Question

A truck blowing a horn of frequency \[500{\text{ }}Hz\] travels towards a vertical mountain and the driver hears an echo of frequency 600Hz. If the velocity of sound in air is \[340m/s\] then the speed of the truck is:
A. \[31m/s\]
B. \[41m/s\]
C. \[51m/s\]
D. \[21m/s\]

Answer 1

Hint: To answer the above question we need to look into the Doppler Effect. The Doppler effect which is also called the Doppler shift is a phenomenon that is observed whenever the source of waves is traveling with respect to the observer. This effect describes the change in frequency of any sound waves or even light waves produced by a moving source with respect to the observer.

Formula used:
Doppler shift’s formula varies with respect to several cases. The general formula of the Doppler shift is given below,
\[f' = \dfrac{{(v \pm {v_0})}}{{(v \pm {v_s})}}f\]
Here, \[f'\] is said to be the observed frequency
\[f\] is the actual frequency
\[v\] is said to be the velocity of the sound waves
\[{v_0}\] is said to be the velocity of the observer
\[{v_s}\] is said to be the velocity of the source.

Complete step-by-step answer:
In our case, the observer is the driver in the truck and the source is the truck. And the exact cause of our question is that the source and the observer both are moving in the direction of the sound. Therefore the formula for the Doppler shift for this case is given by,
\[f' = \dfrac{{(v + {v_0})}}{{(v - {v_s})}}f\]
Since the observer and the source are moving with the same velocity we can write, \[{v_0} = {v_s}\]. Therefore the above equation becomes,
\[f' = \dfrac{{(v + {v_0})}}{{(v - {v_0})}}f\]
Given that the observed frequency that is \[f' = 600Hz\] and the actual frequency that is the frequency with which the truck is blowing is given by \[f = 500Hz\]. And the velocity of the sound that is given by \[v = 340m/s\]. We need to find out \[{v_0}\].
Therefore substituting all the formulas in the above equation we get,
\[600 = \dfrac{{(340 + {v_0})}}{{(340 - {v_0})}}500\]
\[ \Rightarrow \dfrac{6}{5}(340 - {v_0}) = 340 + {v_0}\]
\[ \Rightarrow 2040 - 6{v_0} = 1700 + 5{v_0}\]
\[ \Rightarrow 2040 - 1700 = 5{v_0} + 6{v_0}\]
\[ \Rightarrow 340 = 11{v_0}\]
\[ \Rightarrow \dfrac{{340}}{{11}} = {v_0}\]
Therefore simplifying the above equation we get,
\[{v_0} = 30.9m/s\]
Therefore we can approximate the above value we get,
\[{v_0} = 31m/s\]
Therefore the velocity of the observer is found to \[31m/s\]. Since we already said that the observer and the source are moving at the same velocity we will have the same velocity for the truck. Therefore the speed of the truck is given by, \[31m/s\].

So, the correct answer is “Option A”.

Note: The actual formula for our case is given by,
\[f' = \dfrac{{(v + {v_0})}}{{(v + {v_0})}}f\]
But we have used a negative sign in the denominator because the source and the observer are moving in the same direction. If they both are moving away from each other then we would have used the positive sign. There are also seven other cases in the Doppler shift that have different formulas for each case.