Chirping Bat and Distance to Moth - Sound Waves

AI Thread Summary
A bat emits a chirp at 67 kHz for 1.00 ms while hunting moths, and the echo returns just after the chirp ends. The speed of sound in air at 10.0 degrees Celsius is 343 m/s. Using the formula d = v*t, the distance to the moth is calculated based on the time for the echo, which is 1.0 ms for the outgoing chirp and 1.0 ms for the return trip, making the total time 2.0 ms. The calculation shows that the distance to the moth is approximately 3.43 cm. The discussion confirms that the frequency of the sound does not affect the distance calculation in this context.
dvolpe
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Homework Statement


A bat emits a chirping sounds of frequency 67 kHz while hunting for moths to eat. Suppose the bat emits a chirp that lasts for 1.00 ms adn then is silent while it listens for the echo. If the beginning of the echo returns just after the outgoing chirp is finished, how close to the moth is the bat. Assume it is a cool night with a temperature of 10.0 degrees celsius. (Hint: is the change in distance between the two significant during the 2.0 ms?)


Homework Equations



d=v*t

The Attempt at a Solution


With such a short time interval, I do not see where the frequency matters. Therefore, used d = v*t where v = speed of sound at 343 m/s and the t = 1.0 ms (1/2 the time for the roundtrip echo). d = 34300 cm/s * 1.0 x e-6 s = 3.43 x e-2 cm

Is is this simplistic? Help, Please!
 
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Lookup 'speed of sound in air'
 
The speed of sound in air is 343 m/s per my book and I put that in my information above, so I do not understand your information.
 
Apologies

Seems good, though the time taken for both trips is 1ms(as I understand from the problem), so t is half that.
 

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