Chapter 14: Waves

Ans: For the stationary observer:

Frequency of the sound generated by the whistle, \[\nu =400\text{ }Hz\]

Speed of sound \[=\text{ }340\text{ }m/s\]

Velocity of the wind, \[v=10\text{ }m/s\]

As there is no relative motion between the observer and the source, the sound’s frequency heard by the observer will be  similar to that produced by the source, i.e., \[400\text{ }Hz\]. 

The wind is blowing towards the observer. Hence, the sound’s effective speed increases by \[10\] units, i.e., \[{{v}_{e}}=340+10=350m/s\]

The wavelength \[(\lambda )\] of the sound heard by the observer is given by: 

\[\lambda =\frac{{{v}_{e}}}{\nu }=\frac{350}{400}=0.875m\]

For the running observer: 

Velocity of the observer, \[{{v}_{o}}=10\text{ }m/s\]

The observer is moving towards the source. As a result, the relative motions of the source and the observer, there is a frequency change (\[\nu '\])

 This is given by the relation:

 \[\nu '=\left( \frac{v+{{v}_{o}}}{v} \right)\nu \]

\[\Rightarrow \nu '=\left( \frac{340+10}{340} \right)\times 400=411.76Hz\]

Since the air is still so, the effective speed of sound \[=340+0=340\text{ }m/s\]

The source is at rest. Hence, the sound’s wavelength will not change, i.e., $\lambda $remains \[0.875\text{ }m.\]

Hence, the two given situations are not exactly identical.

Additional Exercise:

Ans: The given harmonic wave is:

\[y(x,t)=7.5\sin \left( 0.0050x+12t+\frac{\pi }{4} \right)\]

For \[x\text{ }=\text{ }1\text{ }cm\] and \[t\text{ }=\text{ }1\text{ }s,\]

\[y(1,1)=7.5\sin \left( 0.0050+12+\frac{\pi }{4} \right)\]

\[\Rightarrow y(1,1)=7.5\sin \left( 12.0050+\frac{\pi }{4} \right)\]

\[\Rightarrow y(1,1)=7.5\sin \theta \]

\[Where,\theta =12.0050+\frac{\pi }{4}=12.0050+\frac{3.14}{4}=12.78rad\]

\[\Rightarrow \theta =\frac{180}{3.14}\times 12.78={{732.81}^{o}}\]

\[\therefore y(1,1)=7.5\sin ({{732.81}^{o}})\]

\[\Rightarrow y(1,1)=7.5\sin (90\times 80+{{12.81}^{o}})=7.5\sin {{12.81}^{o}}\]

\[\Rightarrow y(1,1)=7.5\times 0.2217\]

\[\Rightarrow y(1,1)=1.6629\approx 1.663cm\]

The velocity of the oscillation at a given point and time is given as:

\[v=\frac{d}{dt}y(x,t)=\frac{d}{dt}\left[ 7.5\sin \left( 0.0050x+12t+\frac{\pi }{4} \right) \right]\]

\[\Rightarrow v=7.5\times 12\cos \left( 0.0050x+12t+\frac{\pi }{4} \right)\]

At \[x\text{ }=\text{ }1\text{ }cm\] and \[t\text{ }=\text{ }1s\], 

\[v=y(1,1)=90\cos \left( 12.005+\frac{\pi }{4} \right)\]

\[\Rightarrow v=90\cos ({{732.81}^{o}})=90\cos \left( 90\times 8+{{12.81}^{o}} \right)\]

\[\Rightarrow v=90\cos ({{12.81}^{o}})\]

\[\Rightarrow v=90\times 0.975=87.75cm/s\]

Now, the equation of a propagating wave is given by:

\[y(x,t)=asin(kx+\omega t+\phi )\]

Where, \[k=\frac{2\pi }{\lambda }\]

\[\lambda =\frac{2\pi }{k}\] and \[\omega =2\pi \nu \]

\[\Rightarrow \nu =\frac{\omega }{2\pi }\]

\[Speed,v=\nu \lambda =\frac{\omega }{k}\]

Where, \[\omega =12rad/s\]

\[k=0.0050{{m}^{-1}}\]

\[\Rightarrow v=\frac{12}{0.0050}=2400cm/s\]

Hence, the velocity of the wave oscillation at \[x=1\text{ }cm\] and \[t=1\text{ }s\] is not equal to the wave propagation’s velocity. 

Ans: The relation of propagation constant with wavelength is given by:

\[k=\frac{2\pi }{\lambda }\]

\[\Rightarrow \lambda =\frac{2\pi }{k}=\frac{2\times 3.14}{0.0050}=1256cm=12.56m\]

Therefore, all the points at distances \[n\lambda (n=\pm 1,\pm 2,...and\text{ }so\text{ }on)\], i.e., \[\pm 12.56m,\pm 25.12m,..\ldots \] and so on for \[x=1\text{ }cm\], will have the same displacement as the \[x=1\text{ }cm\] points at \[t=2\text{ }s,\text{ }5\text{ }s,\text{ }and\text{ }11\text{ }s\].

Ans: (a) The sound pulse does not have a constant wavelength or frequency. However, the sound speed vibration remains the same, equal to the sound speed in that medium.

Ans: The short pip produced after every \[20\text{ }s\] does not implies that the frequency of the whistle is \[\frac{1}{20}\] or \[0.05\text{ }Hz\]. It means that \[0.05\text{ }Hz\] is the frequency of the repetition of the pip of the whistle.

Ans: The equation of a travelling wave propagating along the positive y-direction is given by the displacement equation:

\[y(x,t)=a\sin (\omega t-kx)\] ……(i)

Linear mass density, \[\mu =8.0\times {{10}^{-3}}kg{{m}^{-1}}\]

Frequency of the tuning fork, \[\nu =256\text{ }Hz\]

Amplitude of the wave, \[a=5.0\text{ }cm=0.05\text{ }m\]……(ii)

Mass of the pan, \[\text{m}=90\text{ }kg\]

Tension in the string, \[T=mg=90x9.8=882\text{ }N\]

The transverse wave’s velocity is given by:

\[v=\sqrt{\frac{T}{\mu }}\]

\[\Rightarrow v=\sqrt{\frac{882}{8.0\times {{10}^{-3}}}}=332m/s\]

Angular frequency, \[\omega =2\pi \nu \]

\[\Rightarrow w=2\times 3.14\times 256\]

\[\Rightarrow w=1608.5=16\times {{10}^{3}}rad/s\]……(iii)

Wavelength, \[\lambda =\frac{v}{\nu }=\frac{332}{256}m\]

\[\text{Propagation constant},k=\frac{2\pi }{\lambda }\]

\[\Rightarrow k=\frac{2\times 3.14}{{}^{332}/{}_{256}}=4.84{{m}^{-1}}\]……(iv)

Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get,

\[y(x,t)=0.05\sin (1.6\times {{10}^{3}}t-4.84x)m\]

This describes the wave of the string.

Ans: It is provided that,

SONAR system’s operating frequency, \[\nu =40\text{ }kHz\]

Speed of the enemy submarine, \[{{v}_{e}}=360km/h=100\text{ }m/s\]

Sound speed in water, \[v=1450\text{ }m/s\]

The source is at rest and the observer is moving towards it. Hence, the apparent frequency (\[\nu '\]) received and reflected by the submarine is given by:

\[\nu '=\left( \frac{v+{{v}_{e}}}{v} \right)\nu \]

\[\Rightarrow v'=\left( \frac{1450+100}{1450} \right)\times 40=42.76kHz\]

The frequency (\[\nu ''\]) received by the enemy submarine is given by:

\[\nu ''=\left( \frac{v}{v-{{v}_{s}}} \right)\nu '\]

Where, \[{{v}_{s}}=\text{ }100\text{ }m/s\]

\[\Rightarrow \nu ''=\left( \frac{1450}{1450-100} \right)42.72=45.93kHz\]

Ans: Let \[{{v}_{s}}\] and \[{{v}_{p}}\] be the velocities of S and P waves respectively. 

Let L be the distance between the seismograph and the epicentre. 

We have, \[L={{v}_{s}}{{t}_{s}}\]    ……(i)

\[L={{v}_{p}}{{t}_{p}}\]    ……(ii)

Where, \[{{t}_{s}}\] and \[{{t}_{p}}\] are the respective times taken by the S and P waves to reach the seismograph from the epicentre. 

It is given that, \[{{v}_{p}}=8km/s\]

\[{{v}_{s}}=4km/s\]

From equations (i) and (ii), we have:

\[{{v}_{s}}{{t}_{s}}={{v}_{p}}{{t}_{p}}\]

\[4{{t}_{s}}=8{{t}_{p}}\]

\[{{t}_{s}}=2{{t}_{p}}\]   ……(iii)

It is also given that:

\[{{t}_{s}}-{{t}_{p}}=4\min =240s\]

\[\Rightarrow 2{{t}_{p}}-{{t}_{p}}=240\]

\[\Rightarrow {{t}_{p}}=240\]

And \[{{t}_{s}}=2\times 240=480s\]

From equation (ii), we get: 

\[L=8\times 240=1920\text{ }km\]

Clearly, the earthquake occurs at a distance of \[1920\text{ }km\] from the seismograph.

Ans: Bat’s ultrasonic beep frequency, \[\nu =40\text{ }kHz\]

Velocity of the bat, \[{{v}_{b}}=0.03v\]

 Where, v is the velocity of sound in air

 The apparent frequency of the sound striking the wall is given by:

\[\nu '=\left( \frac{v}{v-{{v}_{b}}} \right)\nu \]

\[\Rightarrow \nu '=\left( \frac{v}{v-0.03v} \right)\times 40\]

\[\Rightarrow \nu '=\frac{40}{0.97}kHz\]

This frequency is reflected by the stationary wall (vs =0) toward the bat. 

The frequency (\[\nu ''\]) of the received sound is given by the relation:

\[\nu ''=\left( \frac{v+{{v}_{b}}}{v} \right)\nu '\]

\[\Rightarrow \nu ''=\left( \frac{v+0.03v}{v} \right)\times \frac{40}{0.97}\]

\[\Rightarrow \nu ''=\frac{1.03\times 40}{0.97}=42.47kHz\]

The frequency of the received sound is \[42.47kHz\].