How Long Does It Take to Hear a Stone Hit the Bottom of a Shaft?

  • Thread starter Thread starter ahrog
  • Start date Start date
  • Tags Tags
    Time
Click For Summary
To determine how long it takes to hear a stone hit the bottom of a 122.5 m mine shaft, first calculate the time it takes for the stone to fall, which is approximately 5 seconds. Next, the time for the sound to travel back up the shaft must be calculated using the formula time = distance/velocity. The velocity of sound is given as 343 m/s, allowing for the calculation of the sound travel time. The confusion arises from the need for frequency, which is unnecessary for this calculation as time can be derived directly from distance and velocity. Ultimately, the total time is the sum of the fall time and the sound travel time.
ahrog
Messages
45
Reaction score
0

Homework Statement


Someone drops a stone into a mine shaft 122.5 m deep. How soon after she drops the stone does she hear it hit the bottom of the shaft?


Homework Equations


v= wavelength x frequency


The Attempt at a Solution


I changed the equation to be:
d/t=wf
t=d/wf
t=122.5m/(343m/s x f)

And I'm stuck there...I don't get how to get the frequency

My textbook doesn't even refer to something like this, so I'm extremely confused...It's like asking someone about something they've never learned before...
 
Physics news on Phys.org
First you need to figure the time for it to fall 122.5m, then add the time for the sound to return.

Δt = D/Vsound
 
I got the falling part time to be 5 seconds.

And for the time for the sound to return, how do I figure that out? If the velocity for the sound is wavelength x frequency which I have only the wavelength, how can I find the time?
 
t=122.5m/(343m/s x f)
Here f in not needed. time = distance/velocity.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
2K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
2
Views
2K
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K