Solving Sound Problems: How High & How Loud?

[jrd007]

Okay here goes. I think I am doing the problems right but not the correct answers... :(

(1) A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 3.5 s later. How high is the cliff? Answer: 55 m

My approach: What I did was used the key equation: d = vt since I know both water and air's v. (Water = 1440 m/s & air = 343 m/s) And since I know the change in t = 3.5 s. So I set the equation like so:

{air}v(t - 3.5s) = {water}v(t)
343(t - 3.5s) = 1400(t)
t - 3.5s = (1440/343)(t)
I get the time = -1.09 s

then I use distance(water) = (1440 m/s)(-1.09s) = -1576 m
and for air = -1574 m... what am I doing wrong?
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(2) A person is standing a certain distance from an air plane with four equally noisy jet engines is experiencing a sound level bordering on pain, 120 dB. What sound level would this person experience if the captian shut down all by one engine? (Hint: Add intesties; not dBs) Answer: 114 dB

My approach: Fine the instestiny of all four then divide by three to get what ONE I would be.

So... I = I(intial) = 10^(sound level(B)/(10)
I = (1.0 x 10^ -12)^(120/10)
I = (1.0 x 10^ -12)^(12)
I = 1 and 1/3 = .33... Again what did I do wrong?
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(3) Expensive amplifier A is rated @ 250 W, while the most modest Amp B is rated @ 40 W. Estimate the sound level in decibels you would expect at a point 3.5 m from a loudspeaker connected in turn to each Amp. Answers: 122 dB, 114 dB

No idea how to get the answers. I know that we have Watts but not how to do it.

Thanks to anyone who can help.

 

[Päällikkö]

Okay here goes. I think I am doing the problems right but not the correct answers... :(
(1) A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 3.5 s later. How high is the cliff? Answer: 55 m
My approach: What I did was used the key equation: d = vt since I know both water and air's v. (Water = 1440 m/s & air = 343 m/s) And since I know the change in t = 3.5 s. So I set the equation like so:
{air}v(t - 3.5s) = {water}v(t)
343(t - 3.5s) = 1400(t)
t - 3.5s = (1440/343)(t)
I get the time = -1.09 s
then I use distance(water) = (1440 m/s)(-1.09s) = -1576 m
and for air = -1574 m... what am I doing wrong?

You dropped a stone. Stones rarely travel at the speed of sound and when you drop them, they certainly don't travel at a constant speed. The equation d = vt is therefore wrong.

(2) A person is standing a certain distance from an air plane with four equally noisy jet engines is experiencing a sound level bordering on pain, 120 dB. What sound level would this person experience if the captian shut down all by one engine? (Hint: Add intesties; not dBs) Answer: 114 dB
My approach: Fine the instestiny of all four then divide by three to get what ONE I would be.

Division by 3?

So... I = I(intial) = 10^(sound level(B)/(10)
I = (1.0 x 10^ -12)^(120/10)
I = (1.0 x 10^ -12)^(12)
I = 1 and 1/3 = .33... Again what did I do wrong?

The equation:
L₄ = 10 log (4I / I₀), I₀ = 1 pW/m²

Solve for intensity of one engine: I.

(3) Expensive amplifier A is rated @ 250 W, while the most modest Amp B is rated @ 40 W. Estimate the sound level in decibels you would expect at a point 3.5 m from a loudspeaker connected in turn to each Amp. Answers: 122 dB, 114 dB
No idea how to get the answers. I know that we have Watts but not how to do it.
Thanks to anyone who can help.

Intensity is proportional to power, and inversely proportional to area, as can be seen from the units: = W/m².
What is the area covered in 3.5 meters approximating the speaker as a point source?

 

[ramollari]

So... I = I(intial) = 10^(sound level(B)/(10)
I = (1.0 x 10^ -12)^(120/10)
I = (1.0 x 10^ -12)^(12)
I = 1 and 1/3 = .33... Again what did I do wrong?[/I]

I don't understand why do you raise 10 to the power -12!

 

[jrd007]

Because Io = 1 x 10-12, it is in our book.

Päällikkö, you are suppose to help me with problems 1 and 2 not tell me I am wrong.

 

[Päällikkö]

Because Io = 1 x 10-12, it is in our book.
Päällikkö, you are suppose to help me with problems 1 and 2 not tell me I am wrong.

I helped by telling in which parts you were wrong. I don't think you'd learn much if I posted complete solutions.

1) How fast does the stone drop (what is it's velocity (and/or displacement) as a function of time)?
2) Do the algebra for the equation (solve for I) again. And if you know that 4 cars have the same price, and their combined price is x, then the price of one is x/4, not x/3.
3) I can't help you much more without solving the whole problem. Give it some thought.

 

[daveb]

1) How fast or slow the stone drops is irrelevant, since the problem states the sound is heard after the splash occurs, not after dropping the stone. Sound travels at a specific velocity in air (OK, it also depends on density of air, etc. - assume normal density at 25 celsius). Your equation d=vt is correct - it's simply a matter of plugging in the numbers. You don't really need the speed of sound in water.

 

[DaveC426913]

1) How fast or slow the stone drops is irrelevant, since the problem states the sound is heard after the splash occurs, not after dropping the stone.

No Dave, reread it.

The splash is not seen (as you are inferring), the splash is heard. The operator does not know (by sight) when the splash occurs. i.e. the only way to sense the splash is by its sound.

The only event previous to that is the dropping of the stone. i.e. stone is dropped; 3.5s later the splash is heard.

You don't really need the speed of sound in water.

Quite true, so why do you bring it up? Nobody else did.

 

[daveb]

No Dave, reread it. The splash is not seen (as you are inferring), the splash is heard...

DOH! You're right.

Quite true, so why do you bring it up? Nobody else did.

Because I assume that's what this

(Water = 1440 m/s & air = 343 m/s)

means

 

[Gamma]

The stone falls from the top of the cliff under gravity. Use S= ut + 1/2 at^2 to get the height s in terms of t.

Next step is to find the travel time of sound from water to you.

S = Vs * ts where S is the hill height.

Now use (t+ts) = 3.5 s

You will have a quadratic equation to solve for S.

 

[Gamma]

For the second one,

You don't need to use the value of Io. If the sound intensity of one engine is I, then intensities of four would be 4I

120 = 10 log (4I/Io). From here find I/Io

Now when all but one is turned off (means only one is working now) the new intensity level

is = 10 log (I/Io) = 114 dB

 

[Gamma]

For the third one, intensity I = power / Area. So at a distance R from the source I = Power / area of the sphere of radius R.

I = Power / 4 pi R^2. I think you can go from here.

 

[jrd007]

Thanks Gamma. The equations were useful and I worked them out with a friend at school.