Solving Bat Doppler Effect Homework Problem

[merf]

Homework Statement

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.

Homework Equations

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

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 !

 

[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.

 

[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.

 

[merf]

Thank you !

 

[rwisz]

I would like to commend you on your efforts to solve the homework problem. However, it seems like you have made a few mistakes in your calculations. Let's break down the problem step by step to find the correct solution.

Firstly, we need to understand the concept of the Doppler effect. The Doppler effect is the change in frequency or wavelength of a wave for an observer who is moving relative to the source of the wave. In this case, the bat is the source of the sound wave and it is moving towards a stationary wall.

The equation for the Doppler effect is given by:

f' = f(v ± vr)/ (v ± vs)

Where,
f' is the observed frequency
f is the emitted frequency
v is the speed of sound
vr is the relative speed of the observer (in this case, the bat)
vs is the speed of the source (in this case, the wall)

Now, let's apply this equation to solve the problem. We know that the bat is flying at a constant speed of 19.5 m/s and the speed of sound is 343 m/s. We also know that the bat hears the echo 0.15 seconds after making a clicking noise. This means that the time taken for the sound wave to travel from the bat to the wall and back to the bat is 0.15 seconds.

Using the equation, we can write:

f' = f(v ± vr)/ (v ± vs)

f' = f(v + vr)/ (v - vs)

Where,
f' = observed frequency = 1/0.15 = 6.67 Hz (since the bat hears the echo after 0.15 seconds)
f = emitted frequency = 1/0.15 = 6.67 Hz (since the bat makes a single clicking noise)
v = speed of sound = 343 m/s
vr = relative speed of the observer (bat) = 19.5 m/s
vs = speed of the source (wall) = 0 m/s (since the wall is stationary)

Substituting these values in the equation, we get:

6.67 Hz = 6.67 Hz(343 + 19.5)/ (343 - 0)

Solving for the unknown, we get the emitted frequency, f = 6.67 Hz.

Now, using the equation for speed (speed = distance/time), we can find