You're driving down the road at 30 m/s when you see an ambulance approaching you from behind. If you measure the peak frequency of its siren to be 320 Hz when it's behind you and 300 Hz after it's passed you, how fast is the ambulance driving? Assume the speed of sound in air to be 340 m/s

39.45 m/s

Explanation:

This phenomenon is known as Doppler Effect, and it occurs when there is a source of a wave in motion with respect to an observer. In such situation, the frequency of the wave as perceived by the observer is shifted with respect to the real frequency.

In particular:

- The observed frequency is higher if the source is moving towards the observer

- The observed frequency is lower if the source is moving away from the observer

The formula to calculate the apparent frequency in the Doppler Effect is:

where

f is the real frequency

f' is the observed frequency

v is the speed of the wave

is the velocity of the source, which is positive if the source is moving away, negative if it is moving towards the observer

is the velocity of the observer, which is negative if the observer is moving away, positive if it is moving towards the source

In this problem:

v = 340 m/s is the speed of sound

- When the ambulance is approaching you from behind,

f' = 320 Hz

(car is moving away from ambulance)

So

(1)

- When the ambulance overcomes you and moves away from you,

f' = 300 Hz

(car is moving towards ambulance)

So

(2)

Re-arranging both equations (1) and (2),

1)

2)

Both equations can be rewritten as

And by equating them we find:

no

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