**Q1.**The frequency of a whistle of an engine is 600 cycles/sec is moving with the speed of 30 m/sec towards an observer. The apparent frequency will be (velocity of sound =330 m/s)

Solution

n^'=n(v/(v-v_S ))=600(330/300)=660 cps

n^'=n(v/(v-v_S ))=600(330/300)=660 cps

**Q2.**The second overtone of an open pipe is in resonance with the first overtone of a closed pipe of length 2m. length of the open pipe is

Solution

Second overtone of open pipe of length 𝒍 is v_0=v/2l …….(i) First overtone of a close pipe is v_c=v/4l=v/(4×2)……..(ii) Equating Eqs. (i) and (ii), we get v/2l=v/8⟹l=4m

Second overtone of open pipe of length 𝒍 is v_0=v/2l …….(i) First overtone of a close pipe is v_c=v/4l=v/(4×2)……..(ii) Equating Eqs. (i) and (ii), we get v/2l=v/8⟹l=4m

**Q3.**Find the fundamental frequency of a closed pipe, if the length of the air column is 42 m. (speed of sound in air =332 m/sec)

Solution

For closed pipe n=v/4l⇒n=332/(4×42)=2Hz

For closed pipe n=v/4l⇒n=332/(4×42)=2Hz

**Q4.**Speed of sound at constant temperature depends on

Solution

Speed of sound, doesn’t depend upon pressure and density medium at constant temperature.

Speed of sound, doesn’t depend upon pressure and density medium at constant temperature.

**Q5.**Out of the given waves (1), (2), (3) and (4) y=a sin〖(kx+ωt)〗 …(1) y=a sin〖(ωt-kx)〗 …(2) y=a cos〖(kx+ωt)〗 …(3) y=a cos〖(ωt-kx)〗 …(4) Emitted by four different sources S_1,S_2,S_3 and S_4 respectively, interference phenomena would be observed in space under appropriate conditions when

Solution

For interference, two waves must have a constant phase relationship. Equation '1' and ‘3’ and ‘2’ and ‘4’ have a constant phase relationship of π/2 out of two choices. Only one S_2 emitting ‘2’ and S_4 emitting ‘4’ is given so only (c) option is correct

For interference, two waves must have a constant phase relationship. Equation '1' and ‘3’ and ‘2’ and ‘4’ have a constant phase relationship of π/2 out of two choices. Only one S_2 emitting ‘2’ and S_4 emitting ‘4’ is given so only (c) option is correct

**Q7.**When sound is produced in an aeroplane moving with a velocity of 200 ms^(-1)horizontal its echo is heard after 10√5s. if velocity of sound in air is 300 ms^(-1)the elevation of aircraft is:

Solution

**Q8.**A wave travelling along a string is described by the equation y=a sin(ωt-kx) the maximum particle velocity is

Solution

Given that, the displacement of a particle is Y=a sin (ωt-kx) …(i) The particle velocity v_p=dy/dt …(ii) Now, on differentiating Eq. (i) w.r.t.t, dy/dt=a cos〖(ωt-kx).ω〗 ⟹ dy/dt=aω cos(ωt-kx) From eq. (ii) ⟹ v_p=aω cos(ωt-kx) For maximum particle velocity, cos (ωt-kx)=1 so, v_p=aω×1⟹v_p=aω

Given that, the displacement of a particle is Y=a sin (ωt-kx) …(i) The particle velocity v_p=dy/dt …(ii) Now, on differentiating Eq. (i) w.r.t.t, dy/dt=a cos〖(ωt-kx).ω〗 ⟹ dy/dt=aω cos(ωt-kx) From eq. (ii) ⟹ v_p=aω cos(ωt-kx) For maximum particle velocity, cos (ωt-kx)=1 so, v_p=aω×1⟹v_p=aω

**Q9.**A wave travelling along positive x-axis is given by y=A sin〖(ωt-kx)〗. If it is reflected from rigid boundary such that 80% amplitude is reflected, then equation of reflected wave is

Solution

On getting reflected from a rigid boundary the wave suffers Hence if y_incident=A sin(ωt=kx) Then y_reflected=(0.8A) sin[ωt-k(-x)+π] =-0.8A sin〖(ωt+kx)〗 an additional phase change of π

On getting reflected from a rigid boundary the wave suffers Hence if y_incident=A sin(ωt=kx) Then y_reflected=(0.8A) sin[ωt-k(-x)+π] =-0.8A sin〖(ωt+kx)〗 an additional phase change of π

**Q10.**A wave travelling in positive X-direction with A=0.2m has a velocity of 360 m/sec. If λ=60m, then correct expression for the wave is

Solution

A wave travelling in positive x-direction may be represented as y=A sin〖2π/λ(vt-x)〗. On putting values y=0.2 sin〖2π/60 (360t-x)⇒y=0.2 sin2π(6t-x/60) 〗

A wave travelling in positive x-direction may be represented as y=A sin〖2π/λ(vt-x)〗. On putting values y=0.2 sin〖2π/60 (360t-x)⇒y=0.2 sin2π(6t-x/60) 〗