# Decrease in amplitude with distance from source

1. Feb 3, 2010

1. The problem statement, all variables and given/known data
A stone is thrown into a quiet pool of water. With no fluid friction, the amplitude of the waves falls off with distance r from the impact point as:

A. 1/r3
B. 1/r
C. 1/r1/2
D. 1/r3/2
E. 1/r2

2. Relevant equations
I can't think of any wave function accounting for decreasing amplitude with an increasing x.

3. The attempt at a solution
I think I remembered from high school science classes that the intensity (or something similar) of a sound wave decreases inversely with the square of the distance from the source, so I chose choice E but got it wrong. I do not know what the correct answer is. Any help would be great

2. Feb 3, 2010

### ideasrule

Do you remember why a sound wave decreases inversely with the square of the distance? Imagine a sound source emitting a very short pulse of sound. One second later, the sound energy will be spread across the surface of a sphere. The sphere's surface area is 4pi*r^2, so the intensity (energy per square meter) is P/(4pi*r^2).

Now, for a wave on a pond, all of the energy moves in two dimensions. So intensity is....

3. Feb 3, 2010

Oh, so that's why

I guess the answer is B then, since intensity would be P/(2pi*r). Thank you!

4. Feb 4, 2010

### JPizz

I believe that the answer is not 1/r...

5. Feb 4, 2010

Why not? Since the circumferenc of a circle is 2pi*r, and the energy of the wave would be spread out as the circle gets larger, wouldn't the intensity be Power/circumference, or P/(2pi*r)? Please enlighten me a bit.

6. Feb 4, 2010

### JPizz

I took a quiz online for school and I had put that in as an answer and it said it was wrong. Maybe the quiz I took was incorrect...

7. Feb 4, 2010

### UgOOgU

What you are looking for is the amplitude, not the energy. What you have said about energy is right, now you have to use the relation between energy and amplitude, do you remember it?

8. Feb 4, 2010

The quiz probably was correct... funny thing, I found this question on my online quizzes for school too. Coincedence? :tongue: Do you get a second try at the quiz too?

A search through my book gives me
$$E=\frac{1}{2}\mu\omega^{2}A^{2}\lambda$$

I guess then, $$A\propto\sqrt{E}$$
Okay, I've verified that the correct answer is indeed $$\frac{1}{\sqrt{r}}$$. I had a second try on the quiz and got choice C right.