How Does Intensity Change with Distance from a Vibrating Source?

[Coushander]

Homework Statement

A tiny vibrating source sends waves uniformly in all directions. An area of 3.25 cm2 on a sphere of radius 2.50 m centered on the source receives energy at a rate of 4.20 J/s.
(a) What is the intensity of the waves at 2.50 m from the source?
(b) What is the intensity of the waves at 10.0 m from the source?
(c) At what rate is energy leaving the vibrating source of the waves?

Homework Equations

I = Power/Area

Area (sphere) = 4(3.14)r²

The Attempt at a Solution

a) Power = 4.20 J/s

Radius = 2.5 m

Area = 4(3.14)(2.5)²= 25(pi)

I= 0.0534 W/m²

The actual value of I is 1.29x10⁴ W/m²

If I say Area= 3.25cm² then I get I = 129.23 W/m²

 

[PeterO]

Homework Statement

A tiny vibrating source sends waves uniformly in all directions. An area of 3.25 cm2 on a sphere of radius 2.50 m centered on the source receives energy at a rate of 4.20 J/s.
(a) What is the intensity of the waves at 2.50 m from the source?
(b) What is the intensity of the waves at 10.0 m from the source?
(c) At what rate is energy leaving the vibrating source of the waves?

Homework Equations

I = Power/Area

Area (sphere) = 4(3.14)r²

The Attempt at a Solution

a) Power = 4.20 J/s

Radius = 2.5 m

Area = 4(3.14)(2.5)²= 25(pi)

I= 0.0534 W/m²

The actual value of I is 1.29x10⁴ W/m²

If I say Area= 3.25cm² then I get I = 129.23 W/m²

This is very close to what you should have been doing.
The area of the whole shpere was unimportant, since you were told the powere reaching just 3.25cm² of it.

While there are 100 cm in a metre, I think you overlooked that that means 10000 cm² in each m²

 

[Coushander]

Okay but how does the 2.5m radius factor into that?

 

[gneill]

Okay but how does the 2.5m radius factor into that?

The "area of 3.25 cm²" is located on the r = 2.5m sphere. So it's where it has to be in order to answer question (a). No further need for that radius

 

[PeterO]

Okay but how does the 2.5m radius factor into that?

For the calculation here - it doesn't.

It is common for such a question to be just one part of a bigger situation, and to go on to ask "what is the total power output of the sound source?". That is where the full sphere would come in.

EDIT: just noticed! - that is part (c)

 

[Coushander]

So if the 3.25cm^2 area is only part of the sphere at 2.5m, how can I go about using that to find the area at 10m? Like unless I'm supposed to just times it by four or something I don't know.

 

[PeterO]

So if the 3.25cm^2 area is only part of the sphere at 2.5m, how can I go about using that to find the area at 10m? Like unless I'm supposed to just times it by four or something I don't know.

You should already have studied the effect of distance on intensity of sound waves as they spread out.

Inverse-square-Law should mean something.

EDIT: Alternately, answer part (c) before part (b) this time.

 

[Coushander]

You should already have studied the effect of distance on intensity of sound waves as they spread out.

Inverse-square-Law should mean something.

EDIT: Alternately, answer part (c) before part (b) this time.

It was something that came up on mastering physics only once. So I don't remember how to link 1/d₁² and 1/d₂²

 

[PeterO]

It was something that came up on mastering physics only once. So I don't remember how to link 1/d₁² and 1/d₂²

If you do part (c) you work out the total amount of energy given out each seocd.

Work out the area of the 10m sphere, and that same amount of energy is spread over that new area.

Post your answers

 

[Coushander]

If you do part (c) you work out the total amount of energy given out each seocd.

Work out the area of the 10m sphere, and that same amount of energy is spread over that new area.

Post your answers

But the issue is that I don't know that equation either.

 

[gneill]

But the issue is that I don't know that equation either.

Google: surface area of a sphere

 

[Coushander]

Google: surface area of a sphere

I mean I don't know the equation to find out "total energy given out each second"

 

[gneill]

I mean I don't know the equation to find out "total energy given out each second"

The small area of 3.25 cm² caught a portion of it at its location. What fraction of the total did it intercept?

 

[PeterO]

I mean I don't know the equation to find out "total energy given out each second"

You know how much energy per second was striking each 3.25 cm². how many lots of 3.25 cm² make up an entire sphere of radius 2.5 m [Hint: you worked out how many square metres make up that sphere in your original post; - be careful with cm² and m².]

 

[jambaugh]

This advice may or may not be helpful to you but...

I would solve this problem by first calculating the power per spherical angle.
[Spherical angle in steradians = surface area on the unit sphere cut out by that spherical angle. This gives surface area of a spherical region as A = r^2 \Theta where \Theta is the spherical angle in steradians. Note also that the spherical angle of an entire sphere is 4pi, so spherical angle is 4pi times the percentage of the whole sphere.]

Since the sound is radiating uniformly from a point the power per spherical angle is constant. So figure the power through an area at a given radius and figure the spherical angle of that area (A/r^2) and divide. Use that to solve the other problems.

If you dislike spherical angle, then use instead the percentage of the whole sphere and it works the same way. You'll realized it comes to exactly the same thing when you work out the details.

 

[Coushander]

The small area of 3.25 cm² caught a portion of it at its location. What fraction of the total did it intercept?

C)
Total Area at 2.5m = 4(pi)(2.5m)² = 25(pi)m²

Sub-Area = 3.25cm² = 3.25x10^-4m²

Sub area/total area = 4.14x10^-6

Energy at sub-area = (total energy)(fraction)

total energy = 1.015x10⁶ W

Correct Answer.

B)

I = P/A

P = 1.015x10⁶ W

A = 4(pi)(10m)² = 400(pi)m²

I = 807.69 W/m²

Correct Answer

 

[PeterO]

C)
Total Area at 2.5m = 4(pi)(2.5m)² = 25(pi)m²

Sub-Area = 3.25cm² = 3.25x10^-4m²

Sub area/total area = 4.14x10^-6

Energy at sub-area = (total energy)(fraction)

total energy = 1.015x10⁶ W

Correct Answer.

B)

I = P/A

P = 1.015x10⁶ W

A = 4(pi)(10m)² = 400(pi)m²

I = 807.69 W/m²

Correct Answer

Well done!

btw - the inverse square law tells us that if you increase the radius by a factor of 4, then you reduce the intensity by a factor of 16.
The question setters obviously thought you would do it that way - which could explain why the 10m sphere was part (b) while the total energy was part (c)