MCQ about intensity of sound waves (formula manipulation)

[Originaltitle]

Homework Statement

Homework Equations

The equation that is already given + (maybe) I is proportional to A²f².

The Attempt at a Solution

I took the square root of 8, 8 being the Amplitude and I being proportional to the square of A. (I make the constant of proportionality 1). That's not the answer.

 

[SammyS]

Homework Statement

[ IMG]http://i49.tinypic.com/jts37a.jpg[/PLAIN]

Homework Equations

The equation that is already given + (maybe) I is proportional to A²f².

The Attempt at a Solution

I took the square root of 8, 8 being the Amplitude and I being proportional to the square of A. (I make the constant of proportionality 1). That's not the answer.

If $\ \displaystyle I\propto\frac{1}{x^2}\,,\$ then $\ \displaystyle I=\frac{k}{x^2}\,,\$ where k is the constant of proportionality.

This means that $I\cdot x^2=k\,,\ \text{ a constant}\ .$

Added in Edit:

This is wrong. See correct response below.

 

[Originaltitle]

Obviously, but we can't find k here because we don't have a numerical value for x OR I.

 

[ap123]

You don't need to calculate k.

At point P, we have I=k/r^2
Write a similar expression for the intensity at Q and compare it with the above expression.

 

[SammyS]

You are given an amplitude, A, at point P.

Also, $\ \displaystyle I\propto\frac{1}{x^2}\,,\$ and $\ \displaystyle I\propto{A^2}\ .\$ This means that $\ \displaystyle A\propto\frac{1}{x}\,,\$ assuming they're both positive. Therefore, Ax = k, where k is some constant.

You don't need a numerical result for k.

Ax at P

is equal to

Ax at Q .
@ P, A = 8μm and x = r .

@ Q, x = 2r , what is A ?

 

[Originaltitle]

A is inversely proportional to x. So if A is doubled, x will be two times less. So if A at x is 8, A at 2x will be 8/2 which is 4.

Thanks!