Chirping Bat and Distance to Moth - Sound Waves

[dvolpe]

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!

 

[blacknovember]

Lookup 'speed of sound in air'

 

[dvolpe]

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.

 

[blacknovember]

Apologies

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

 

[rwisz]

Your solution is correct. The frequency of the bat's chirping does not affect the calculation of the distance to the moth. The speed of sound is the key factor in determining the distance. However, the temperature can affect the speed of sound, so it is important to consider it when calculating the distance. Overall, your approach is correct and your solution is valid.