It is an everyday experience that the pitch (or frequency) of the whistle of a fast moving train decreases as it recedes away. When we approach a stationary source of sound with high speed, the pitch of the sound heard appears to be higher than that of the source. As the
observer recedes away from the source, the observed pitch (or frequency) becomes lower than that of the source. This motion-related frequency change is called Doppler effect.

The Austrian physicist Johann Christian Doppler first proposed the effect in 1842. Buys Ballot in Holland tested it experimentally in 1845.  Doppler effect is a wave phenomenon, it holds not only for sound waves but also for electromagnetic waves. However, here we shall consider only sound waves. We shall analyse changes in frequency under three different situations:

1. The observer is stationary but the source is moving,
2. The observer is moving but the source is stationary, and
3. Both the observer and the source are moving.

Source Moving; Observer Stationary

Let us choose the convention to take the direction from the observer to the source as the positive direction of velocity. Consider a source \(\mathrm{S}\) moving with velocity \(v_{\mathrm{s}}\) and an observer who is stationary in a frame in which the medium is also at rest. Let the speed of a wave of angular frequency \(\omega\) and period \(T_o\), both measured by an observer at rest with respect to the medium, be \(v\). We assume that the observer has a detector that counts every time a wave crest reaches it. As shown in Fig. 15. 17, at time \(t=0\) the source is at point \(\mathrm{S}_1\), located at a distance \(L\) from the observer, and emits a crest. This reaches the observer at time \(t_1=L / v\). At time \(t=T_0\) the source has moved a distance \(v_s T_0\) and is at point \(\mathrm{S}_2\), located at a distance \(\left(L+v_s T_{\mathrm{o}}\right)\) from the observer. At \(\mathrm{S}_2\), the source emits a second crest. This reaches the observer at
\(
t_2=T_0+\frac{\left(L+v_{\mathrm{s}} T_0\right)}{v}
\)

At time \(n T_0\), the source emits its \((n+1)^{\text {th }}\) crest and this reaches the observer at time
\(
t_{n+1}=n T_0+\frac{\left(L+n v_{\mathrm{s}} T_0\right)}{v}
\)
Hence, in a time interval
\(
\left[n T_0+\frac{\left(L+n v_s T_0\right)}{v}-\frac{L}{v}\right]
\)
the observer’s detector counts \(n\) crests and the observer records the period of the wave as \(T\) given by
\(
\begin{aligned}
T & =\left[n T_0+\frac{\left(L+n v_{\mathrm{s}} T_0\right)}{v}-\frac{L}{v}\right] / n \\
& =T_0+\frac{v_{\mathrm{s}} T_0}{v} \\
& =T_0\left(1+\frac{v_{\mathrm{s}}}{v}\right) \dots(15.49)
\end{aligned}
\)
Equation (15.49) may be rewritten in terms of the frequency \(f_o\) that would be measured if the source and observer were stationary, and the frequency \(f\) observed when the source is moving, as
\(
f=f_0\left(1+\frac{v_{\mathrm{s}}}{v}\right)^{-1} \dots(15.50)
\)
If \(v_s\) is small compared with the wave speed \(v\), taking binomial expansion to terms in first order in \(v_s / v\) and neglecting higher power, Eq. (15.50) may be approximated, giving
\(
f=f_0\left(1-\frac{v_s}{v}\right) \dots(15.51)
\)
For a source approaching the observer, we replace \(v_s\) by \(-v_s\) to get
\(
f=f_0\left(1+\frac{v_s}{v}\right) \dots(15.52)
\)
The observer thus measures a lower frequency when the source recedes from him than he does when it is at rest. He measures a higher frequency when the source approaches him.

Observer Moving; Source Stationary

Now to derive the Doppler shift when the observer is moving with velocity \(v_o\) towards the source and the source is at rest, we have to proceed in a different manner. We work in the reference frame of the moving observer. In this reference frame the source and medium are approaching at speed \(v_o\) and the speed with which the wave approaches is \(v_o+v\). Following a similar procedure as in the previous case, we find that the time interval between the arrival of the first and the \((n+1)\) th crests is
\(
t_{n+1}-t_1=n T_0-\frac{n v_0 T_0}{v_0+v}
\)
The observer thus, measures the period of the wave to be
\(
\begin{aligned}
& =T_0\left(1-\frac{v_0}{v_0+v}\right) \\
& =T_0\left(1+\frac{v_0}{v}\right)^{-1}
\end{aligned}
\)
giving
\(
f=f_0\left(1+\frac{v_0}{v}\right) \dots(15.53)
\)
If \(\frac{v_0}{v}\) is small, the Doppler shift is almost same whether it is the observer or the source moving since Eq. (15.53) and the approximate relation Eq. (15.51) are the same.

Both Source and Observer Moving

As before, let us take the direction from the observer to the source as the positive direction. Let the source and the observer be moving with velocities \(v_s\) and \(v_o\) respectively as shown in Fig. 15. 18. Suppose at time \(t=0\), the observer is at \(\mathrm{O}_1\) and the source is at \(\mathrm{S}_1, \mathrm{O}_1\) being to the left of \(\mathrm{S}_1\). The source emits a wave of velocity \(v\), of frequency \(f\) and period \(T_0\) all measured by an observer at rest with respect to the medium. Let \(L\) be the distance between \(\mathrm{O}_1\) and \(\mathrm{S}_1\) at \(t=0\), when the source emits the first crest. Now, since the observer is moving, the velocity of the wave relative to the observer is \(v+v_0\). Therefore, the first crest reaches the observer at time \(t_1=L /\) \(\left(v+v_0\right)\). At time \(t=T_0\), both the observer and the source have moved to their new positions \(\mathrm{O}_2\) and \(\mathrm{S}_2\) respectively. The new distance between the observer and the source, \(\mathrm{O}_2 \mathrm{~S}_2\), would be \(\left.L+\left(v_{\mathrm{s}}-v_0\right) T_0\right]\). At \(S_2\), the source emits a second crest.

This reaches the observer at time.
\(
\left.t_2=T_o+\left[L+\left(v_s-v_o\right) T_o\right)\right] /\left(v+v_o\right)
\)
At time \(n T_o\) the source emits its \((n+1)\) th crest and this reaches the observer at time
\(
\left.t_{n+1}=n T_o+\left[L+n\left(v_s-v_o\right) T_o\right)\right] /\left(v+v_o\right)
\)
Hence, in a time interval \(t_{n+1}-t_1\), i.e.,
\(
\left.n T_o+\left[L+n\left(v_s-v_o\right) T_o\right)\right] /\left(v+v_o\right)-L /\left(v+v_o\right),
\)
the observer counts \(n\) crests and the observer records the period of the wave as equal to \(T\) given by
\(
T=T_0\left(1+\frac{v_s-v_o}{v+v_0}\right)=T_0\left(\frac{v+v_s}{v+v_0}\right) \dots(15.54)
\)
The frequency \(f\) observed by the observer is given by
\(
{f}={f}_o\left(\frac{v+v_o}{v+v_s}\right) \dots(15.55)
\)

Consider a passenger sitting in a train moving on a straight track. Suppose she hears a whistle sounded by the driver of the train. What frequency will she measure or hear? Here both the observer and the source are moving with the same velocity, so there will be no shift in frequency and the passenger will note the natural frequency. But an observer outside who is stationary with respect to the track will note a higher frequency if the train is approaching him and a lower frequency when it recedes from him.

Note that we have defined the direction from the observer to the source as the positive direction. Therefore, if the observer is moving towards the source, \(v_0\) has a positive (numerical) value whereas if \(\mathrm{O}\) is moving away from \(\mathrm{S}, v_0\) has a negative value. On the other hand, if \(S\) is moving away from \(\mathrm{O}, v_{\mathrm{s}}\) has a positive value whereas if it is moving towards \(\mathrm{O}, v_{\mathrm{s}}\) has a negative value. The sound emitted by the source travels in all directions. It is that part of sound coming towards the observer which the observer receives and detects. Therefore, the relative velocity of sound with respect to the observer is \(v+v_0\) in all cases.

Example 1: A rocket is moving at a speed of \(200 \mathrm{~m} \mathrm{~s}^{-1}\) towards a stationary target. While moving, it emits a wave of frequency \(1000 \mathrm{~Hz}\). Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket.

Solution:

(1) The observer is at rest and the source is moving with a speed of \(200 \mathrm{~m} \mathrm{~s}^{-1}\). Since this is comparable with the velocity of sound, \(330 \mathrm{~m} \mathrm{~s}^{-1}\), we must use Eq. (15.50) and not the approximate Eq. (15.51). Since the source is approaching a stationary target, \(v_o=0\), and \(v_s\) must be replaced by \(-v_s\). Thus, we have
\(
\begin{aligned}
& f=f_0\left(1-\frac{v_s}{v}\right)^{-1} \\
v & =1000 \mathrm{~Hz} \times\left[1-200 \mathrm{~m} \mathrm{~s}^{-1} / 330 \mathrm{~m} \mathrm{~s}^{-1}\right]^{-1} \\
\simeq & 2540 \mathrm{~Hz}
\end{aligned}
\)
(2) The target is now the source (because it is the source of echo) and the rocket’s detector is now the detector or observer (because it detects echo). Thus, \(v_s=0\) and \(v_o\) has a positive value. The frequency of the sound emitted by the source (the target) is \(f\), the frequency intercepted by the target and not \(f_{o}\) Therefore, the frequency as registered by the rocket is
\(
\begin{aligned}
& f^{\prime}=f\left(\frac{v+v_0}{v}\right) \\
& =2540 \mathrm{~Hz} \times\left(\frac{200 \mathrm{~m} \mathrm{~s}^{-1}+330 \mathrm{~m} \mathrm{~s}^{-1}}{330 \mathrm{~m} \mathrm{~s}^{-1}}\right) \\
& \simeq 4080 \mathrm{~Hz}
\end{aligned}
\)

Example 2: Two waves, each having a frequency of \(100 \mathrm{~Hz}\) and a wavelength of \(2.0 \mathrm{~cm}\), are travelling in the same direction on a string. What is the phase difference between the waves (a) if the second wave was produced \(0.015 \mathrm{~s}\) later than the first one at the same place, (b) if the two waves were produced at the same instant but the first one was produced a distance \(4.0 \mathrm{~cm}\) behind the second one? (c) If each of the waves has an amplitude of \(2.0 \mathrm{~mm}\), what would be the amplitudes of the resultant waves in part (a) and (b)?

Solution:

Given Two waves have same frequency \((f)\), which is \(100 \mathrm{~Hz}\).
Wavelength \((\lambda)=2.0 \mathrm{~cm}\)
\(
=2 \times 10^{-2} m
\)
Wave speed, \(v=f \times \lambda=100 \times 2 \times 10^{-2} \mathrm{~m} / \mathrm{s}\)
\(
=2 m / s
\)
(a) First wave will travel the distance in \(0.015 \mathrm{~s}\).
\(
\begin{aligned}
& \Rightarrow x=0.015 \times 2 \\
& =0.03 \mathrm{~m}
\end{aligned}
\)
This will be the path difference between the two waves.
So, the corresponding phase difference will be as follows:
\(
\begin{aligned}
& \phi=\frac{2 \pi x}{\lambda} \\
& =\left\{\frac{2 \pi}{\left(2 \times 10^{-2}\right)}\right\} \times 0.03=3 \pi
\end{aligned}
\)
(b) Path difference between the two waves, \(x=4 \mathrm{~cm}=0.04 \mathrm{~m}\)
So, the corresponding phase difference will be as follows:
\(
\begin{aligned}
& \Rightarrow \phi=\frac{2 \pi x}{\lambda} \\
& =\left\{\frac{2 \pi}{\left(2 \times 10^{-2}\right)} \times 0.04\right\} \\
& =4 \pi
\end{aligned}
\)
(c) The waves have the same frequency, same wavelength, and same amplitude.
Let the wave equation for the two waves be as follows:
\(y_1=r \sin \omega t\)
\(A n d, y_2=r \sin (\omega t+\phi)\)
By using the principle of superposition:
\(
\begin{aligned}
& y=y_1+y_2 \\
& =r[\sin \omega t+(\omega t+\phi)] \\
& =2 r \sin \left(\omega t+\frac{\phi}{2}\right) \cos \left(\frac{\phi}{2}\right)
\end{aligned}
\)
\(\therefore\) Resultant amplitude
\(
=2 r \cos \frac{\phi}{2}
\)
So, when \(\phi=3 x\) :
\(
\begin{aligned}
& \Rightarrow r=2 \times 10^{-3} \mathrm{~m} \\
& R_{\text {resultant }}=2 \times\left(2 \times 10^{-3}\right) \cos \left(\frac{3 \pi}{2}\right) \\
& =0 \\
& \text { Again, when } \phi=4 \pi: \\
& R_{\text {resultant }}=2 \times\left(2 \times 10^{-3}\right) \cos \left(\frac{4 \pi}{2}\right) \\
& =4 \times 10^{-3} \times 1 \\
& =4 \mathrm{~mm}
\end{aligned}
\)

Example 3: A \(660 \mathrm{~Hz}\) tuning fork sets up a vibration in a string clamped at both ends. The wave speed for a transverse wave on this string is \(220 \mathrm{~m} \mathrm{~s}^{-1}\) and the string vibrates in three loops. (a) Find the length of the string. (b) If the maximum amplitude of a particle is \(0.5 \mathrm{~cm}\), write a suitable equation describing the motion.

Solution:

Given Frequency \((f)=660 \mathrm{~Hz}\)
Wave speed \((v)=220 \mathrm{~m} / \mathrm{s}\)
Wavelength, \(\lambda=\frac{v}{f}=\frac{220}{660}=\frac{1}{3} m\)
(a) No. of loops, \(n=3\)
\(
\begin{aligned}
& \therefore L=\frac{n}{2} \lambda \\
& \Rightarrow L=\frac{3}{2} \times \frac{1}{3} \\
& \Rightarrow L=\frac{1}{2} m=50 \mathrm{~cm}
\end{aligned}
\)
(b) Equation of resultant stationary wave can be given by:
\(
\begin{aligned}
& y=2 A \cos \left(\frac{2 \pi x}{\lambda}\right) \sin \left(\frac{2 \pi v t}{\lambda}\right) \\
& \Rightarrow y=0.5 \cos \left(\frac{2 \pi x}{\frac{1}{3}}\right) \sin \left(\frac{2 \pi \times 220 \times t}{\frac{1}{3}}\right) \\
& \Rightarrow y=0.5 \mathrm{~cm} \cos \left(6 \pi x \mathrm{~m}^{-1}\right) \sin \left(1320 \pi t \mathrm{~s}^{-1}\right)
\end{aligned}
\)

Example 4: A particular guitar wire is \(30.0 \mathrm{~cm}\) long and vibrates at a frequency of \(196 \mathrm{~Hz}\) when no finger is placed on it. The next higher notes on the scale are \(220 \mathrm{~Hz}, 247 \mathrm{~Hz}\), \(262 \mathrm{~Hz}\) and \(294 \mathrm{~Hz}\). How far from the end of the string must the finger be placed to play these notes?

Solution:

Given:
Length of the guitar wire \(\left(L_1\right)=30.0 \mathrm{~cm}=0.30 \mathrm{~m}\)
Frequency, when no finger is placed on it, \(\left(f_1\right)=196 \mathrm{~Hz}\)
And \(\left(f_2\right)=220 \mathrm{~Hz},\left(f_3\right)=247 \mathrm{~Hz},\left(f_4\right)=262 \mathrm{~Hz}\) and \(\left(f_5\right)=294 \mathrm{~Hz}\)
The velocity is constant for a medium.
We have:
\(
\begin{aligned}
& f \propto\left(\frac{1}{L}\right) \\
& \Rightarrow \frac{f_1}{f_2}=\frac{L_2}{L_1} \\
& \Rightarrow \frac{196}{220}=\frac{L_2}{0.3} \\
& \Rightarrow L_2=\frac{196 \times 0.3}{220}=0.267 \mathrm{~m} \\
& \Rightarrow L_2=26.7 \mathrm{~cm}
\end{aligned}
\)
Again,
\(
\begin{aligned}
& f_3=247 \mathrm{~Hz} \\
& \Rightarrow \frac{f_3}{f_1}=\frac{L_1}{L_3} \\
& \Rightarrow \frac{247}{196}=\frac{0.3}{L_3} \\
& \Rightarrow L_3=196 \times \frac{0.3}{247}=0.238 \mathrm{~m}
\end{aligned}
\)
Similarly, \(L_4=196 \times \frac{0.3}{262}=0.224 m\)
\(
\Rightarrow L_4=22.4 \mathrm{~cm}
\)
And, \(L_5=20 \mathrm{~cm}\)

Example 5: Three resonant frequencies of a string are 90,150 and \(210 \mathrm{~Hz}\). (a) Find the highest possible fundamental frequency of vibration of this string. (b) Which harmonics of the fundamental are the given frequencies? (c) Which overtones are these frequencies? (d) If the length of the string is \(80 \mathrm{~cm}\), what would be the speed of a transverse wave on this string?

Solution:

Given, Let the three resonant frequencies of a string be
\(
\begin{aligned}
& f_1=90 \mathrm{~Hz} \\
& f_2=150 \mathrm{~Hz} \\
& f_3=210 \mathrm{~Hz}
\end{aligned}
\)
(a) So, the highest possible fundamental frequency of the string is \(f=30 \mathrm{~Hz}\) because \(f_1, f_2\) and \(f_3\) are the integral multiples of
\(
30 \mathrm{~Hz} .
\)
(b) So, these frequencies can be written as follows:
\(
\begin{aligned}
& f_1=3 f \\
& f_2=5 f \\
& f_3=7 f
\end{aligned}
\)
Hence, \(f_1, f_2\), and \(f_3\) are the third harmonic, the fifth harmonic and the seventh harmonic, respectively.
(c) The frequencies in the string are \(f, 2 f, 3 f, 4 f, 5 f \ldots\)
\(\therefore 3 f=2\) nd overtone and 3rd harmonic
\(5 f=4\) th overtone and 5 th harmonic
7th \(=6\) th overtone and 7th harmonic
(d) Length of the string \((L)=80 \mathrm{~cm}=0.8 \mathrm{~m}\)
Let the speed of the wave be \(v\).
So, the frequency of the third harmonic is given by:
\(
\begin{aligned}
& f_1=\left(\frac{3}{2 \times L}\right) v \\
& \Rightarrow 90=\left\{\frac{3}{(2 \times 80)}\right\} \times v \\
& \Rightarrow v=\frac{(90 \times 2 \times 80)}{3} \\
& =30 \times 2 \times 80 \\
& =4800 \mathrm{~cm} / \mathrm{s} \\
& \Rightarrow v=48 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Example 6: A string of length \(L\) fixed at both ends vibrates in its fundamental mode at a frequency \(v\) and a maximum amplitude \(A\). (a) Find the wavelength and the wave number \(k\). (b) Take the origin at one end of the string and the \(X\)-axis along the string. Take the \(Y\)-axis along the direction of the displacement. Take \(t=0\) at the instant when the middle point of the string passes through its mean position and is going towards the positive \(y\)-direction. Write the equation describing the standing wave.

Solution:

Given, Length of the string \(=\mathrm{L}\)
The velocity of wave is given as:
\(
v=\sqrt{\frac{T}{m}}
\)
(a) Wavelength, \(\lambda=\frac{\text { Velocity }}{\text { Frequency }}\)
\(
\Rightarrow \lambda=\frac{\sqrt{\frac{T}{m}}}{\frac{1}{2 L} \sqrt{\frac{T}{m}}}=2 L
\)
Wave number, \(K=\frac{2 \pi}{\lambda}=\frac{2 \pi}{2 L}=\frac{\pi}{L}\)
(b) Equation of the stationary wave:
\(
\begin{aligned}
& y=A \cos \left(\frac{2 \pi x}{\lambda}\right) \sin \left(\frac{2 \pi v t}{\lambda}\right) \\
& =A \cos \left(\frac{2 \pi x}{2 L}\right) \sin \left(\frac{2 \pi v t}{2 L}\right)\left[\because f=\left(\frac{v}{2 L}\right)\right] \\
& =A \cos \left(\frac{\pi x}{L}\right) \sin (2 \pi f t)
\end{aligned}
\)

Example 7: A \(2 \mathrm{~m}\)-long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is \(200 \mathrm{~m} \mathrm{~s}^{-1}\) and the amplitude is \(0.5 \mathrm{~cm}\). (a) Find the wavelength and the frequency. (b) Write the equation giving the displacement of different points as a function of time. Choose the \(X\)-axis along the string with the origin at one end and \(t=0\) at the instant when the point \(x=50 \mathrm{~cm}\) has reached its maximum displacement.

Solution:

Given, Length of the string \((L)=2.0 \mathrm{~m}\)
Wave speed on the string in its first overtone \((v)=200 \mathrm{~m} / \mathrm{s}\)
Amplitude \((A)=0.5 \mathrm{~cm}\)
(a) Wavelength and frequency of the string when it is vibrating in its \(1^{\text {st }}\) overtone \((n=2)\) :
\(
\begin{aligned}
& L=\frac{n \lambda}{2} \\
& \Rightarrow \lambda=L=2 m \\
& \Rightarrow f=\frac{\nu}{\lambda}=\frac{200}{2}=100 \mathrm{~Hz}
\end{aligned}
\)
(b) The stationary wave equation is given by:
\(
\begin{aligned}
& y=2 A \sin \frac{2 \pi x}{\lambda} \cos \frac{2 \pi v t}{\lambda} \\
& =0.5 \sin \frac{2 \pi x}{2} \cos \frac{2 \pi \times 200 t}{2} \\
& =(0.5 \mathrm{~cm}) \sin \left[\left(\pi m^{-1}\right) x\right] \cos \left[\left(200 \pi s^{-1}\right) t\right]
\end{aligned}
\)

Example 8: The equation for the vibration of a string, fixed at both ends vibrating in its third harmonic, is given by
\(
y=(0.4 \mathrm{~cm}) \sin \left[\left(0.314 \mathrm{~cm}^{-1}\right) x\right] \cos \left[\left(600 \pi \mathrm{s}^{-1}\right) t\right] .
\)
(a) What is the frequency of vibration? (b) What are the positions of the nodes? (c) What is the length of the string? (d) What is the wavelength and the speed of two travelling waves that can interfere to give this vibration?

Solution:

Given, the stationary wave equation of a string vibrating in its third harmonic is given by \(y=(0.4 \mathrm{~cm}) \sin \left[\left(0.314 \mathrm{~cm}^{-1}\right) x\right] \cos \left[\left(.600 \pi \mathrm{s}^{-1}\right) t\right]\)
By comparing with the standard equation,
\(y=A \sin (k x) \cos (w t)\)
(a) From the above equation, we can infer the following:
\(
\begin{aligned}
& \omega=600 \pi \\
& \Rightarrow 2 \pi f=600 \pi \\
& \Rightarrow f=300 \mathrm{~Hz}
\end{aligned}
\)
Wavelength, \(\lambda=\frac{2 \pi}{0.314}=\frac{(2 \times 3.14)}{0.314}\)
\(
\Rightarrow \lambda=20 \mathrm{~cm}
\)
\(
\text { (b) Therefore, the nodes are located at } 0 \mathrm{~cm}, 10 \mathrm{~cm}, 20 \mathrm{~cm}, 30 \mathrm{~cm} \text {. }
\)

(c) Length of the string, \(l=n \frac{\lambda}{2}\)
\(
\Rightarrow l=\frac{3 \lambda}{2}=\frac{3 \times 20}{2}=30 \mathrm{~cm}
\)
(d) \(y=0.4 \sin (0.314 x) \cos (600 \pi t)\)
\(
=0.4 \sin \left\{\left(\frac{\pi}{10}\right) x\right\} \cos (600 \pi t)
\)
\(\lambda\) and \(v\) are the wavelength and velocity of the waves that interfere to give this vibration.
\(
\begin{aligned}
& \lambda=20 \mathrm{~cm} \\
& v=\frac{\omega}{k}=\frac{600 \pi}{\left(\frac{\pi}{10}\right)}=6000 \mathrm{~cm} / \mathrm{s} \\
& \Rightarrow v=60 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Example 9: A \(2.00 \mathrm{~m}\)-long rope, having a mass of \(80 \mathrm{~g}\), is fixed at one end and is tied to a light string at the other end. The tension in the string is \(256 \mathrm{~N}\). (a) Find the frequencies of the fundamental and the first two overtones. (b) Find the wavelength in the fundamental and the first two overtones.

Solution:

\(l=\) length of rope \(=2 \mathrm{~m}\)
\(M=\) mass \(=80 \mathrm{gm}=0.8 \mathrm{~kg}\)
mass per unit length \(=m=0.08 / 2=0.04 \mathrm{~kg} / \mathrm{m}\)
Tension \(\mathrm{T}=256 \mathrm{~N}\)
Velocity, \(v=\sqrt{T / \mathrm{m}}=80 \mathrm{~m} / \mathrm{s}\)
For fundamental frequency,
\(
\begin{aligned}
& l=\lambda / 4 \Rightarrow \lambda=4 \mathrm{I}=8 \mathrm{~m} \\
& \Rightarrow \mathrm{f}=80 / 8=10 \mathrm{~Hz}
\end{aligned}
\)
a) Therefore, the frequency of \(1^{\text {st }}\) two overtones are \(1^{\text {st }}\) overtone \(=3 \mathrm{f}=30 \mathrm{~Hz}\)
\(
2^{\text {nd }} \text { overtone }=5 f=50 \mathrm{~Hz}
\)
\(
\text { b) } \begin{aligned}
& \lambda_1=4 l=8 \mathrm{~m} \\
& \lambda_1=v / f_1=2.67 \mathrm{~m} \\
& \lambda_2=v / f_2=1.6 \mathrm{m}
\end{aligned}
\)
so, the wavelengths are \(8 \mathrm{~m}, 2.67 \mathrm{~m}\) and \(1.6 \mathrm{~m}\) respectively.

Example 10: Figure below shows two wave pulses at \(t=0\) travelling on a string in opposite directions with the same wave speed \(50 \mathrm{~cm} \mathrm{~s}^{-1}\). Find the shape of the string at \(t=4 \mathrm{~ms}, 6 \mathrm{~ms}, 8 \mathrm{~ms}\), and \(12 \mathrm{~ms}\).

Solution:

The distance travelled by the pulses are shown below.
\(
\begin{array}{ll}
\mathrm{t}=4 \mathrm{~ms}=4 \times 10^{-3} \mathrm{~s} & \mathrm{~s}=\mathrm{vt}=50 \times 10 \times 4 \times 10^{-3}=2 \mathrm{~mm} \\
\mathrm{t}=8 \mathrm{~ms}=8 \times 10^{-3} \mathrm{~s} & \mathrm{~s}=\mathrm{vt}=50 \times 10 \times 8 \times 10^{-3}=4 \mathrm{~mm} \\
\mathrm{t}=6 \mathrm{~ms}=6 \times 10^{-3} \mathrm{~s} & \mathrm{~s}=3 \mathrm{~mm} \\
\mathrm{t}=12 \mathrm{~ms}=12 \times 10^{-3} \mathrm{~s} & \mathrm{~s}=50 \times 10 \times 12 \times 10^{-3}=6 \mathrm{~mm}
\end{array}
\)

Example 11: Given below are some functions of \(x\) and \(t\) to represent the displacement of an elastic wave.
(a) \(y=5 \cos (4 x) \sin (20 t)\)
(b) \(y=4 \sin (5 x-t / 2)+3 \cos (5 x-t / 2)\)
(c) \(y=10 \cos [(252-250) \pi t] \cos [(252+250) \pi t]\)
(d) \(y=100 \cos (100 \pi t+0.5 x)\)
State which of these represent
(i) a travelling wave along \(-x\) direction
(ii) a stationary wave
(iii) beats
(iv) a travelling wave along \(+x\) direction.
Given reasons for your answers.

Solution:

\(
\text { (a) (ii), (b) (iv), (c) (iii), (d) (i). }
\)

(a) The equation \(y=5 \cos (4 x) \sin (20 t)\) represents a stationary wave because it contains \(\sin\), \(\cos\) terms i.e., the combination of two progressive waves.
(b) As the equation, \(y=4 \sin (5 x-t / 2)+3 \cos (5 x-t / 2)\) involves a negative sign with \(x\), have if represents a travelling wave along the \(x\)-direction.
(c) As the equation \(y=10 \cos [(252-250) \pi t]-\cos [(252+250) \pi t]\) involving the sum and difference of two nearby frequencies 252 and 250 this equation represents beats formation.
(d) The equation \(y=100 \cos (100 \pi t+0.5 x)\) is representing a travelling wave along the \(x \text {-direction. }\)

Example 12: In the given progressive wave
\(
y=5 \sin (100 \pi t-0.4 \pi x)
\)
where \(y\) and \(x\) are in \(\mathrm{m}, t\) is in \(\mathrm{s}\). What is the
(a) amplitude
(b) wavelength
(c) frequency
(d) wave velocity
(e) particle velocity amplitude.

Solution:

Standard progressive wave-
\(
\begin{aligned}
& y=\operatorname{asin}(\omega t-k x+\phi) \\
& y=5 \sin (100 \pi t-0.4 \pi x+0)
\end{aligned}
\)

(a) \(\text { Amplitude, a = } 5 \mathrm{~m}\)

(b) Wavelength, \(\lambda\)
\(\mathrm{k}=\frac{2 \pi}{\lambda}\) or \(\lambda=\frac{2 \pi}{k}\)
\(k=0.4 \pi\)
\(\lambda=\frac{2 \pi}{0.4 \pi}=5 \mathrm{~m}\)

(c) Frequency f,
\(
\omega=2 \pi f
\)
Or \(f=\frac{w}{2 \pi}\)
\(
\because \omega=100 \pi
\)
\(
f=\frac{100 \pi}{2 \pi}=50 \mathrm{~Hz}
\)

(d) Wave velocity, \(v=f \lambda=50 \times 5\)
\(
=250 \frac{\mathrm{m}}{\mathrm{s}}
\)

(e)
\(
\begin{aligned}
& y=5 \sin (100 \pi t-0.4 \pi x) \\
& \frac{d y}{d x}=5 \times 100 \pi \cos (100 \pi t-0.4 \pi x)
\end{aligned}
\)
For maximum velocity at mean position –
\(
\cos (100 \pi t-0.4 \pi x)=1
\)
Or \(100 \pi t-0.4 \pi x=0\)
\(
\begin{aligned}
& \therefore\left(\frac{d y}{d x}\right)_{\text {maximum }}=5 \times 100 \pi \times 1 \\
& {v}_{\max }=500 \pi \mathrm{m} / \mathrm{s}
\end{aligned}
\)

Example 13: For the harmonic travelling wave \(y=2 \cos 2 \pi(10 t-0.0080 x+3.5)\) where \(x\) and \(y\) are in \(\mathrm{cm}\) and \(t\) is second. What is the phase difference between the oscillatory motion at two points separated by a distance of
(a) \(4 \mathrm{~m}\)
(b) \(0.5 \mathrm{~m}\)
(c) \(\frac{\lambda}{2}\)
(d) \(\frac{3 \lambda}{4}\) (at a given instant of time)

Solution:

The equation for a travelling harmonic wave is given as:
\(
\begin{aligned}
y(x, t) & =2.0 \cos 2 \pi(10 t-0.0080 x+3.5) \\
& =2.0 \cos (20 \pi t-0.016 \pi x+0.70 \pi)
\end{aligned}
\)
where,
Propagation constant, \(k=0.0160 \pi\)
Amplitude, \(a=2 \mathrm{~cm}\)
Angular frequency, \(\omega=20 \pi \mathrm{rad} / \mathrm{s}\)
Phase difference is given by the relation,
\(
\phi=k x=\frac{2 \pi}{\lambda}
\)

(a) For \(x=4 \mathrm{~m}=400 \mathrm{~cm}\)
\(
\begin{aligned}
\phi & =0.016 \pi \times 400 \\
& =6.4 \pi \mathrm{rad}
\end{aligned}
\)

(b) For \(0.5 \mathrm{~m}=50 \mathrm{~cm}\),
\(
\begin{aligned}
\phi & =0.016 \pi \times 50 \\
& =0.8 \pi \mathrm{rad}
\end{aligned}
\)

(c) For \(x=\lambda / 2\)
\(
\phi=2 \pi / \lambda \times \lambda / 2=\pi \mathrm{~rad}
\)

(d) For \(x=3 \lambda / 4\)
\(
\phi=2 \pi / \lambda \times 3 \lambda / 4=1.5 \pi \mathrm{~rad} .
\)

Example 14: Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops, and 4 loops, the frequencies are in the ratio \(1: 2: 3: 4\)

Solution:

\(
\text { Let a string fixed at both ends vibrate with } n \text { loops. }
\)
We know the length of one loop is \(\frac{\lambda}{2}\).
Now, we can write
\(
L=\frac{n \lambda}{2} \Rightarrow \lambda=\frac{2 L}{n} \quad \text { [for } n \text { loops] }
\)
\(
\begin{aligned}
& \frac{v}{f}=\frac{2 L}{n} \quad[\because v=f \lambda] \\
& f=\frac{n}{2 L} v=\frac{n}{2 L} \sqrt{\frac{T}{\mu}}
\end{aligned}
\)
As velocity of transverse waves in string \(v=\sqrt{\frac{T}{\mu}}\) \(\Rightarrow \quad f \propto n \quad[\because\) length and speed are constants \(]\) \(f_1: f_2: f_3: f_4=n_1: n_2: n_3: n_4\)
Hence, \(f_1: f_2: f_3: f_4=1: 2: 3: 4\)