling on a string in opposite directions. Mass density of the string is \(\rho\) and area of cross-section is s. Find the total mechanical energy between two adjacent nodes on the string.

Solution: The distance between two adjacent nodes is \(\frac{\lambda}{2}\) or \(\frac{\pi}{k}\).
\(\therefore\) Volume of string between two nodes will be
\(
\begin{aligned}
V & =(\text { area of cross-section }) \times(\text { distance between two nodes }) \\
& =(S)\left(\frac{\pi}{k}\right)
\end{aligned}
\)
Energy density (energy per unit volume) of a travelling wave is given by
\(
u=\frac{1}{2} \rho A^2 \omega^2
\)
A standing wave is formed by two identical waves travelling in opposite directions. Therefore, the energy stored between two nodes in a standing wave.
\(
\begin{aligned}
E & \left.=2 \text { [energy stored in a distance of } \frac{\pi}{k} \text { of a travelling wave }\right] \\
& =2 \text { (energy density) (volume) } \\
& =2\left(\frac{1}{2} \rho A^2 \omega^2\right)\left(\frac{\pi S}{k}\right)
\end{aligned}
\)
\(
E=\frac{\rho A^2 \omega^2 \pi S}{k}
\)

Resonance Tube

To observe the resonance phenomenon in an open-ended cylindrical tube. The resonance is a standing wave phenomenon in the air column and occurs when the column length is:
\(
\lambda / 4,3 \lambda / 4,5 \lambda / 4
\)
where \(\lambda\) is the sound wavelength.

Wavelength \(\quad \lambda=2\left(\ell_2-\ell_1\right)\)
End correction \(e=\frac{\ell_2-3 \ell_1}{2}\)

Intensity of sound in decibels

Sound level, \(\mathrm{SL}=10 \log _{10}\left(\frac{\mathrm{I}}{\mathrm{I}_0}\right)\)
Where \(\mathrm{I}_0=\) threshold of human ear \(=10^{-12} \mathrm{~W} / \mathrm{m}^2\)