Question

An ambulance is traveling towards the scene of an accident at 40 m/s. Since this is a rather high speed the ambulance is blaring its siren that has a frequency of 325 Hz. The speed of sound is around 340 m/s

Part 1. Find the wavelength of the sound waves in front of the ambulance. Give your answer to 3 significant digits

m

Part 2. Due to the Doppler Effect what is the effective frequency of this sound to the listener f_L? Give your answer to 3 significant digits

Hz

Part 3. Use this new frequency and that the amplitude of the sound waves are 3.85 cm to find what the intensity of the sound wave is to an observer standing at the crash site when the ambulance is 1000 m away?

Give your answer to 3 significant digits

W/m²

Part 4. What is the intensity when the ambulance is parked 10 m away from the crash site? Give your answer to 3 significant digits

W/m²

Part 5. What is the intensity level for both? Give your answer to the nearest decibel

1000 m: dB

10 m: dB

Answer 1

1) frequency f= 325 hz ; speed = 340 m/s

wave length in front of the ambulance   = v/f = 340/325 = 1.046 m

2)

Doppler effect

f_o = f_s * v/ (v - v_s)

here source is moving and observer is stationary, source approaching the observer

f_o = 325 * 340 /(340-40) = 368.33 Hz , observer frequency

wave length   = 340/368.88 = 0.923 m

2) frequency (observer) = 368.33 Hz

3) Amplitude of the sound wave = 3.85 cm

Intensity of sound

  

s_o = 3.85 displacement amplitude

I = 340 * 2(3.14*368.33)² (3.85e-2)² *1.275

= 1.27 e+6 w , total power emitted in all 4 direction

intensity = power /unit Area

at 1000 m

I = 1.27 e+6 /4 1000² = 0.10 w/sq.m , when the ambulance is 1000 m away from the crash site.

4) Intensity when the ambulance is 10m away

I = 1.27 e+6 /4 10² = 1000 w/sq.m

5) Intensity in db

1000 m : 10 log (0.1/10^-12 ) = 110 db

10 m : 10 log (1000/10^-12 ) = 150 db