Kinemetic and waves and sound

[anna sung]

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

 

[berkeman]

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?

 

[anna sung]

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.

 

[berkeman]

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?

 

[anna sung]

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...

 

[berkeman]

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?

 

[anna sung]

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?

 

[berkeman]

...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?

 

[anna sung]

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 accleration? I am so sorry but its so confusing.

 

[berkeman]

is it accleration? 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.

 

[anna sung]

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?

 

[berkeman]

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?

 

[anna sung]

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?

 

[berkeman]

d= vf(t)-1/2a(t^2)
this equation?

That's close, but I don't know what "df(t)" is...

 

[anna sung]

That's close, but I don't know what "df(t)" is...

sorry? df(t)? you mean vf(t)? velocity final multiply by time??

 

[berkeman]

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?

 

[anna sung]

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?

 

[berkeman]

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.

 

[anna sung]

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

 

[berkeman]

okay thank you so much
oh i got the answer it so easy...omg..thank you so much

Great!