An acoustic burglar alarm consists of a source emitting waves of frequency 28.0 kHz. What is the beat frequency between the source waves and the waves reflected from an intruder walking at an average speed of 0.095 m/s directly away from the alarm?
To find the beat frequency between the source waves and the waves reflected from the intruder, we first need to calculate the frequency shift caused by the Doppler effect.
The Doppler effect describes the change in frequency of a wave due to the relative motion between the source of the wave and the observer.
The formula for the Doppler effect with a moving source and a moving observer is given by:
f' = ((v + vo)/(v + vs)) * f
Where:
f' is the observed frequency
f is the source frequency
v is the speed of sound in air (assumed to be 343 m/s)
vo is the speed of the observer (in this case, 0.095 m/s away from the alarm)
vs is the speed of the source (assumed to be zero for a fixed alarm)
v is the velocity of the sound wave
Let's plug in the values:
f' = ((343 + 0.095)/(343 + 0)) * 28000
Simplifying further:
f' = (343.095/343) * 28000
f' ≈ 28171.8 Hz
Now, the beat frequency is the absolute difference between the observed frequency and the source frequency:
beat frequency = |f' - f|
beat frequency = |28171.8 - 28000|
beat frequency ≈ 171.8 Hz
Therefore, the beat frequency between the source waves and the waves reflected from the intruder is approximately 171.8 Hz.