Question

Two police cars have identical sirens that produce sound of frequency of 570 Hz. A stationary listener is standing between two cars. One car is parked and the other is approaching the listener and both have their sirens on. The listener measures 9.0 beats per second. Find the speed of the approaching police car. The speed of sound is 340 m/s.

Answer 1

Answer:

The speed of approaching police car is 5.35 m/s.

Explanation:

Given :

Police car actual frequency = 570 Hz

No of beats heard by stationary observer = 9

The speed of sound = 340 m/s

Beats is nothing but the difference of the frequency that heard by the observer.

According to the doppler equation,

⇒  $f = f' (v-vo)/(v-vs)$

Where $f$= observed frequency, $f'$ = actual frequency, $v$ = speed of sound,

$vo$ = speed of observer, $vs$ = speed of source.

One car approach stationary listener ($vo$ = $0$ )

∴ $f1$ = $570$×$(340)/(340-vs)$

One car parked, so ( $vs = 0$) and $vo=0$

∴ $f2$ = $570$

From beats,

∴ $f1 -f2=$ no. of beats.

Where $f1$ = frequency of approaching car, $f2$ = frequency of stationary car.

∴ $570$×$340/(340-vs)$ $- 570$ = $9$

While solving above equation,

∴ $vs =$ 5.35 m/s.

Therefore, the speed of approaching police car is 5.35m/s.