Question

From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequency that is 0.959 times as small as the frequency emitted by the car when it is stationary. The speed of sound is 343 m/s. What is the speed of the car?

Answer 1

Approximately 15 m/s is the speed of the car.

Explanation:

Given:

speed of sound - 343 m/s

You detect a frequency that is 0.959 times as small as the frequency emitted by the car when it is stationary. So, it can be written as,

$f^{\prime}=0.959 f$

$\frac{f^{\prime}}{f}=0.959$

If there is relative movement between an observer and source, the frequency heard by an observer differs from the actual frequency of the source. This changed frequency is called the apparent frequency. This variation in frequency of sound wave due to motion is called the Doppler shift (Doppler effect). In general,

                     $f^{\prime}=\left(\frac{v+v_{0}}{v-v_{s}}\right) \times f$

Where,

$f^'$ - Observed frequency

f – Actual frequency

v – Velocity of sound waves

$v_0$ – Velocity of observer

$v_s$ - velocity of source

When source moves away from an observer at rest ($v_{0} = 0$), the equation would be

                        $f^{\prime}=\left(\frac{v}{v-\left(-v_{s}\right)}\right) \times f$

                        $f^{\prime}=\left(\frac{v}{v+v_{s}}\right) \times f$

                        $\frac{f^{\prime}}{f}=\left(\frac{v}{v+v_{s}}\right)$

By substituting the known values, we get

            $0.959=\left(\frac{343}{343+v_{s}}\right)$

           $0.959\left(343+v_{s}\right)=343$

           $0.959(343)+0.959\left(v_{s}\right)=343$

           $328.937+0.959 v_{s}=343$

           $0.959 v_{s}=343-328.937=14.063$

           $v_{s}=\frac{14.063}{0.959}=14.66 \mathrm{m} / \mathrm{s}$

Approximately 15 m/s is the speed of the car.