# Bat Doppler Effect

1. Sep 22, 2010

### merf

1. The problem statement, all variables and given/known data
A bat, flying at a constant speed of 19.5 m/s in a straight line toward a vertical cave wall, makes a single clicking noise and hears the echo 0.15 s later. Assuming that she continued flying at her original speed, how close was she to the wall when she received the echo? Assume a speed of 343 m/s for the speed of sound.

2. Relevant equations

The only equation we were given is speed= change of d/time

3. The attempt at a solution

i took the speed of sound 343*.15s to get 51.45
then i took the speed of the bat 19.5*.15s to get 2.93

then i subtracted the two and got 48.52

this doesn't not seem right any suggestions?

Thank you !

2. Sep 22, 2010

### Delphi51

Welcome to PF, Merf!
Clever of you to figure out that the bat goes 2.9 m and the sound goes 51.5m. I didn't see that and did something much more complicated for nothing.

But you have oversimplified the last step.
Say the initial distance to the wall is x. Then the sound goes distance x on the way to the wall plus x - 2.9 on the way back. If you total that up and set it equal to 51.5, you'll be able to find x and then have only a small step to get the answer, which is about half the answer you had.

3. Sep 22, 2010

### rl.bhat

In the given time ( 0.15 s) the noise travels x m in forward direction and y m in reverse direction. So
0.15 = (x+y)/Vs. where Vs is the velocity of the sound.
The bat travels (x-y) m. So
0.15 = (x-y)/Vb, where Vb is the velocity of the bat.

Now solve for x and y. Required answer is y.

4. Sep 23, 2010

Thank you !!