# Waves and Sound - Bat Chirp/Echo?

1. Dec 7, 2013

### harujina

1. The problem statement, all variables and given/known data

When near an object, a bat decreases the duration of its chirps and the time interval between chirps.
If the chirps last 3 x 10-4 s, what is the minimum distance for an object at which the first part of the echo overlaps the ending of the chirp? Note: the speed of sound in air is 344 m/s.

2. Relevant equations

d = vav x t ?

3. The attempt at a solution

I don't think I have the right idea with this problem. My teacher told me to use the above equation but what does that determine? length?

I don't understand: ...what is the minimum distance for an object at which the first part of the echo overlaps the ending of the chirp?
I drew a diagram (required) of chirps to an object and then the echo off of the object, but still can't figure out how I would find minimum distance.

This is what I tried:
d = vav x t; = 344 m/s x (3 x 10-4 s); = 0.1032 m
I don't even know if that's correct, but I'm stuck at that...

2. Dec 7, 2013

### BOYLANATOR

I guess it means that the bat receives the first part of the echo just as it finishes the burst.

3. Dec 7, 2013

### harujina

Okay... I kind of understand it now, but I'm still not sure how I could figure it out.
My teacher also told me something about it having 1.5 total wavelength or something?
I don't know what that means though.

4. Dec 7, 2013

### BOYLANATOR

In 3 x 10-4 secs the sound must have travelled to the object and then back to the bat. Can you work out how far the object must have been?

Your teacher was probably talking about constructive/destructive interference of waves but if we have interpreted the question properly this is not relevant.

5. Dec 7, 2013

### harujina

I thought the 3 x 10-4 s was just the chirp going to the object, and not back?

d = vav x t; = 344 m/s x (3 x 10-4 s); = 0.1032 m
Wouldn't this be the distance at which the object is at?

6. Dec 7, 2013

### BOYLANATOR

No, the time given is the duration of each chirp he makes. But if the first part of the echo arrives back as the bat finishes its chirp, then it must have travelled there and back in that time.

7. Dec 8, 2013

### harujina

Okay that makes sense.
Since the time given is the time measured when it travels to the object and back, is it correct if I divide the distance I calculated by 2?

d = vav x t; = 344 m/s x (3 x 10-4 s); = 0.1032 m
0.1032 m / 2 = 0.0516 m

8. Dec 8, 2013

### BOYLANATOR

That looks good to me.