A car alarm is emitting sound waves of frequency 520 Hz. You are on a motorcycle, traveling directly away from the parked car. How fast must you be traveling if you detect a frequency of 490 Hz?

Given data:

Given that the stationary car's sound frequency is 520 Hz and the frequency detected by the moving listener is 490 Hz, the velocity at which the listener is moving away from the car must be determined in order to detect that frequency.

Therefore, the formula to detect frequency by the listener;

$f_L=f_s\left(\frac{v\pm v_L}{v+v_S}\right)$

Here,

The speed of sound in air $v=344m/s$

Speed listener $v_L$

The source is stationary $v_S=0m/s$

Frequency of the sound emitted by the car localid="1664339995390" $f_S=520\mathrm{Hz}$

Frequency of the sound detected by the listener $f_L=490Hz$

Minus sign will be used in the numerator as the listener is moving away from the car.

$$\begin{gathered} \frac{f_L}{f_S}=\left(\frac{v-v_L}{v+0}\right) \\ v-v_L=\frac{v\times f_L}{f_S} \\ {v}_L=v-\frac{v\times f_L}{f_S} \end{gathered}$$

Putting the values;

$$\begin{gathered} {\mathrm{v}}_{\mathrm{L}}=\left(344\mathrm{m}/\mathrm{s}\right)-\frac{\left(344\mathrm{m}/\mathrm{s}\right)\times \left(490\mathrm{Hz}\right)}{520\mathrm{Hz}} \\ {\mathrm{v}}_{\mathrm{L}}=19.8\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the velocity will be 19.8m/s.