# Sound waves problem

1. May 14, 2006

### FlipStyle1308

I am clueless on how to approach the following problem, and was wondering if anyone would be able to help me. Thanks!

2. May 14, 2006

For the 1st part of the question, it may be helpful to note that $$Intensity\propto (Amplitude)^2$$

3. May 14, 2006

### FlipStyle1308

Just to double check, does that symbol mean indirectly proportional?

4. May 15, 2006

### Hootenanny

Staff Emeritus
No, it mean directly proportional to. If you wish to form an equality you must add in a constant $I = kAmp^2$

~H

5. May 15, 2006

### FlipStyle1308

So would this be the answer...

The intensity of Y is greater than that of X by 10,000 times. Soundwave Y is 100 dB louder than soundwave X. ?

6. May 15, 2006

### nrqed

I don't see how you got 10000.

sound waves are abit tricky because there are two ways to represent them: in terms of a *displacement wave* or in terms of a *pressure wave*.

If I recall correctly (don't take my word for it!), the formula are
$I= {1 \over 2} \rho v \omega^2 s_{max}^2$ in terms of the displacement amplitude or $I = {1\over 2} {\Delta P_{max}^2 \over \rho v}$ ( I am quoting from memory, I don't have my books here so caveat emptor). You must have seen these equations?

So I am a bit confused. If the two waves have the same *pressure* amplitude, they would have the same intensity (after checking that they travel at the same speed and they do..as they must ).

If you were given the *displacement* amplitude to be the same, I could see how the different frequencies would influence the result.

So I am :uhh: confused.

7. May 15, 2006

### FlipStyle1308

I used the equation that Hootenanny gave and plugged in the amplitudes of 0.01Pa and 1 Pa, getting inertias of 0.0001 and 1, respectively. 1 is 10,000 times bigger than 0.0001, and that is how I got my answer.

8. May 16, 2006

### Hootenanny

Staff Emeritus
Decibels is given in terms of logarithms,

$$I(dB) = 10\log \left( \frac{I}{I_0} \right)$$

and can be use to compare increases in intensity.

Apologies for the confusion, but my answer is post #4 was a direct answer to your question in #3.

~H

9. May 16, 2006

### FlipStyle1308

Okay, I am confused with all these equations. Are you guys able to provide me a step-by-step process for solving this equation?

10. May 16, 2006

I think there is some confusion among the posters because there may be something you left out in your first post. In it, you wrote "Sound waves X and Y have wavelengths of 1 m and 2 m, and amplitudes of 0.01 Pa, respectively". Does the value of 0.01 refer to X, Y, or both? If I am not mistaken, X has an amplitude of 0.01 Pa and Y's is 1 Pa, as you mentioned in one of your later posts that "I used the equation that Hootenanny gave and plugged in the amplitudes of 0.01Pa and 1 Pa".

Perhaps you would like to clarify?

Last edited: May 16, 2006
11. May 16, 2006

### FlipStyle1308

Sound waves X and Y have wavelengths of 1 m and 2 m, and amplitudes of 0.01 Pa and 1 Pa, respectively. The frequency of X is 343 Hz, and the frequency of Y is 171.5 Hz. The intensity of Y is greater than that of X by ___ times. Soundwave Y is ___ dB louder than soundwave X.

Fixed. I didn't notice that! Sorry! Is my answer still correct?

12. May 16, 2006

Does intensity vary with wavelength and frequency? If no, then I think your answer of 10,000 for the 1st part of the question should be correct.

For the 2nd part, use the equation that Hootenanny gave in post 8 (note that the logarithm has base 10). You may not know the absolute values of the intensities, but this should not be a problem. Just let the intensity of X be a and the intensity of Y be 10,000a and you can go from there. Also, remember that $$I_{0}$$ in the equation given is a constant.

13. May 16, 2006

### FlipStyle1308

So I solve for:

10log(10,000)? How do I plug these intensities in?

14. May 16, 2006

I don't really get what you mean by "plugging the intensities in". Perhaps you would like to show some working?

15. May 16, 2006

### FlipStyle1308

I got it correct, thanks for the help!

Last edited: May 16, 2006
16. May 16, 2006

Good! But take note of the fact that $$I_{0}$$ is a constant value (the threshold of hearing, if I understand it correctly). The $$log_{10}(I_{0})$$ term should cancel off when you are doing your working.