Waves and Sound - Bat Chirp/Echo?

  • Thread starter Thread starter harujina
  • Start date Start date
  • Tags Tags
    Sound Waves
harujina
Messages
77
Reaction score
1

Homework Statement



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.

Homework Equations



d = vav x t ?

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...
 
Physics news on Phys.org
I guess it means that the bat receives the first part of the echo just as it finishes the burst.
 
BOYLANATOR said:
I guess it means that the bat receives the first part of the echo just as it finishes the burst.

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.
 
In 3 x 10-4 secs the sound must have traveled 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.
 
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?
 
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 traveled there and back in that time.
 
BOYLANATOR said:
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 traveled there and back in that time.
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
 
That looks good to me.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
9K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
3
Views
4K
Replies
10
Views
5K
Replies
6
Views
4K
  • · Replies 27 ·
Replies
27
Views
6K