Doppler Effect & Beat Frequency

[dcramps]

Homework Statement

Horseshoe bats use the Doppler Effect to determine their
location. A horseshoe bat flies toward a wall at a speed of 15 m/s while emitting a sound of
frequency 35 kHz. What is the beat frequency between the emission frequency and the
echo? Assume that the air temperature is 20C.

Homework Equations

Doppler effect
f₀=[(1-v₀/v)/(1-v_s/v)]f_s
Beat Frequency
f_beat = Δf

The Attempt at a Solution

f₀=[(1-(-343m/s)/(15m/s))/(1-(343m/s)/(15m/s))]*35kHz
f₀=-38.20121951kHz

f_beat = Δf = f_s-f₀
f_beat = 35kHz-(-38.20121951kHz)
f_beat = 73.2kHz

I'm not sure that I have solved this correctly. It seems strange to me that the echo has more than double the frequency of the original sound. edit; or should I say, that the beat frequency is more than double that of the original sound.

 

[dcramps]

Also, I assumed v₀ was negative because the sound is traveling in the opposite direction of the source at that point. Is this correct?

 

[simondupnik]

Well, the answering 'tutor's' fomulas may be approximately OK, especially if lifted from some random handbook, but his understanding of beat frequency is rubbish.

Instead mechanically using the signs that come out of these arbitrary formulas, you should find the difference between absoulte values (modulus), ie compare both frequencies without the minus sign. When you do this you will have computed approx 3kHz, which could be a more reasonable answer.

Frequency here is a scalar, not a vector. You complare them just like you compare the lengths of two sticks: you subtract from the longer one the length of the shorter, no matter in which direction the sticks may be pointing.