Let us consider a sinusoidal travelling wave. For convenience, we shall take the wave to be transverse so that if the position of the constituents (particles) of the medium is denoted by \(\mathrm{x}\), the displacement from the equilibrium position may be denoted by \(\mathrm{y}\). A sinusoidal travelling wave is then described by:
\(
y=a \sin (k x-\omega t+\phi) \dots(15.2)
\)
Where
\(y\) is the displacement of the particle from the equilibrium position
\(a\) is the amplitude of a wave
\(\omega\) is the angular frequency of the wave
\(\mathrm{k}\) is the angular wave number
\((k x-\omega t+\phi)\) is the initial phase angle
The term \(\phi\) in the equation means that we are considering a linear combination of sine and cosine functions:
\(
y=A \sin (k x-\omega t)+B \cos (k x-\omega t) \dots(15.3)
\)
From Equations (15.2) and (15.3),
\(
a=\sqrt{A^2+B^2} \text { and } \phi=\tan ^{-1}\left(\frac{B}{A}\right)
\)
To understand why Equation (15.2) represents a sinusoidal travelling wave, take a fixed instant, say \(t=t_0\). Then, the argument of the sine function in Equation (15.2) 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 (15.2) 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. (15.2) represents a sinusiodal (harmonic) wave travelling along the positive direction of the \(x\)-axis. On the other hand, the equation for a wave travelling in the negative direction is given as
\(
y(x, t)=a \sin (k x+\omega t+\phi) \dots(15.4)
\)
F1g. 15.6 shows the plots of Eq. (15.2) for different values of time differing by equal intervals of time. In a wave, the crest is the point of maximum positive displacement, the trough is the point of maximum negative displacement. To see how a wave travels, we can fix attention on a crest and see how it progresses with time. In the figure, this is shown by a cross \((\times)\) on the crest. In the same manner, we can see the motion of a particular constituent of the medium at a fixed location, say at the origin of the \(x\)-axis. This is shown by a solid \(\operatorname{dot}(\bullet)\). The plots of Fig. 15.6 show that with time, the solid \(\operatorname{dot}(\bullet)\) at the origin moves periodically, 1.e., the particle at the origin oscillates about its mean position as the wave progresses. This is true for any other location also. We also see that during the time the solid dot \((\bullet)\) has completed one full oscillation, the crest has moved further by a certain distance. Using the plots of Fig. 15.6, we now define the various quantities of Eq. (15.2).

Amplitude

The amplitude of a wave is the maximum displacement of the particles of the medium from their mean position.

In Eq. (15.2), 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 (as shown in the above waveform), 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.

Phase

Given the amplitude \(a\), the phase determines the displacement of the wave at any position and at any instant.

The quantity \((k x-\omega t+\phi)\) appearing as the argument of the sine function in Eq. (15.2) is called the phase of the wave. 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. (15.2) with \(\phi=0\).

Phase Difference

Phase difference (also called phase or phase shift) describes how much one sine wave is shifted relative to another. Sine waves that are perfectly aligned peak to peak are called in phase. If one wave is shifted by half a wavelength (relative to the other), the troughs of one wave are aligned with the peaks of the other, and the waves are called perfectly (or completely) out of phase. 

For waves that are neither perfectly aligned nor perfectly out of phase, phase is usually described as an angle: zero degrees for perfectly in phase and 180 degrees out of phase for a shift of half a wavelength (aka perfectly out of phase). Notice that a phase shift of 360 degrees is the same thing as no phase shift at all- shifting a sine wave by a full wavelength gives the same wave back again.

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(as shown in the waveform above). The wavelength is then the distance between two consecutive crests or troughs in a wave. Taking \(\phi=0\) in Eq. (15.2), the displacement at

\(t=0\) is given by
\(
y(x, 0)=a \sin k x \dots(15.5)
\)
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
\(\lambda=\frac{2 \pi}{k} \quad\) or \(\quad k=\frac{2 \pi}{\lambda} \dots(15.6)\)
\(k\) 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

F1g. 15.7 shows again a sinusoidal plot. It describes not the shape of the wave at a certain instant but the displacement of an element (at any fixed location) of the medium as a function of time. We may for, simplicity, take Eq. (15.2) with \(\phi=0\) and monitor the motion of the element say at \(x=0\). We then get
\(
\begin{aligned}
y(0, t) & =a \sin (-\omega t) \\
& =-a \sin \omega t
\end{aligned}
\)

Time Period

The time period of oscillation of the wave is the time it takes for an element to complete one full oscillation. That is

\(
\begin{aligned}
-a \sin \omega t & =-a \sin \omega(t+T) \\
& =-a \sin (\omega t+\omega T)
\end{aligned}
\)
Since sine function repeats after every \(2 \pi\),
\(
\omega \mathrm{T}=2 \pi \text { or } \omega=\frac{2 \pi}{\mathrm{T}} \dots(15.7)
\)
\(\omega\) is called the angular frequency of the wave. Its SI unit is \(\mathrm{rad} \mathrm{~s}^{-1}\). The frequency \(f\) is the number of oscillations per second. Therefore,
\(
f=\frac{1}{\mathrm{~T}}=\frac{\omega}{2 \pi} \dots(15.8)
\)
\(f\) is usually measured in hertz.

In the discussion above, reference has always been made to a wave travelling along a string or a transverse wave. In a longitudinal wave, the displacement of an element of the medium is parallel to the direction of propagation of the wave. In Eq. (15.2), the displacement function for a longitudinal wave is written as,
\(
s(x, t)=a \sin (k x-\omega t+\phi) \dots(15.9)
\)
where \(s(x, t)\) is the displacement of an element of the medium in the direction of propagation of the wave at position \(x\) and time \(t\). In Eq. (15.9), \(a\) is the displacement amplitude; other quantities have the same meaning as in the case of a transverse wave except that the displacement function \(y(x, t)\) is to be replaced by the function \(s(x, t)\).

Example 15.2: A wave travelling along a string is described by,
\(y(x, t)=0.005 \sin (80.0 x-3.0 t)\),
in which the numerical constants are in SI units \(\left(0.005 \mathrm{~m}, 80.0 \mathrm{~rad} \mathrm{~m}^{-1}\right.\), and \(3.0 \mathrm{~rad} \mathrm{~s}^{-1}\) ). Calculate
(a) the amplitude,
(b) the wavelength, and
(c) the period and frequency of the wave. Also, calculate the displacement \(y\) of the wave at a distance \(x=30.0 \mathrm{~cm}\) and time \(t=20 \mathrm{~s} ?\)

Solution:

On comparing this displacement equation with Eq. (15.2)
\(
y(x, t)=a \sin (k x-\omega t)
\)
we find
(a) the amplitude of the wave is \(0.005 \mathrm{~m}=5 \mathrm{~mm}\).
(b) the angular wave number \(k\) and angular frequency \(\omega\) are
\(
k=80.0 \mathrm{~m}^{-1} \text { and } \omega=3.0 \mathrm{~s}^{-1}
\)
We, then, relate the wavelength \(\lambda\) to \(k\) through Eq. (15.6)
\(
\begin{aligned}
\lambda & =2 \pi / k \\
& =\frac{2 \pi}{80.0 \mathrm{~m}^{-1}} \\
& =7.85 \mathrm{~cm}
\end{aligned}
\)
(c) Now, we relate \(T\) to \(\omega\) by the relation
\(
\begin{aligned}
T & =2 \pi / \omega \\
& =\frac{2 \pi}{3.0 \mathrm{~s}^{-1}} \\
& =2.09 \mathrm{~s}
\end{aligned}
\)
and frequency, \(v=1 / T=0.48 \mathrm{~Hz}\)
The displacement \(y\) at \(x=30.0 \mathrm{~cm}\) and time \(t=20 \mathrm{~s}\) is given by
\(
\begin{aligned}
y & =(0.005 \mathrm{~m}) \sin (80.0 \times 0.3-3.0 \times 20) \\
& =(0.005 \mathrm{~m}) \sin (-36+12 \pi) \\
& =(0.005 \mathrm{~m}) \sin (1.699) \\
& =(0.005 \mathrm{~m}) \sin \left(97^{\circ}\right) \simeq 5 \mathrm{~mm}
\end{aligned}
\)

Example 15.3: Consider the wave \(y=(5 \mathrm{~mm}) \sin \left[\left(1 \mathrm{~cm}^{-1}\right) x-\left(60 \mathrm{~s}^{-1}\right) t\right]\). Find
(a) the amplitude,
(b) the wave number,
(c) the wavelength,
(d) the frequency,
(e) the time period and ( \(f)\) the wave velocity.

Solution:

Comparing the given equation with equation (15.2)\(y=A \sin (k x-\omega t)\), we find
(a) amplitude \(A=5 \mathrm{~mm}\)
(b) wave number \(k=1 \mathrm{~cm}^{-1}\)
(c) wavelength \(\lambda=\frac{2 \pi}{k}=2 \pi \mathrm{cm}\)
(d) frequency \(f=\frac{\omega}{2 \pi}=\frac{60}{2 \pi} \mathrm{Hz}=\frac{30}{\pi} \mathrm{Hz}\)
(e) time period \(T=\frac{1}{f}=\frac{\pi}{30} \mathrm{~s}\)
(f) wave velocity \(v=f \lambda=\lambda / T=60 \mathrm{~cm} \mathrm{~s}^{-1}\).

Example 15.4: A travelling harmonic wave on a string is described by
\(
y(x, t)=7.5 \sin (0.0050 x+12 t+\pi / 4)
\)
(a)what are the displacement and velocity of oscillation of a point at \(x=1 \mathrm{~cm}\), and \(t=1 \mathrm{~s}\)? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse displacements and velocity as the \(x=1 \mathrm{~cm}\) point at \(t=2 \mathrm{~s}, 5 \mathrm{~s}\) and \(11 \mathrm{~s}\).

Solution:

a) The given harmonic wave is
\(
\begin{aligned}
& y(x, t)=7.5 \sin \left(0.0050 x+12 t+\frac{\pi}{4}\right) \\
& \text { For } {x}=1 \mathrm{~cm} \text { and } {t}=1 \mathrm{~s} \\
& y=(1,1)=7.5 \sin \left(0.0050+12+\frac{\pi}{4}\right) \\
& =7.5 \sin \left(12.0050+\frac{\pi}{4}\right) \\
& =7.5 \sin \theta \\
& \text { Where, theta }=12.0050+\mathrm{pi} / 4=12.0050+3.14 / 4=12.79 \\
& =\frac{180}{3.14} \times 12.79=732.81^{\circ} \\
& \therefore y=(1,1)=7.5 \sin \left(732.81^{\circ}\right) \\
& =7.5 \sin \left(90 \times 8+12.81^{\circ}\right)=7.5 \sin 12.81^{\circ} \\
& =7.5 \times 0.2217 \\
& =1.6629 \approx 1.663 \mathrm{~cm}
\end{aligned}
\)
The velocity of the oscillation at a given point and time is given as:
\(
\begin{aligned}
& v=\frac{d}{d t} y(x, t)=\frac{d}{d t}\left[7.5 \sin \left(0.0050 x+12 t+\frac{\pi}{4}\right)\right] \\
& =7.5 \times 12 \cos \left(0.0050 x+12 t+\frac{\pi}{4}\right) \\
& \text { At } x=1 \mathrm{~cm} \text { and } \mathrm{t}=1 \mathrm{~s} \\
& v=y(1,1)=90 \cos \left(12.005+\frac{\pi}{4}\right) \\
& =90 \cos \left(732.81^{\circ}\right)=90 \cos \left(90 \times 8+12.81^{\circ}\right) \\
& =90 \cos \left(12.81^{\circ}\right) \\
& =90 \times 0.975=87.75 \mathrm{~cm} / \mathrm{s}
\end{aligned}
\)
Now, the equation of a propagating wave is given by:
\(
y(x, t)=a \sin (k x+\omega t+\phi)
\)
where,
\(
\begin{aligned}
& \mathrm{k}=2 \pi / \lambda \\
& \therefore \lambda=2 \pi / \mathrm{k} \\
& \text { and, } \omega=2 \pi f \\
& \therefore f=\omega / 2 \pi
\end{aligned}
\)
Speed, \(v=f \lambda=\omega / \mathrm{k}\)
Where,
\(
\begin{aligned}
& \omega=12 \mathrm{rad} / \mathrm{s} \\
& \mathrm{k}=0.0050 \mathrm{~m}^{-1} \\
& \therefore \mathrm{v}=12 / 0.0050=24 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Hence, the velocity of the wave oscillation at \({x}=1 \mathrm{~cm}\) and \({t}=1 \mathrm{~s}\) is not equal to the velocity of the wave propagation.
b) Propagation constant is related to wavelength as:
\(
\begin{aligned}
& k=\frac{2 \pi}{\lambda} \\
& \therefore \lambda=2 \frac{\pi}{k}=\frac{2 \times 3.14}{0.0050} \\
& =1256 \mathrm{~cm}=12.56 \mathrm{~m}
\end{aligned}
\)
Therefore, all the points at distances \(n \lambda, n= \pm 1, \pm 2 \ldots\), i.e. \(\pm 12.56 \mathrm{~m}, \pm 25.12 \mathrm{~m}, \ldots\) and so on for \(x=1 \mathrm{~cm}\), will have the same displacement as the \(x=1 \mathrm{~cm}\), point \(t=2 \mathrm{~s}, 5 \mathrm{~s}\), and \(11 \mathrm{~s}\).

Example 15.5: A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation? (b) If the pulse rate is 1 after every \(20 \mathrm{~s}\), (that is the whistle is blown for a split of second after every \(20 \mathrm{~s}\)), is the frequency of the note produced by the whistle equal to \(1 / 20\) or \(0.05 \mathrm{~Hz}\)?

Solution:

(a) In a non-dispersive medium, the wave propagates with definite speed but its wavelength of the frequency is not definite.
(b) No, the frequency of the note is not \(1 / 20\) or \(0.50 \mathrm{~Hz} .0 .005 \mathrm{~Hz}\) is only the frequency ‘ of repetition of the pip of the whistle.

Example 15.6: One end of a long string of linear mass density \(8.0 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}\) is connected to an electrically driven tuning fork of frequency \(256 \mathrm{~Hz}\). The other end passes over a pulley and is tied to a pan containing a mass of \(90 \mathrm{~kg}\). The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At \(t=0\), the left end (fork end) of the string \(x=0\) has zero transverse displacement \((y=0)\) and is moving along positive \(y\)-direction. The amplitude of the wave is 5.0 \(\mathrm{cm}\). Write down the transverse displacement \(y\) as a function of \(x\) and \(t\) that describes the wave on the string.

Solution:

Here, mass per unit length, \(\mathrm{g}=\) linear mass density \(=8 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}\);
Tension in the string, \(T=90 \mathrm{~kg}=90 \times 9.8 \mathrm{~N}=882 \mathrm{~N}\);
Frequency, \(\quad f=256 \mathrm{~Hz}\)
and amplitude, \(A=5.0 \mathrm{~cm}=0.05 \mathrm{~m}\)
As the wave propagating along the string is a transverse travelling wave, the velocity of the wave,
\(
v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{882}{8 \times 10^{-3}}} \mathrm{~ms}^{-1}=3.32 \times 10^2 \mathrm{~ms}^{-1}
\)
Now,
Also,
\(
\begin{aligned}
& \omega=2 \pi f=2 \times 3.142 \times 256=1.61 \times 10^3 \mathrm{rad} \mathrm{s}^{-1} \\
& v=f \lambda \text { or } \lambda=\frac{v}{f}=\frac{3.32 \times 10^2}{256} \mathrm{~m}
\end{aligned}
\)
Propagation constant, \(k=\frac{2 \pi}{\lambda}=\frac{2 \times 3.142 \times 256}{3.32 \times 10^2}\)
\(
=4.84 \mathrm{~m}^{-1}
\)
\(\therefore\) The equation of the wave is，
\(
\begin{aligned}
v(x, t) & =\mathrm{A} \sin (\omega t-k x) \\
& =0.05 \sin \left(1.61 \times 10^3 t-4.84 x\right)
\end{aligned}
\)
Here, \(x, y\) are in metre and \(t\) is in second.