So far we considered waves propagating in an unbounded medium. What happens if a pulse or a wave meets a boundary? If the boundary is rigid, the pulse or wave gets reflected. The phenomenon of echo is an example of reflection by a rigid boundary. If the boundary is not completely rigid or is an interface between two different elastic media, the situation is some what complicated. A part of the incident wave is reflected and a part is transmitted into the second medium. If a wave is incident obliquely on the boundary between two different media the transmitted wave is called the refracted wave. The incident and refracted waves obey Snell’s law of refraction, and the incident and reflected waves obey the usual laws of reflection.

Fig. 15.11 shows a pulse travelling along a stretched string and being reflected by the boundary. Assuming there is no absorption of energy by the boundary, the reflected wave has the same shape as the incident pulse but it suffers a phase change of \(\pi\) or \(180^{\circ}\) on reflection. This is because the boundary is rigid and the disturbance must have zero displacement at all times at the boundary. By the principle of superposition, this is possible only if the reflected and incident waves differ by a phase of \(\pi\), so that the resultant displacement is zero. This reasoning is based on boundary condition on a rigid wall. As the pulse arrives at the wall, it exerts a force on the wall. By Newton’s Third Law, the wall exerts an equal and opposite force on the string generating a reflected pulse that differs by a phase of \(\pi\).

If on the other hand, the boundary point is not rigld but completely free to move (such as in the case of a string tied to a freely moving ring on a rod), the reflected pulse has the same phase and amplitude (assuming no energy dissipation) as the incident pulse. The net maximum displacement at the boundary is then twice the amplitude of each pulse. An example of non- rigid boundary is the open end of an organ pipe.
To summarise, a travelling wave or pulse suffers a phase change of \(\pi\) on reflection at a rigid boundary and no phase change on reflection at an open boundary. To put this mathematically, let the incident travelling wave be
\(
y_2(x, t)=a \sin (k x-\omega t)
\)
At a rigid boundary, the reflected wave is given by
\(
\begin{aligned}
y_r(x, t) & =a \sin (k x-\omega t+\pi) . \\
& =-a \sin (k x-\omega t) \dots(15.35)
\end{aligned}
\)
At an open boundary, the reflected wave is given by
\(
\begin{aligned}
y_r(x, t) & =a \sin (k x-\omega t+0) . \\
& =a \sin (k x-\omega t) \dots(15.36)
\end{aligned}
\)
Clearly, at the rigid boundary, \(y=y_2+y_r=0\) at all times.

The rule about the inversion at reflection may be stated in terms of the wave velocity. The wave velocity is smaller for the heavier string \((v=\sqrt{F / \mu})\) (\(F\) is here called the tension on spring) and larger for the lighter string. The above observation may be stated as follows.

If a wave enters a region where the wave velocity is smaller, the reflected wave is inverted. If it enters a region where the wave velocity is larger, the reflected wave is not inverted. The transmitted wave is never inverted.

Standing Waves and Normal Modes

In a string, for example, a wave travelling in one direction will get reflected at one end, which in turn will travel and get reflected from the other end. This will go on until there is a steady wave pattern set up on the string. Such wave patterns are called standing waves or stationary waves. To see this mathematically, consider a wave travelling along the positive direction of \(x\)-axis and a reflected wave of the same amplitude and wavelength in the negative direction of \(x\)-axis. The equations of the two waves are given by
\(
\begin{aligned}
& y_1(x, t)=a \sin (k x-\omega t) \\
& y_2(x, t)=a \sin (k x+\omega t+\phi)
\end{aligned}
\)
These waves interfere to produce what we call standing waves. To understand these waves, let us discuss the special case when \(phi=0\).
The resultant wave on the string is, according to the principle of superposition:
\(
y(x, t)=y_1(x, t)+y_2(x, t)
\)
\(
=a[\sin (k x-\omega t)+\sin (k x+\omega t)]
\)
Using the familiar trignometric identity \(\operatorname{Sin}(A+B)+\operatorname{Sin}(A-B)=2 \sin A \cos B\)
we get, \(y(x, t)=2 a \sin k x \cos \omega t \dots(15.37)\)
Note that the terms in eq. 15.37 \(\mathrm{kx}\) and \(\omega t\) appear separately, not in the combination \(k x-\omega t\). The amplitude of this wave is \(2 a \sin k x\). Thus, in this wave pattern, the amplitude varies from point to point, but each element of the string oscillates with the same angular frequency \(\omega\) or time period. There is no phase difference between oscillations of different elements of the wave. The string as a whole vibrates in phase with differing amplitudes at different points. The wave pattern is neither moving to the right nor to the left. Hence, they are called standing or stationary waves. The amplitude is fixed at a given location but, as remarked earlier, it is different at different locations.

Nodes: The points at which the amplitude is zero (i.e., where there is no motion at all) are nodes.

Antinodes: The points at which the amplitude is the largest are called antinodes.

Fig. 15.12 shows a stationary wave pattern resulting from the superposition of two travelling waves in opposite directions.

Normal Modes

The most significant feature of stationary waves is that the boundary conditions constrain the possible wavelengths or frequencies of vibration of the system. 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. Let us determine these normal modes for a stretched string fixed at both ends.

First, from Eq. (15.37), the positions of nodes (where the amplitude is zero) are given by \(\sin k x=0\). which implies
\(
k x=n \pi ; n=0,1,2,3, \ldots
\)
Since, \(k=2 \pi / \lambda\), we get
\(
x=\frac{n \lambda}{2} ; n=0,1,2,3, \ldots(15.38)
\)
Clearly, the distance between any two successive nodes is \(\frac{\lambda}{2}\). In the same way, the positions of antinodes (where the amplitude is the largest) are given by the largest value of \(\sin\) \(k x\) :
\(|\sin k x|=1\)
which implies
\(k x=(n+1 / 2) \pi ; n=0,1,2,3, \ldots\)
With \(k=2 \pi / \lambda\), we get
\(
x=(n+1 / 2) \frac{\lambda}{2} ; n=0,1,2,3, \ldots(15.39)
\)
Again the distance between any two consecutive antinodes is \(\frac{\lambda}{2}\). Eq. (15.38) can be applied to the case of a stretched string of length \(L\) fixed at both ends. Taking one end to be at \(x=0\), the boundary conditions are that \(x=0\) and \(x=L\) are positions of nodes. The \(x=0\) condition is already satisfied. The \(x=L\) node condition requires that the length \(L\) is related to \(\lambda\) by
\(
L=n \frac{\lambda}{2} ; \quad n=1,2,3, \ldots(15.40)
\)
Thus, the possible wavelengths of stationary waves are constrained by the relation
\(
\lambda=\frac{2 L}{n}; \quad n=1,2,3, \ldots(15.41)
\)
with corresponding frequencies
\(
f=\frac{n v}{2 \mathrm{~L}}=\frac{n}{2 \mathrm{~L}} \sqrt{F / \mu}, \text { for } n=1,2,3, \ldots(15.42)
\)
We have thus obtained the natural frequencies – the normal modes of oscillation of the system. The lowest possible natural frequency of a system is called its fundamental mode or the first harmonic.

\(
f_0=\frac{1}{2 L} \sqrt{F / \mu}
\)

This is called the fundamental frequency of the string. The other possible frequencies of standing waves are integral multiples of the fundamental frequency. The frequencies given by equation (15.42) are called the natural frequencies, normal frequencies, or resonant frequencies. Here \(v\) is the speed of the wave determined by the properties of the medium. The \(n=2\) frequency is called the second harmonic; \(n=3\) is the third harmonic and so on. We can label the various harmonics by the symbol \(f_n(n=1\), \(2, \ldots)\).

The other natural frequencies with which standing waves can be formed on the string are
1st overtone, or 2nd harmonic:
\(
f_1=2 f_0=\frac{2}{2 L} \sqrt{F / \mu}
\)
2nd overtone, or 3rd harmonic:
\(
f_2=3 f_0=\frac{3}{2 L} \sqrt{F / \mu}
\)
3rd overtone, or 4th harmonic:
\(
f_3=4 f_0=\frac{4}{2 L} \sqrt{F / \mu}
\) etc.

An integral multiple of a frequency is called it’s harmonic. Thus, for a string fixed at both ends, all the overtones are harmonics of the fundamental frequency and all the harmonics of the fundamental frequency are overtones.

Example 1: A \(50 \mathrm{~cm}\) long wire of mass \(20 \mathrm{~g}\) supports a mass of \(1.6 \mathrm{~kg}\) as shown in figure below. Find the fundamental frequency of the portion of the string between the wall and the pulley. Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).

Solution:

The tension in the string is \(F=(1.6 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)=16 \mathrm{~N} \text {. }\)
The linear mass density is \(\mu=\frac{20 \mathrm{~g}}{50 \mathrm{~cm}}=0.04 \mathrm{~kg} \mathrm{~m}^{-1}\).
The fundamental frequency is
\(
\begin{gathered}
f_0=\frac{1}{2 L} \sqrt{F / \mu} \\
=\frac{1}{2 \times\left(0.4 \mathrm{~m}\right)} \sqrt{\frac{16 \mathrm{~N}}{0.04 \mathrm{~kg} \mathrm{~m}^{-1}}}=25 \mathrm{~Hz}
\end{gathered}
\)

Harmonics of vibrations of a stretched string

Fig. 15.13 shows the first six harmonics of a stretched string fixed at either end. Generally, the vlbration of a string will be a superposition of different modes; some modes may be more strongly excited and some less. Musical instruments like the sitar or violin are based on this principle. Where the string is plucked or bowed, determines which modes are more prominent than others.

Normal modes of an air column open at one end and closed at other end

Let us next consider normal modes of oscillation of an air column with one end closed and the other open. A glass tube partially filled with water illustrates this system. The end in contact with water is a node, while the open end is an antinode. At the node, the pressure changes are the largest, while the displacement is minimum (zero). At the open end – the antinode, it is just the other way – least pressure change and maximum amplitude of displacement. Taking the end in contact with water to be \(x=0\), the node condition (Eq. 15.38) is already satisfied. If the other end \(x=L\) is an antinode, Eq. (15.39) gives

\(L=\left(n+\frac{1}{2}\right) \frac{\lambda}{2}\), for \(n=0,1,2,3, \ldots\)
The possible wavelengths are then restricted by the relation :
\(
\lambda=\frac{2 L}{(n+1 / 2)}, \text { for } n=0,1,2,3, \ldots(15.43)
\)
The normal modes – the natural frequencies of the system are
\(
f=\left(n+\frac{1}{2}\right) \frac{v}{2 L} ; n=0,1,2,3, \ldots(15.44)
\)
The fundamental frequency corresponds to \(n=0\), \(L=\left(n+\frac{1}{2}\right) \frac{\lambda}{2}\), for \(n=0,1,2,3, \ldots\)
The possible wavelengths are then restricted by the relation :
\(
\lambda=\frac{2 L}{(n+1 / 2)}, \text { for } n=0,1,2,3, \ldots(15.43)
\)
The normal modes – the natural frequencies of the system are
\(
f=\left(n+\frac{1}{2}\right) \frac{v}{2 L} ; n=0,1,2,3, \ldots(15.44)
\)
The fundamental frequency corresponds to \(n=0\), and is given by \(f_0=\frac{v}{4 L}\). The higher frequencies are odd harmonics, i.e., odd multiples of the fundamental frequency : \(f_1=3 \frac{v}{4 L} =3f_0, f_2=5 \frac{v}{4 L}=5f_0\), etc.
Fig. 15. 14 shows the first six odd harmonics of the air column with one end closed and the other open. For a pipe open at both ends, each end is an antinode.

It is then easily seen that an open-air column at both ends generates all harmonics (See Fig. 15.15). The systems above, strings, and air columns can also undergo forced oscillations. If the external frequency is close to one of the natural frequencies, the system shows resonance.

Example 2: A pipe, \(30.0 \mathrm{~cm}\) long, is open at both ends. Which harmonic mode of the pipe resonates a \(1.1 \mathrm{~kHz}\) source? Will resonance with the same source be observed if one end of the pipe is closed? Take the speed of sound in air as \(330 \mathrm{~m} \mathrm{~s}^{-1}\).

Solution:

The first harmonic frequency is given by
\(
v_1=\frac{v}{\lambda_1}=\frac{v}{2 L} \quad \text { (open pipe) }
\)
where \(L\) is the length of the pipe. The frequency of its nth harmonic is:
\(
f_n=\frac{n v}{2 L}, \text { for } n=1,2,3, \ldots \text { (open pipe) }
\)
The first few modes of an open pipe are shown in Fig. 15. 15.
For \(L=30.0 \mathrm{~cm}, v=330 \mathrm{~m} \mathrm{~s}^{-1}\),
\(
f_{\mathrm{n}}=\frac{n 330\left(\mathrm{~m} \mathrm{~s}^{-1}\right)}{0.6(\mathrm{~m})}=550 \mathrm{~n} \mathrm{s}^{-1}
\)
Clearly, a source of frequency \(1.1 \mathrm{~kHz}\) will resonate at \(v_2\), i.e. the second harmonic.
Now if one end of the pipe is closed (Fig. 15.15), then the fundamental frequency is
\(
f_1=\frac{v}{\lambda_1}=\frac{v}{4 L} \text { (pipe closed at one end) }
\)
and only the odd-numbered harmonics are present:
\(f_3=\frac{3 v}{4 L}, f_5=\frac{5 v}{4 L}\), and so on.
For \(L=30 \mathrm{~cm}\) and \(v=330 \mathrm{~m} \mathrm{~s}^{-1}\), the fundamental frequency of the pipe closed at one end is \(275 \mathrm{~Hz}\) and the source frequency corresponds to its fourth harmonic. Since this harmonic is not a possible mode, no resonance will be observed with the source, the moment one end is closed.

Laws Of Transverse Vibrations Of a String: Sonometer

The fundamental frequency of vibration of a string fixed at both ends is given by the equation
\(
f_0=\frac{v}{2 L}=\frac{1}{2 L} \sqrt{F / \mu} .
\)
From this equation, one can immediately write the following statements known as “Laws of transverse vibrations of a string”.

Law of length:

The fundamental frequency of vibration of a string (fixed at both ends) is inversely proportional to the length of the string provided its tension and its mass per unit length remain the same. \(f \propto 1 / L\) if \(F\) and \(\mu\) are constants.

Law of tension:

The fundamental frequency of a string is proportional to the square root of its tension provided its length and the mass per unit length remain the same.
\(
f \propto \sqrt{F} \text { if } L \text { and } \mu \text { are constants. }
\)

Law of mass:

The fundamental frequency of \(a\) string is inversely proportional to the square root of the linear mass density, i.e., mass per unit length, provided the length and the tension remain the same.
\(f \propto \frac{1}{\sqrt{\mu}}\) if \(L\) and \(F\) are constants.

Example 3: In a sonometer experiment, resonance is obtained when the experimental wire has a length of \(21 \mathrm{~cm}\) between the bridges, and the vibrations are excited by a tuning fork of frequency \(256 \mathrm{~Hz}\). If a tuning fork of frequency \(384 \mathrm{~Hz}\) is used, what should be the length of the experimental wire to get the resonance?

Solution:

\(
\text { By the law of length, } l_1 v_1=l_2 v_2
\)
or,
\(
l_2=\frac{v_1}{v_2} l_1=\frac{256}{384} \times 21 \mathrm{~cm}=14 \mathrm{~cm} .
\)

Polarization Of Waves

Suppose a stretched string goes through a slit made in a cardboard which is placed perpendicular to the string (figure above ). If we take the \(X\)-axis along the string, the cardboard will be in \(Y\) – \(Z\) plane. Suppose the particles of the string are displaced in \(y\)-direction as the wave passes. If the slit in the cardboard is also along the \(Y\)-axis, the part of the string in the slit can vibrate freely in the slit and the wave will pass through the slit. However, if the cardboard is rotated by \(90^{\circ}\) in its plane, the slit will point along the \(Z\)-axis. As the wave arrives at the slit, the part of the string in it tries to move along the \(Y\)-axis but the contact force by the cardboard does not allow it. The wave is not able to pass through the slit. If the slit is inclined to the \(Y\)-axis at some other angle, only a part of the wave is transmitted and in the transmitted wave the disturbance is produced parallel to the slit.

If the disturbance produced is always along a fixed direction, we say that the wave is linearly polarized in that direction. The waves considered in this chapter are linearly polarized in \(y\)-direction. Similarly, if a wave produces displacement along the \(z\)-direction, its equation is given by \(z=A \sin \omega(t-x / v)\) and it is a linearly polarized wave, polarized in \(z\)-direction. Linearly polarized waves are also called plane polarized.

If each particle of the string moves in a small circle as the wave passes through it, the wave is called circularly polarized. If each particle goes in an ellipse, the wave is called elliptically polarized.

Finally, if the particles are randomly displaced in the plane perpendicular to the direction of propagation, the wave is called unpolarized.

A circularly polarized or unpolarized wave passing through a slit does not show the change in intensity as the slit is rotated in its plane. But the transmitted wave becomes linearly polarized in the direction parallel to the slit.

Example 4: The displacement of a particle of a string carrying a travelling wave is given by
\(
y=(3.0 \mathrm{~cm}) \sin 6.28(0.50 x-50 t),
\)
where \(x\) is in centimetre and \(t\) in second. Find (a) the amplitude, (b) the wavelength, (c) the frequency and (d) the speed of the wave.

Solution:

Comparing with the standard wave equation
\(
\begin{aligned}
y & =A \sin (k x-\omega t) \\
& =A \sin 2 \pi\left(\frac{x}{\lambda}-\frac{t}{T}\right)
\end{aligned}
\)
we see that,
amplitude \(=A=3.0 \mathrm{~cm}\),
wavelength \(=\lambda=\frac{1}{0.50} \mathrm{~cm}=2.0 \mathrm{~cm}\),
and the frequency \(=f=\frac{1}{T}=50 \mathrm{~Hz}\).
The speed of the wave is \(v=f \lambda\)
\(
\begin{aligned}
& =\left(50 \mathrm{~s}^{-1}\right)\left(2^{\cdot} 0 \mathrm{~cm}\right) \\
& =100 \mathrm{~cm} \mathrm{~s}^{-1}
\end{aligned}
\)

Example 5: The equation for a wave travelling in \(x\)-direction on a string is
\(
y=(3 \cdot 0 \mathrm{~cm}) \sin \left[\left(3 \cdot 14 \mathrm{~cm}^{-1}\right) x-\left(314 \mathrm{~s}^{-1}\right) t\right] .
\)
(a) Find the maximum velocity of a particle of the string.
(b) Find the acceleration of a particle at \(x=6.0 \mathrm{~cm}\) at time \(t=0.11 \mathrm{~s}\).

Solution:

(a) The velocity of the particle at \(x\) at time \(t\) is
\(
\begin{aligned}
v=\frac{\partial y}{\partial t} & =(3 \cdot 0 \mathrm{~cm})\left(-314 \mathrm{~s}^{-1}\right) \cos \left[\left(3^{\cdot} 14 \mathrm{~cm}^{-1}\right) x-\left(314 \mathrm{~s}^{-1}\right) t\right] \\
& =\left(-9 \cdot 4 \mathrm{~m} \mathrm{~s}^{-1}\right) \cos \left[\left(3 \cdot 14 \mathrm{~cm}^{-1}\right) x-\left(314 \mathrm{~s}^{-1}\right) t\right]
\end{aligned}
\)
The maximum velocity of a particle will be
\(
v=9.4 \mathrm{~m} \mathrm{~s}^{-1}
\)
(b) The acceleration of the particle at \(x\) at time \(t\) is
\(
\begin{aligned}
a=\frac{\partial v}{\partial t} & =-\left(9 \cdot 4 \mathrm{~m} \mathrm{~s}^{-1}\right)\left(314 \mathrm{~s}^{-1}\right) \sin \left[\left(3 \cdot 14 \mathrm{~cm}^{-1}\right) x-\left(314 \mathrm{~s}^{-1}\right) t\right] \\
& =-\left(2952 \mathrm{~m} \mathrm{~s}^{-2}\right) \sin \left[\left(3 \cdot 14 \mathrm{~cm}^{-1}\right) x-\left(314 \mathrm{~s}^{-1}\right) t\right] .
\end{aligned}
\)
The acceleration of the particle at \(x=6.0 \mathrm{~cm}\) at time \(t=0 \cdot 11 \mathrm{~s}\) is \(a=-\left(2952 \mathrm{~m} \mathrm{~s}^{-2}\right) \sin [6 \pi-11 \pi]=0\).

Example 6: A long string having a cross-sectional area \(0.80 \mathrm{~mm}^2\) and density \(12.5 \mathrm{~g} \mathrm{~cm}^{-3}\), is subjected to a tension of \(64 \mathrm{~N}\) along the \(X\)-axis. One end of this string is attached to a vibrator moving in transverse direction at a frequency of \(20 \mathrm{~Hz}\). At \(t=0\), the source is at a maximum displacement \(y=1.0 \mathrm{~cm}\). (a) Find the speed of the wave travelling on the string. (b) Write the equation for the wave. (c) What is the displacement of the particle of the string at \(x=50 \mathrm{~cm}\) at time \(t=0.05 \mathrm{~s}\)? (d) What is the velocity of this particle at this instant?

Solution:

(a) The mass of \(1 \mathrm{~m}\) long part of the string is
\(
\begin{aligned}
m & =\left(0.80 \mathrm{~mm}^2\right) \times(1 \mathrm{~m}) \times\left(12.5 \mathrm{~g} \mathrm{~cm}^{-3}\right) \\
& =\left(0.80 \times 10^{-6} \mathrm{~m}^3\right) \times\left(12.5 \times 10^3 \mathrm{kgm}^{-3}\right) \\
& =0.01 \mathrm{~kg} .
\end{aligned}
\)
The linear mass density is \(\mu=0.01 \mathrm{~kg} \mathrm{~m}^{-1}\). The wave speed is \(v=\sqrt{F / \mu}\)
\(
=\sqrt{\frac{64 \mathrm{~N}}{0.01 \mathrm{~kg} \mathrm{~m}^{-1}}}=80 \mathrm{~m} \mathrm{~s}^{-1} .
\)
(b) The amplitude of the source is \(A=1.0 \mathrm{~cm}\) and the frequency is \(v=20 \mathrm{~Hz}\). The angular frequency is \(\omega=2 \pi v=40 \pi \mathrm{s}^{-1}\). Also at \(t=0\), the displacement is equal to its amplitude, i.e., at \(t=0, y=A\). The equation of motion of the source is, therefore,
\(
y=(1.0 \mathrm{~cm}) \cos \left[\left(40 \pi \mathrm{s}^{-1}\right) t\right] \dots(i)
\)
The equation of the wave travelling on the string along the positive \(X\)-axis is obtained by replacing \(t\) with \(t-x / v\) in equation (i). It is, therefore,
\(
\begin{aligned}
y & =(1 \cdot 0 \mathrm{~cm}) \cos \left[\left(40 \pi \mathrm{s}^{-1}\right)\left(t-\frac{x}{v}\right)\right] \\
& =(1 \cdot 0 \mathrm{~cm}) \cos \left[\left(40 \pi \mathrm{s}^{-1}\right) t-\left(\frac{\pi}{2} \mathrm{~m}^{-1}\right) x\right], \dots(ii)
\end{aligned}
\)
where the value of \(v\) has been put from part (a).
(c) The displacement of the particle at \(x=50 \mathrm{~cm}\) at time
\(t=0.05 \mathrm{~s}\) is by equation (ii),
\(
\begin{aligned}
y & =(1.0 \mathrm{~cm}) \cos \left[\left(40 \pi \mathrm{s}^{-1}\right)(0.05 \mathrm{~s})-\left(\frac{\pi}{2} \mathrm{~m}^{-1}\right)(0.5 \mathrm{~m})\right] \\
& =(1.0 \mathrm{~cm}) \cos \left[2 \pi-\frac{\pi}{4}\right] \\
& =\frac{1.0 \mathrm{~cm}}{\sqrt{ } 2}=0.71 \mathrm{~cm} .
\end{aligned}
\)
(d) The velocity of the particle at position \(x\) at time \(t\) is, by equation (ii),
\(
v=\frac{\partial y}{\partial t}=-(1 \cdot 0 \mathrm{~cm})\left(40 \pi \mathrm{s}^{-1}\right) \sin \left[\left(40 \pi \mathrm{s}^{-1}\right) t-\left(\frac{\pi}{2} \mathrm{~m}^{-1}\right) x\right] \text {. }
\)
Putting the values of \(x\) and \(t\),
\(
\begin{aligned}
v & =-\left(40 \pi \mathrm{cm} \mathrm{s}^{-1}\right) \sin \left(2 \pi-\frac{\pi}{4}\right) \\
& =\frac{40 \pi}{\sqrt{2}} \mathrm{~cm} \mathrm{~s}^{-1} \approx 89 \mathrm{~cm} \mathrm{~s}^{-1} .
\end{aligned}
\)

Example 7: The speed of a transverse wave, going on a wire having a length \(50 \mathrm{~cm}\) and mass \(5^{\circ} 0 \mathrm{~g}\), is \(80 \mathrm{~m} \mathrm{~s}^{-1}\). The area of the cross-section of the wire is \(1.0 \mathrm{~mm}^2\) and its Young modulus is \(16 \times 10^{11} \mathrm{Nm}^{-2}\). Find the extension of the wire over its natural length.

Solution:

The linear mass density is
\(
\mu=\frac{5 \times 10^{-3} \mathrm{~kg}}{50 \times 10^{-2} \mathrm{~m}}=1^{\cdot} \cdot \times 10^{-2} \mathrm{~kg} \mathrm{~m}^{-1} .
\)
The wave speed is \(v=\sqrt{F / \mu}\).
Thus, the tension is \(F=\mu v^2\)
\(
=\left(1 \cdot 0 \times 10^{-2} \mathrm{~kg} \mathrm{~m}^{-1}\right) \times 6400 \mathrm{~m}^2 \mathrm{~s}^{-2}=64 \mathrm{~N} .
\)
The Young modulus is given by
\(
Y=\frac{F / A}{\Delta L / L}
\)
The extension is, therefore,
\(
\begin{aligned}
\Delta L & =\frac{F L}{A Y} \\
& =\frac{(64 \mathrm{~N})(0.50 \mathrm{~m})}{\left(1.0 \times 10^{-6} \mathrm{~m}^2\right) \times\left(16 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\right)}=0.02 \mathrm{~mm} .
\end{aligned}
\)

Example 8: A uniform rope of length \(12 \mathrm{~m}\) and mass \(6 \mathrm{~kg}\) hangs vertically from a rigid support. A block of mass \(2 \mathrm{~kg}\) is attached to the free end of the rope. A transverse pulse of wavelength \(0.06 \mathrm{~m}\) is produced at the lower end of the rope. What is the wavelength of the pulse when it reaches the top of the rope?

Solution:

As the rope is heavy, its tension will be different at different points. The tension at the free end will be \((2 \mathrm{~kg}) g\) and that at the upper end it will be \((8 \mathrm{~kg}) g\).
We have, \(v=f \lambda\)
or,
\(\sqrt{F / \mu}=f \lambda\)
or,
\(\sqrt{F} / \lambda=f \sqrt{\mu} \dots(i)\).
The frequency of the wave pulse will be the same everywhere on the rope as it depends only on the frequency of the source. The mass per unit length is also the same throughout the rope as it is uniform. Thus, by (i), \(\frac{\sqrt{ } F}{\lambda}\) is constant.
Hence, \(\quad \frac{\sqrt{(2 \mathrm{~kg}) g}}{0 \cdot 06 \mathrm{~m}}=\frac{\sqrt{(8 \mathrm{~kg}) g}}{\lambda_1}\),
where \(\lambda_1\) is the wavelength at the top of the rope. This gives \(\lambda_1=0.12 \mathrm{~m}\).

Example 9: Two waves passing through a region are represented by
\(
y=(1.0 \mathrm{~cm}) \sin \left[\left(3.14 \mathrm{~cm}^{-1}\right) x-\left(157 \mathrm{~s}^{-1}\right) t\right]
\)
\(
\text { and } \quad y=(1.5 \mathrm{~cm}) \sin \left[\left(1.57 \mathrm{~cm}^{-1}\right) x-\left(314 \mathrm{~s}^{-1}\right) t\right] \text {. }
\)
Find the displacement of the particle at \(x=4.5 \mathrm{~cm}\) at time \(t=5.0 \mathrm{~ms}\)

Solution:

According to the principle of superposition, each wave produces its disturbance independent of the other and the resultant disturbance is equal to the vector sum of the individual disturbances. The displacements of the particle at \(x=4.5 \mathrm{~cm}\) at time \(t=5.0 \mathrm{~ms}\) due to the two waves are,
\(
\begin{aligned}
& y_1=(1.0 \mathrm{~cm}) \sin [\left(3.14 \mathrm{~cm}^{-1}\right)(4.5 \mathrm{~cm}) \\
&\left.-\left(157 \mathrm{~s}^{-1}\right)\left(5.0 \times 10^{-3} \mathrm{~s}\right)\right]
\end{aligned}
\)
\(
=(1.0 \mathrm{~cm}) \sin \left[4.5 \pi-\frac{\pi}{4}\right]
\)
\(
=(1.0 \mathrm{~cm}) \sin [4 \pi+\pi / 4]=\frac{1.0 \mathrm{~cm}}{\sqrt{ } 2}
\)
and
\(
\begin{aligned}
& y_2=(1.5 \mathrm{~cm}) \sin [\left(1.57 \mathrm{~cm}^{-1}\right)(4.5 \mathrm{~cm}) \\
&\left.-\left(314 \mathrm{~s}^{-1}\right)\left(5.0 \times 10^{-3} \mathrm{~s}\right)\right]
\end{aligned}
\)
\(
\begin{aligned}
& =(1.5 \mathrm{~cm}) \sin \left[2 \cdot 25 \pi-\frac{\pi}{2}\right] \\
& =(1.5 \mathrm{~cm}) \sin [2 \pi-\pi / 4] \\
& =-(1.5 \mathrm{~cm}) \sin \frac{\pi}{4}=-\frac{1.5 \mathrm{~cm}}{\sqrt{2}} .
\end{aligned}
\)
The net displacement is
\(
y=y_1+y_2=\frac{-0.5 \mathrm{~cm}}{\sqrt{2}}=-0.35 \mathrm{~cm} .
\)

Example 10: The vibrations of a string fixed at both ends are described by the equation
\(
y=(5.00 \mathrm{~mm}) \sin \left[\left(1.57 \mathrm{~cm}^{-1}\right) x\right] \sin \left[\left(314 \mathrm{~s}^{-1}\right) t\right]
\)
(a) What is the maximum displacement of the particle at \(x=5.66 \mathrm{~cm}\)? (b) What are the wavelengths and the wave speeds of the two transverse waves that combine to give the above vibration? (c) What is the velocity of the particle at \(x=5.66 \mathrm{~cm}\) at time \(t=2.00 \mathrm{~s}\)? (d) If the length of the string is \(10.0 \mathrm{~cm}\), locate the nodes and the antinodes. How many loops are formed in the vibration?

Solution:

(a) The amplitude of the vibration of the particle at position \(x\) is
\(
A=\left|(5.00 \mathrm{~mm}) \sin \left[\left(1.57 \mathrm{~cm}^{-1}\right) x\right]\right|
\)
For
\(
\begin{aligned}
& x=5.66 \mathrm{~cm}, \\
& A=\left|(5.00 \mathrm{~mm}) \sin \left[\frac{\pi}{2} \times 5.66\right]\right|
\end{aligned}
\)
\(
\begin{aligned}
& =\left|(5.00 \mathrm{~mm}) \sin \left(2.5 \pi+\frac{\pi}{3}\right)\right| \\
& =\left|(5.00 \mathrm{~mm}) \cos \frac{\pi}{3}\right|=2.50 \mathrm{~mm} .
\end{aligned}
\)
(b) From the given equation, the wave number \(k=1.57 \mathrm{~cm}^{-1}\) and the angular frequency \(\omega=314 \mathrm{~s}^{-1}\). Thus, the wavelength is
\(
\lambda=\frac{2 \pi}{k}=\frac{2 \times 3.14}{1.57 \mathrm{~cm}^{-1}}=4.00 \mathrm{~cm}
\)
\(
\text { and the frequency is } f=\frac{\omega}{2 \pi}=\frac{314 \mathrm{~s}^{-1}}{2 \times 3 \cdot 14}=50 \mathrm{~s}^{-1} \text {. }
\)
\(
\text { The wave speed is } v=f \lambda=\left(50 \mathrm{~s}^{-1}\right)(4.00 \mathrm{~cm})=2.00 \mathrm{~m} \mathrm{~s}^{-1} \text {. }
\)
(c) The velocity of the particle at position \(x\) at time \(t\) is given by
\(
\begin{aligned}
v & =\frac{\partial y}{\partial t}=(5 \cdot 00 \mathrm{~mm}) \sin \left[\left(1 \cdot 57 \mathrm{~cm}^{-1}\right) x\right] \\
& {\left[314 \mathrm{~s}^{-1} \cos \left(314 \mathrm{~s}^{-1}\right) t\right] } \\
& =\left(157 \mathrm{~cm} \mathrm{~s}^{-1}\right) \sin \left(1 \cdot 57 \mathrm{~cm}^{-1}\right) x \cos \left(314 \mathrm{~s}^{-1}\right) t .
\end{aligned}
\)
Putting \(x=5.66 \mathrm{~cm}\) and \(t=2.00 \mathrm{~s}\), the velocity of this particle at the given instant is
\(
\begin{aligned}
& \left(157 \mathrm{~cm} \mathrm{~s}^{-1}\right) \sin \left(\frac{5 \pi}{2}+\frac{\pi}{3}\right) \cos (200 \pi) \\
= & \left(157 \mathrm{~cm} \mathrm{~s}^{-1}\right) \times \cos \frac{\pi}{3} \times 1=78.5 \mathrm{~cm} \mathrm{~s}^{-1} .
\end{aligned}
\)
(d) The nodes occur where the amplitude is zero, i.e.,
or,
\(
\begin{aligned}
\sin \left(1 \cdot 57 \mathrm{~cm}^{-1}\right) x & =0 . \\
\left(\frac{\pi}{2} \mathrm{~cm}^{-1}\right) x & =n \pi
\end{aligned}
\)
where \(n\) is an integer.
Thus, \(\quad x=2 n \mathrm{~cm}\).
The nodes, therefore, occur at \(x=0,2 \mathrm{~cm}, 4 \mathrm{~cm}, 6 \mathrm{~cm}\), \(8 \mathrm{~cm}\) and \(10 \mathrm{~cm}\). Antinodes occur in between them, i.e., at \(x=1 \mathrm{~cm}, 3 \mathrm{~cm}, 5 \mathrm{~cm}, 7 \mathrm{~cm}\) and \(9 \mathrm{~cm}\). The string vibrates in 5 loops.

Example 11: A guitar string is \(90 \mathrm{~cm}\) long and has a fundamental frequency of \(124 \mathrm{~Hz}\). Where should it be pressed to produce a fundamental frequency of \(186 \mathrm{~Hz}\)?

Solution:

The fundamental frequency of a string fixed at both ends is given by
\(
f=\frac{1}{2 L} \sqrt{\frac{F}{\mu}}
\)
As \(F\) and \(\mu\) are fixed, \(\frac{f_1}{f_2}=\frac{L_2}{L_1}\) or,
\(
L_2=\frac{f_1}{f_2} L_1=\frac{124 \mathrm{~Hz}}{186 \mathrm{~Hz}}(90 \mathrm{~cm})=60 \mathrm{~cm} \text {. }
\)
Thus, the string should be pressed at \(60 \mathrm{~cm}\) from an end.

Example 12: A sonometer wire has a total length of \(1 \mathrm{~m}\) between the fixed ends. Where should the two bridges be placed below the wire so that the three segments of the wire have their fundamental frequencies in the ratio \(1: 2: 3\)?

Solution:

Suppose the lengths of the three segments are \(L_1, L_2\), and \(L_2\) respectively. The fundamental frequencies are
\(
\begin{aligned}
& f_1=\frac{1}{2 L_1} \sqrt{F / \mu} \\
& f_2=\frac{1}{2 L_2} \sqrt{F / \mu} \\
& f_3=\frac{1}{2 L_3} \sqrt{F / \mu}
\end{aligned}
\)
\(
\text { so that } \quad f_1 L_1=f_2 L_2=f_3 L_3 \dots(i)
\)
As \(f_1: f_2:f_3=1: 2: 3\), we have
\(f_2=2 f_1\) and \(f_3=3 f_1\) so that by (i)
\(
L_2=\frac{f_1}{f_2} L_1=\frac{L_1}{2}
\)
\(
\text { and } \quad L_3=\frac{f_1}{f_3} L_1=\frac{L_1}{3}
\)
As \(L_1+L_2+L_3=1 \mathrm{~m}\), we get \(L_1\left(1+\frac{1}{2}+\frac{1}{3}\right)=1 \mathrm{~m}\) or,
\(
\begin{aligned}
& L_1=\frac{6}{11} \mathrm{~m} \\
& L_2=\frac{L_1}{2}=\frac{3}{11} \mathrm{~m}
\end{aligned}
\)
and
\(
L_3=\frac{L_1}{3}=\frac{2}{11} \mathrm{~m}
\)
One bridge should be placed at \(\frac{6}{11} \mathrm{~m}\) from one end and the other should be placed at \(\frac{2}{11} \mathrm{~m}\) from the other end.

Example 13: A wire having a linear mass density \(5.0 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}\) is stretched between two rigid supports with a tension of \(450 \mathrm{~N}\). The wire resonates at a frequency of \(420 \mathrm{~Hz}\). The next higher frequency at which the same wire resonates is \(490 \mathrm{~Hz}\). Find the length of the wire.

Solution:

Suppose the wire vibrates at \(420 \mathrm{~Hz}\) in its \(n\)th harmonic and at \(490 \mathrm{~Hz}\) in its \((n+1)\) th harmonic.
and \(\quad 490 \mathrm{~s}^{-1}=\frac{(n+1)}{2 L} \sqrt{F / \mu} \dots(i)\).
This gives \(\frac{490}{420}=\frac{n+1}{n} \dots(ii)\)
or, \(\quad n=6\).
Putting the value in (i),
\(
\begin{aligned}
420 \mathrm{~s}^{-1} & =\frac{6}{2 L} \sqrt{\frac{450 \mathrm{~N}}{5 \cdot 0 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}}}=\frac{900}{L} \mathrm{~m} \mathrm{~s}^{-1} \\
\text { or, } \quad L & =\frac{900}{420} \mathrm{~m}=2 \cdot 1 \mathrm{~m} .
\end{aligned}
\)