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Question: A band playing music at a frequency f is moving towards a wall at a speed \({V_b}\). A motorist is f...

A band playing music at a frequency f is moving towards a wall at a speed Vb{V_b}. A motorist is followed by the band with a speed Vm{V_m}. If VV is the speed of the sound, the expression for the beat frequency heard by the motorist is:
A) V+VmV+Vbf\dfrac{{V + {V_m}}}{{V + {V_b}}}f
B) V+VmV−Vbf\dfrac{{V + {V_m}}}{{V - {V_b}}}f
C) 2Vb(V+Vm)V2−Vb2f\dfrac{{2{V_b}(V + {V_m})}}{{{V^2} - V_b^2}}f
D) 2Vm(V+Vb)V2−Vm2f\dfrac{{2{V_m}(V + {V_b})}}{{{V^2} - V_m^2}}f

Explanation

Solution

When the source of the waves is moving with respect to an observer, then Doppler’s effect can be observed. It is the change in the frequency of a wave. Since the Doppler’s effect depends on the moving objects, it can also be used to determine the speed or motion of the object.

Complete step by step answer:
Step I:
Given that the band is playing a music that moves towards the ball with velocity Vb{V_b}. Then a motorist is following the band with a speed Vm{V_m}. Therefore, the band is the source and the motorist is the observer.

Step II:
So, in this case, according to Doppler’s effect, the observed frequency is given by
f′=f(V+Vm)V+Vb)f' = f(\dfrac{{V + {V_m})}}{{V + {V_b}}})----(i)
f′f' is the observer frequency of sound
ff is actual frequency of sound waves
VV is the speed of sound waves
Vm{V_m} is the observer velocity
Vb{V_b} is the source velocity
Step III:
When the sound waves of the music band strikes the wall, the wall acts as an observer and the band acts as a source. The observer is stationary and not moving so its velocity will be zero.
Again using Doppler’s effect, when the source is approaching towards the observer, the frequency is given as
f′′=f(VV−Vb)f'' = f(\dfrac{V}{{V - {V_b}}})---(ii)
Step IV:
Now when the sound waves get reflected from the wall, they approach the motorist. Therefore, the wall will be the source and the motorist will be the observer. The wall is stationary and acting as a source, so its velocity will be zero. When the observer is approaching the source, the frequency is
f′′′=f′′(V+VmV)f''' = f''(\dfrac{{V + {V_m}}}{V})---(iii)
Step V:
Substitute the value from equation (ii) in equation (iii),
f′′′=f(VV−Vb)(V+VmV)f''' = f(\dfrac{V}{{V - {V_b}}})(\dfrac{{V + {V_m}}}{V})
⇒f′′′=f(V+VmV−Vb)\Rightarrow f''' = f(\dfrac{{V + {V_m}}}{{V - {V_b}}})---(iv)
Step VI:
The sound heard by the motorist can be written by using beat frequency. The beat frequency is equal to the difference of the frequency of two waves. Therefore,
fb=f′′′−f′{f_b} = f''' - f'
⇒fb=f(V+VmV−Vb)−f(V+VmV+Vb)\Rightarrow {f_b} = f(\dfrac{{V + {V_m}}}{{V - {V_b}}}) - f(\dfrac{{V + {V_m}}}{{V + {V_b}}})
⇒fb=f(V+Vm)(1V−Vb−1V+Vb)\Rightarrow {f_b} = f(V + {V_m})(\dfrac{1}{{V - {V_b}}} - \dfrac{1}{{V + {V_b}}})
⇒fb=f(V+Vm)(V+Vb−V+VbV2−Vb2)\Rightarrow {f_b} = f(V + {V_m})(\dfrac{{V + {V_b} - V + {V_b}}}{{{V^2} - V_b^2}})
⇒fb=f(V+Vm)(2VbV2−Vb2)\Rightarrow {f_b} = f(V + {V_m})(\dfrac{{2{V_b}}}{{{V^2} - V_b^2}})

∴\therefore The frequency of the band heard by the motorist is f(V+Vm)(2VbV2−Vb2)f(V + {V_m})(\dfrac{{2{V_b}}}{{{V^2} - V_b^2}}). Hence option C is the right answer.

Note:
It is important to note that the Doppler’s effect does not result because of an actual frequency change of the source. It can be observed for many types of waves like sound, water or light waves. Also sound waves require a medium to propagate. Doppler’s effect is asymmetric for sound waves.