Question

At an auto race, a member of the pit crew stands beside the track. A car approaches him at 100 m/s and emits a sound at frequency 1100Hz. The air is still and the speed of sound is 340mls. What frequency will the pit crew member hear?

A.1560Hz
B.1420Hz
C.1640Hz
D.850Hz

Answer 1

Answer:

A. 1560 Hz

Explanation:

The Doppler effect occur when a source of a wave (like sound) is moving relative to an observer (or when the observer is moving relative to the source). When this happens, the apparent frequency of the sound as heard by the observer is shifted with respect to the real frequency, according to the formula:

$f'=\frac{v}{v-v_s}f$

where

f' is the apparent frequency

v is the speed of the wave

vs is the velocity of the source relative to the observer (positive if the source is moving towards the observer, negative if it is moving away from the observer)

f is the real frequency of the wave

In this problem, we have the following data:

f = 1100 Hz is the original frequency of the sound

v = 340 m/s is the speed of the sound wave

$v_s = 100 m/s$ is the velocity of the car approaching the pit crew

Substituting these numbers into the formula, we can find the frequency heard by the pit crew:

$f'=\frac{340 m/s}{340 m/s-100 m/s}(1100 Hz)=1558 Hz \sim 1560 Hz$

Answer 2

Answer:  option A. 1560 Hz

Explanation:

Doppler effect is the phenomenon wherein the frequency of the source of the sound is different from what is received. This happens when either of the source or observer or both are relatively in motion.

In the given question, the observer is stationary and the source of sound is moving at the speed of $v_s = 100 m/s$ towards the observer. The frequency of the sound emitted is $f_s = 1100 Hz$. The frequency heard by the pit crew member would be given by:

$f_o = \frac{V}{V-v_s}f_s$

$f_o = \frac{340 m/s}{340m/s - 100 m/s}\times 1100 Hz = 1558.3 Hz = 1560 Hz (approx)$

Thus, the frequency heard by the pit crew would be 1560 Hz. Correct option is A.