Class 11th

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 $*{10}^{-3}$ kg/m and frequency of the tuning fork,  $\nu$ = 256 Hz

Amplitude of the wave, a= 5.0 cm = 0.05 m …. (ii)

Mass of the pan, m = 90 kg and tension of the string, T = mg = 90 $*9.8$ = 882 N

The velocity of the transverse wave, v is given by the relation:

v = $\sqrt{\frac{T}{\mu }}$ = $\sqrt{\frac{882}{8.0\mathrm{}*{10}^{-3}}}$ = 332 m/s

Angular frequency,  $\omega$ = 2 $\pi \nu$ = 2 $\pi *256$ = 1608.5 rad/s = 1.6 $*{10}^3$ rad/s

Wavelength,  $\lambda$ = $\frac{v}{\nu }$ = $\frac{332}{256}$ = 1.297 m

Propagation constant, k = $\frac{2\pi }{\lambda }$ = $\frac{2\pi }{1.297}$ = 4.844 /m

Substituting these values in the displacement equation, we get

y (x, t) = a sin ( $\omega t-kx)$

y (x, t) = 0.05 sin (1.6 $*{10}^3$ t – 4.844 x) m

(a) The given harmonic wave is

y(x, t) = 7.5 sin (0.0050x +12t + $\pi$ /4)

For x = 1 cm and t = 1 s

y(1, 1) = 7.5 sin (0.0050 +12+ $\mathrm{\pi }$ /4)

= 7.5 sin( 12.0050 + $\frac{\pi }{4}$ )

= 7.5 sin (12.0050 + 0.7854)

= 7.5 sin (732.84 $°$ )

= 7.5 sin (90 $*8+$ 12.81) $°$

= 7.5 sin 12.81 $°$

=1.6629 cm

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

v = $\frac{d}{dt}$ y(x,t) = $\frac{d}{dt}\left[7.5\mathrm{}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{}(0.0050x+12t+\pi /4)\right]$

= 7.5 $*12\mathrm{c}\mathrm{o}\mathrm{s}?(0.0050x+12t+\pi /4$ )

At x= 1 cm and t= 1 s

v= y (1,1) = 90 cos (12.005 + $\frac{\pi }{4}$ ) = 90 cos(12.81 $°$ ) = 87.75 cm/s

Now the equation of a propagating wave is given by

Y(x, t) = a sin (kx + $\omega$ t + $\phi )$ , where

k = $\frac{2\pi }{\lambda }$ or $\lambda$ = $\frac{2\pi }{k}$ and $\omega =2\pi v$

Speed, v = $\frac{\omega }{k}$ , where $\omega$ = 12 rad/s and k = 0.0050 /m

v = $\frac{12}{0.0050}$ = 2400 cm/s

Hence, the velocity of the wave oscillation at x=1 cm and t=1s is not equal to the velocity of the wave propagation.

(b) Propagation constant is related to wavelength as:

$\lambda$ = $\frac{2\pi }{k}$ = $\frac{2\pi }{0.0050}$ = 1256 cm = 12.56 m

Therefore, all the points at distances n $\lambda$ (n = $\pm 1,\pm 2\dots \dots .\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{}\mathrm{o}\mathrm{n})$ i.e. $\pm 12.56m,\pm 25.12m\dots \dots .$ and so on for x=1 cm, will have the same displacement as the x=1 cm points at t= 2s, 5s and 11s.

(a) For the stationary observer:

Frequency of the sound produced by the whistle,  = 400 Hz

Speed of sound = 340 m/s

Velocity of wind, v = 10 m/s

As there is no relative motion between the source and the observer, the frequency of the sound heard by the observer will be the same as that produced by the source, i.e. 400 Hz. The wind is blowing towards the observer. Hence, the effective speed of the sound increases by 10 units, i.e.

Effective speed of the sound, $v_e$ = 340 + 10 = 350 m/s

The wavelength ( $\lambda )\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}$ heard by the observer is given by the relation:

$\lambda =\frac{v_e}{\nu }$ = $\frac{350}{400}$ = 0.875 m

(b) For the running Observer:

$\mathrm{V}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{b}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{v}\mathrm{e}\mathrm{r}$ , $v_o$ = 10 m/s

$\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{b}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{v}\mathrm{e}\mathrm{r}$ is moving towards the source. As a result of the relative motions of the source and the observer, there is a change in frequency (f''')

$\mathrm{T}\mathrm{h}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{b}\mathrm{y}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}:$ f''' = ( $\frac{v+v_o}{v})\nu$ = ( $\frac{340+10}{340})*400=$ 411.76 Hz

Since the air is still, the effective speed of sound = 340 + 0 = 340 m/s

The source is at rest. Therefore, the wavelength of the sound will not change, i.e. $\lambda$ remains 0.875 m.

Hence, the given two situations are not exactly identical.

Frequency of the whistle,  $\nu$ = 400 Hz

Speed of the train,  $v_r$ = 10 m/s

Speed of sound, v = 340 m/s

The apparent frequency (f') of the whistle as the train approaches the platform is given by the relation:

f' = ( $\frac{v}{v-v_r})\nu$ = ( $\frac{340}{340-10})*400=$ 412.12 Hz

The apparent frequency (f”) of the whistle as the train recedes from the platform is given by the relation:

f” = ( $\frac{v}{v+v_r})\nu$ = ( $\frac{340}{340+10})*400=$ 388.57 Hz

The apparent change in the frequency of sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of sound.

Therefore, the speed of sound in air in both the cases remains the same, i.e. 340 m/s.

(a) A node is a point where the amplitude of vibration is the minimum and the pressure is maximum. On the other hand, an antinode is a point where the amplitude of vibration is maximum and the pressure is minimum. Therefore, a displacement node is nothing but a pressure antinode and vice versa.

(b) Bats emit very high-frequency ultrasonic sound waves. These waves get reflected back toward them by obstacles. A bat receives a reflected wave (frequency) and estimates the distance, direction, nature and size of the obstacle with the help of its brain senses.

(c) The overtone produced by a sitar and a violin, and the strengths of these overtones, are different. Hence, one can distinguish between the notes produced by a sitar and a violin even if they have the same frequency of vibration.

(d) Solids have shear modulus. They can sustain shearing stresses. Since fluids do not have any definite shape, they yield to shearing stress. The propagation of a transverse waves is such that it produces shearing stress in a medium. The propagation of such wave is possible only in solids, and not in gases. Both solids and fluids have their respective bulk moduli. They can sustain compressive stress. Hence, longitudinal waves can propagate through solids and fluids.

(e) A pulse is actually being a combination of waves having different wavelengths. These waves travel in a dispersive medium with different velocities, depending on the nature of the medium. This results in the distortion of the shape of a wave pulse.