JEE Physics Waves MCQs Set 25

Answer: (d) (a), (b) and (c) only

Explanation:
1. The surface (observer) approaches: it receives \( f(c+v)/c \): (a) true. As a moving source it re-emits \( f'=f\dfrac{c+v}{c-v} \): (c) true.
2. \( \lambda'=\dfrac c{f'}=\dfrac{c(c-v)}{f(c+v)} \): (b) true. Beats \( =f'-f=\dfrac{2vf}{c-v} \), not \( \dfrac{vf}{c-v} \): (d) false.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (a) \( T_2 \) was decreased, or \( T_1 \) was increased

Explanation:
1. \( f\propto\sqrt T \): the string with \( T_1 \) has the higher frequency, the difference is 6 Hz.
2. A slight change keeps the difference at 6 Hz only if the gap between the two frequencies is not reduced to a smaller value: that is, \( T_1 \) is increased or \( T_2 \) is decreased (the gap would shrink if \( T_1 \) were decreased or \( T_2 \) increased).

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (b) 5

Explanation:
1. Each walker is 13 m away (horizontal 5 m, vertical 12 m): \( \cos\theta=\dfrac5{13} \). The front man recedes at \( 1\times\dfrac5{13} \) and the rear man approaches at \( 2\times\dfrac5{13} \).
2. \( f_2-f_1=1430\left[\dfrac{330}{330-10/13}-\dfrac{330}{330+5/13}\right]\approx5.0 \) Hz.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (c) 6 Hz

Explanation:
1. Direct sound: source and passenger move together, so he hears 200 Hz.
2. Echo: \( f=200\dfrac{342+5}{342-5}=205.9 \) Hz. Beats \( \approx6 \) Hz.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (d) zero

Explanation:
1. The direct wave reaching him (source approaching) has frequency \( f\dfrac c{c-v} \).
2. The reflected wave from the stationary wall carries the same frequency \( f\dfrac{c}{c-v} \). Equal frequencies give no beats.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (a) 6

Explanation:
1. The car receives \( f_0\dfrac{330+2}{330} \) and re-emits it as a moving source.
2. Frequency back at the source \( =492\dfrac{330+2}{330-2}=498 \) Hz, so the beat frequency is 6 Hz.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (b) 3

Explanation:
1. Sum of components: \( \cos \)-part \( 1+\tfrac12-\tfrac12-1=0 \); \( \sin \)-part \( 0+\tfrac{\sqrt3}2+\tfrac{\sqrt3}2+0=\sqrt3 \).
2. Resultant amplitude \( =\sqrt3A \), so intensity \( =3I_0 \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (c) 30 m/s

Explanation:
1. \( \dfrac{f_a}{f_r}=\dfrac{v+u}{v-u}=\dfrac{2.2}{1.8}=\dfrac{11}9 \).
2. \( 9v+9u=11v-11u\Rightarrow u=\dfrac v{10}=30 \) m/s.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (d) \( 1.0069\times10^5\ \text{Hz} \)

Explanation:
1. \( v_w=\sqrt{B/\rho}=1445 \) m/s, so the transmitter frequency is \( f_0=\dfrac{1445}{0.01445}=10^5 \) Hz.
2. Source speed relative to water is 8 m/s, wave speed relative to the receiver is \( 1445+2 \): \( f=10^5\dfrac{1447}{1445-8}\approx1.0069\times10^5 \) Hz.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (a) \( 1.0304\times10^5\ \text{Hz} \)

Explanation:
1. Relative to the air the source moves at \( 10+5=15 \) m/s towards the receiver; sound travels at \( 344-5 \) relative to the ground (against the wind).
2. \( f=10^5\dfrac{344-5}{344-15}\approx1.0304\times10^5 \) Hz.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (b) \( \dfrac{2fv_b(v+v_m)}{v^2-v_b^2} \)

Explanation:
1. Direct sound (band recedes from motorist): \( f_1=f\dfrac{v+v_m}{v+v_b} \). Echo: \( f_2=f\dfrac{v}{v-v_b}\cdot\dfrac{v+v_m}{v} \).
2. \( f_2-f_1=f(v+v_m)\left[\dfrac1{v-v_b}-\dfrac1{v+v_b}\right]=\dfrac{2fv_b(v+v_m)}{v^2-v_b^2} \).

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (c) 403.3 Hz to 484 Hz

Explanation:
1. Speed of the whistle \( u=\omega r=30 \) m/s.
2. \( f_{max}=440\dfrac{330}{330-30}=484 \) Hz; \( f_{min}=440\dfrac{330}{330+30}=403.3 \) Hz.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (d) 1.5 m/s

Explanation:
1. Beat frequency \( \approx\dfrac{2uf}{v}=3 \).
2. \( u=\dfrac{3\times340}{2\times340}=1.5 \) m/s.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (a) 0.75 m/s

Explanation:
1. \( \mu=0.01 \) kg/m, \( v=80 \) m/s, \( f=\dfrac{80}{2\times0.1}=400 \) Hz.
2. Receding fork: \( \Delta f\approx\dfrac{uf}v=1\Rightarrow u=\dfrac{300}{400}=0.75 \) m/s.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (b) 27.04 N

Explanation:
1. Pipe: \( f=\dfrac{320}{4(0.4)}=200 \) Hz. Lowering the tension reduces the beats, so the string is at 208 Hz.
2. String first overtone: \( f=\dfrac{v_s}{L}\Rightarrow v_s=52 \) m/s, \( \mu=0.01 \) kg/m, \( T=\mu v_s^2=27.04 \) N.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (c) 7.87

Explanation:
1. Wall receives \( 256\dfrac{330}{325} \); the moving observer hears \( 256\dfrac{330}{325}\cdot\dfrac{335}{330}=263.9 \) Hz.
2. The source itself is still at 256 Hz, so beats \( \approx7.87 \) per second.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (d) \( \pi\ \text{rad (amplitude difference 0)} \)

Explanation:
1. \( \lambda=\dfrac{300}{25}=12 \) m, so \( \Delta\phi=\dfrac{2\pi}\lambda\Delta x=\dfrac{2\pi}{12}\times6=\pi \).
2. The two points have the same amplitude \( 2.5\times10^{-5} \) m, so the difference in amplitudes is zero.

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (a) bends away from the normal

Explanation:
1. Snell's law: \( \dfrac{\sin i}{\sin r}=\dfrac{v_{air}}{v_{water}} \).
2. Since \( v_{water}>v_{air} \), \( r>i \): the wave bends away from the normal.

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (b) 7

Explanation:
1. Reflected frequency \( \approx f_0\left(1+\dfrac{2u}v\right) \), so \( \Delta f=\dfrac{2f_0\Delta u}v=0.012f_0 \).
2. \( \Delta u=0.006\times330=1.98 \) m/s \( \approx7.1 \) km/h \( \approx7 \) km/h.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (c) \( 6.28\times10^{-3}\ \text{s} \)

Explanation:
1. Resultant amplitude \( 2A\cos\dfrac{\Delta\omega t}2 \), so \( I\propto4A^2\cos^2\dfrac{\Delta\omega t}{2}=2A^2(1+\cos\Delta\omega t) \).
2. Maxima repeat with period \( \dfrac{2\pi}{\Delta\omega}=6.28\times10^{-3} \) s.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (d) \( 3.14\times10^{-3}\ \text{s} \)

Explanation:
1. \( I=2A^2(1+\cos\Delta\omega t)\ge2A^2 \) requires \( \cos\Delta\omega t\ge0 \): the detector works for half of each cycle.
2. Idle time \( =\dfrac12\cdot\dfrac{2\pi}{\Delta\omega}=\dfrac\pi{10^3}=3.14\times10^{-3} \) s.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (a) \( \dfrac{2\pi}a,\ \dfrac b{2\pi} \)

Explanation:
1. Comparing with \( \cos(kx+\omega t) \): \( k=a,\ \omega=b \).
2. \( \lambda=\dfrac{2\pi}a \), \( f=\dfrac{\omega}{2\pi}=\dfrac b{2\pi} \).

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (b) \( y_r=-0.8A\cos(ax-bt) \)

Explanation:
1. Intensity \( \propto A^2 \): amplitude ratio \( \sqrt{0.64}=0.8 \), so the amplitude is 0.8A and the wave travels in +x: \( \cos(ax-bt) \).
2. A node at x = 0 needs a phase change of \( \pi \) (reflection from a rigid obstacle): \( y_r=-0.8A\cos(ax-bt) \).

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (c) \( 1.8Ab\ \text{and}\ 0 \)

Explanation:
1. \( y=-1.6A\sin ax\sin bt+0.2A\cos(ax+bt) \): at antinodes the two travelling parts add in magnitude \( (1+0.8)Ab \).
2. Maximum speed \( 1.8Ab \) occurs where \( \sin ax=\pm1 \). The minimum value is zero, since every particle is momentarily at rest at its turning points.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (d) 599.33 Hz

Explanation:
1. The train is at rest relative to the air (speed 40 km/h equals the wind speed), so the wavelength is \( \lambda=\dfrac{1200}{580} \) in the air frame.
2. Speed of sound relative to the hill \( =1200+40 \): \( f=\dfrac{1240}{1200/580}=599.33 \) Hz.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (a) 0.935 km, 620 Hz

Explanation:
1. Echo travels back at \( 1200-40=1160 \) km/h. Condition \( 1160\left(t-\dfrac1{1240}\right)=1-40t\Rightarrow t=1.613\times10^{-3} \) h, giving distance \( 1-40t=0.935 \) km.
2. Echo frequency: the driver (40 km/h towards the hill) receives \( 599.33\times\dfrac{1160+40}{1160}\approx620 \) Hz.

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (b) \( z_1\ \text{and}\ z_2;\ x=(2n+1)\dfrac\pi{2k} \)

Explanation:
1. \( z_1+z_2=2A\cos kx\cos\omega t \), a standing wave, and it is zero for all t where \( \cos kx=0 \).
2. \( kx=(2n+1)\dfrac\pi2\Rightarrow x=(2n+1)\dfrac\pi{2k} \).

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (c) \( z_1\ \text{and}\ z_3;\ x-y=(2n+1)\dfrac\pi k \)

Explanation:
1. \( z_1+z_3=2A\cos\dfrac{k(x-y)}2\cos\left[\dfrac{k(x+y)}2-\omega t\right] \): travelling along the direction \( x+y \), i.e. 45°.
2. Zero where \( \cos\dfrac{k(x-y)}2=0\Rightarrow x-y=(2n+1)\dfrac\pi k \).

Teacher's Note:
1. In Doppler problems write f' = f (v +/- v_observer)/(v -/+ v_source) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.