Chirping Bat and Distance to Moth - Sound Waves

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Homework Help Overview

The problem involves a bat emitting a chirping sound at a frequency of 67 kHz while attempting to determine the distance to a moth based on the echo of the sound. The scenario is set at a temperature of 10.0 degrees Celsius, and the bat's chirp lasts for 1.00 ms.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relevance of frequency in the context of the problem and explore the calculation of distance using the speed of sound and time intervals. There are questions regarding the interpretation of the time taken for the sound to travel to the moth and back.

Discussion Status

Some participants have provided insights into the calculations and assumptions made regarding the time intervals involved. There is an ongoing exploration of whether the initial approach is too simplistic, and clarification on the time taken for the sound's round trip is being discussed.

Contextual Notes

Participants are considering the implications of the short time interval for the echo and whether the change in distance during this time is significant. There is also a mention of the speed of sound being consistent with known values.

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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