Question

A car is moving at a velocity $$17\text{m/s}$$ towards an approaching bus that blows horn at a frequency of $$640\text{Hz}$$ on a straight track. The frequency of this horn appears to be $$640\text{Hz}$$ to the car driver. If the velocity of sound in air is $$340\text{m/s}$$, then the velocity of the approaching bus is, in $$\text{m/s}$$
A. 2
B. 17
C. 8
D. 10

Answer 1

Hint: Use the formula for the apparent frequency heard by the observer when the observer and the source of sound are approaching each other. This formula gives the relation between the apparent frequency of the sound, original frequency of sound, velocity of sound, velocity of the source of sound and velocity of the observer. Substitute all the values in this equation and calculate the velocity of the bus.

Formula used:
The frequency $$f^{\prime }$$ of the sound heard by the observer when the source of sound and the observer are approaching towards each other is
$$f^{\prime }=\left({\displaystyle \frac{v+v_o}{v-v_s}}\right)f$$ …… (1)
Here, $$v$$ is velocity of the sound, $$v_s$$ is the velocity of the source of sound, $$v_o$$ is velocity of the observer and $$f$$ is actual frequency of the source of sound.

Complete step by step answer:
We have given that the velocity of the car which is the velocity of the driver of the car is $$17\text{m/s}$$. This is the velocity of the observer (car driver).
$$v_o=17\text{m/s}$$
The given frequency of the sound from the bus horn is $$640\text{Hz}$$ and the frequency of the sound heard by the observer that is car driver is also $$640\text{Hz}$$.
$$f^{\prime }=640\text{Hz}$$
$$\Rightarrow f=640\text{Hz}$$
The velocity of the sound in air is $$340\text{m/s}$$.
$$v=340\text{m/s}$$
We have asked to calculate the velocity of the bus which is the velocity of source of sound.We can calculate the velocity of the bus using equation (1).

Substitute $$640\text{Hz}$$ for $$f^{\prime }$$, $$340\text{m/s}$$ for $$v$$, $$17\text{m/s}$$ for $$v_o$$ and $$640\text{Hz}$$ for $$f$$ in equation (1).
$$\left(640\text{Hz}\right)=\left({\displaystyle \frac{\left(340\text{m/s}\right)+\left(17\text{m/s}\right)}{\left(340\text{m/s}\right)-v_s}}\right)\left(640\text{Hz}\right)$$
$$\Rightarrow 1=\left({\displaystyle \frac{\left(340\text{m/s}\right)+\left(17\text{m/s}\right)}{\left(340\text{m/s}\right)-v_s}}\right)$$
$$\Rightarrow \left(340\text{m/s}\right)-v_s=\left(340\text{m/s}\right)+\left(17\text{m/s}\right)$$
$$\therefore v_s=-17\text{m/s}$$
Therefore, the velocity of the approaching bus is $$17\text{m/s}$$. The negative sign indicates that the direction of the bus is opposite to that of the car and the bus is approaching the car.

Hence, the correct option is B.

Note:The students may think that how can we have the same velocity of the bus as that of the car driver. But the students should keep in mind that the frequency of the sound from the horn of the bus and the frequency of the sound heard by the car driver is the same. If the apparent frequency is different from the frequency of sound then we would have obtained some other value for velocity of the bus. Hence, the velocity of the bus is the same as that of the car.