How Deep Is the Well When Measuring with Sound and Kinematics?

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Homework Help Overview

The discussion revolves around determining the depth of a well using sound and kinematics. A scenario is presented where a stone is dropped into a well, and the time taken to hear the splash is measured. The problem involves understanding the relationship between the time of fall, the speed of sound, and the depth of the well.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to consider two different times and velocities in the measurement process. Questions arise about the roles of acceleration and the kinematic equations in solving the problem. There is also an exploration of the sequence of events from dropping the stone to hearing the splash.

Discussion Status

Participants are actively engaging with the problem, asking clarifying questions and attempting to outline the necessary equations. Some guidance has been provided regarding the use of kinematic equations and the distinction between the time taken for the stone to fall and the time for the sound to travel back.

Contextual Notes

There is an emphasis on understanding the process of measuring the depth, including the need to account for both the falling stone and the sound traveling back, which introduces complexity in the timing and calculations involved.

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Homework Statement



A person explains that the well in his house is 500m deep. the person decides to examine herself and drops a stone from rest and measure the time interval until you hear the splash of the stone striking the water. you find it to be 6.0s. you assume the speed of sound in air to be 343m/s. how deep is the well?

Homework Equations


since this question is in chapter waves and sound, i thought i only had to use
D= vt
and its now giving me the answer
the answer is less than 176m deep
 
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anna sung said:

Homework Statement



A person explains that the well in his house is 500m deep. the person decides to examine herself and drops a stone from rest and measure the time interval until you hear the splash of the stone striking the water. you find it to be 6.0s. you assume the speed of sound in air to be 343m/s. how deep is the well?

Homework Equations


since this question is in chapter waves and sound, i thought i only had to use
D= vt
and its now giving me the answer
the answer is less than 176m deep

Remember that you have two different times and two different sets of velocities. What are the two parts of the process of measuring the depth this way?
 
berkeman said:
Remember that you have two different times and two different sets of velocities. What are the two parts of the process of measuring the depth this way?

what...can you please elaborate more i cannot understand.
 
anna sung said:
what...can you please elaborate more i cannot understand.

You listed an equation, D=vt.

I assume the D is the depth of the well. What are v and t? Describe the process of dropping the rock and listening for the splash. What-all happens during that process?
 
berkeman said:
You listed an equation, D=vt.

I assume the D is the depth of the well. What are v and t? Describe the process of dropping the rock and listening for the splash. What-all happens during that process?

well since the question gave us the speed of sound which is V and they also gave us the time interval. is acceleration necessary in this question? i know that the kinemetic equations should be used...though...
 
anna sung said:
well since the question gave us the speed of sound which is V and they also gave us the time interval. is acceleration necessary in this question? i know that the kinemetic equations should be used...though...

The time interval is from the time the stone is dropped until the time you hear the splash. The spash sound is not traveling for that whole amount of time. What else is going on first?
 
berkeman said:
The time interval is from the time the stone is dropped until the time you hear the splash. The spash sound is not traveling for that whole amount of time. What else is going on first?

...uh..what else is going on..first..uh...echo?
 
anna sung said:
...uh..what else is going on..first..uh...echo?

Nope.

If you could see the stone hit the water, and count the time off from there, you would only have to use the constant speed of sound to figure out the depth of the well.

But you can't see the water in this question. You can only see the time counter start when you drop the stone. What happens during the time between when you drop the stone, and it hits the water?
 
berkeman said:
Nope.

If you could see the stone hit the water, and count the time off from there, you would only have to use the constant speed of sound to figure out the depth of the well.

But you can't see the water in this question. You can only see the time counter start when you drop the stone. What happens during the time between when you drop the stone, and it hits the water?

is it acceleration? I am so sorry but its so confusing.
 
  • #10
anna sung said:
is it acceleration? I am so sorry but its so confusing.

Yes! You need to use the kinematic equations of motion for the first part, while the stone is dropping and increasing its speed. And then use the equation you listed with the constant speed of sound for the second part of the process.

You will end up with two equations and two unknowns (the depth is unknown, and the portion of the total 6 seconds that it takes for the stone to hit the water is the second unknown). Write the two equations and solve them simultaneously.
 
  • #11
berkeman said:
Yes! You need to use the kinematic equations of motion for the first part, while the stone is dropping and increasing its speed. And then use the equation you listed with the constant speed of sound for the second part of the process.

You will end up with two equations and two unknowns (the depth is unknown, and the portion of the total 6 seconds that it takes for the stone to hit the water is the second unknown). Write the two equations and solve them simultaneously.

wait. so i have to use the acclecration equation, a=(vf-vi)/t and after i get a i can plug it into the distance equation?
 
  • #12
anna sung said:
wait. so i have to use the acclecration equation, a=(vf-vi)/t and after i get a i can plug it into the distance equation?

That equation "a=(vf-vi)/t" is not the one I would use for the falling stone portion. What other kinematic equation has vertical distance traveled for an object that is falling under the constant acceleration of gravity?
 
  • #13
berkeman said:
That equation "a=(vf-vi)/t" is not the one I would use for the falling stone portion. What other kinematic equation has vertical distance traveled for an object that is falling under the constant acceleration of gravity?

d= vf(t)-1/2a(t^2)
this equation?
 
  • #14
anna sung said:
d= vf(t)-1/2a(t^2)
this equation?

That's close, but I don't know what "df(t)" is...
 
  • #15
berkeman said:
That's close, but I don't know what "df(t)" is...

sorry? df(t)? you mean vf(t)? velocity final multiply by time??
 
  • #16
anna sung said:
sorry? df(t)? you mean vf(t)? velocity final multiply by time??

Sorry, my typo. Yes, I was asking about vf(t). But final velocity vf multiplied by time is not part of the kinematic equations of motion, is it?

http://en.wikipedia.org/wiki/Equations_of_motion

I see an initial velocity Vi listed multiplying time, but what is Vi in this question?
 
  • #17
berkeman said:
Sorry, my typo. Yes, I was asking about vf(t). But final velocity vf multiplied by time is not part of the kinematic equations of motion, is it?

http://en.wikipedia.org/wiki/Equations_of_motion

I see an initial velocity Vi listed multiplying time, but what is Vi in this question?
0.0m/s? because it starts at rest?
 
  • #18
anna sung said:
0.0m/s? because it starts at rest?

Yes, good. Now you're ready to write the two equations and solve.

I need to bail for a few hours. I'll try to check in later if I can, but you should be able to solve this now on your own.
 
  • #19
berkeman said:
Yes, good. Now you're ready to write the two equations and solve.

I need to bail for a few hours. I'll try to check in later if I can, but you should be able to solve this now on your own.

okay thank you so much
oh i got the answer it so easy...omg..thank you so much
 
  • #20
anna sung said:
okay thank you so much
oh i got the answer it so easy...omg..thank you so much

Great!
 

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