Solving Sound Waves Problem: X vs Y

[FlipStyle1308]

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

Sound waves X and Y have wavelengths of 1 m and 2 m, and amplitudes of 0.01 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.

 

[pizzasky]

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

 

[FlipStyle1308]

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

 

[Hootenanny]

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

~H

 

[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. ?

 

[nrqed]

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. ?

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 confused.

 

[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.

 

[Hootenanny]

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.

more information is given here; http://hyperphysics.phy-astr.gsu.edu/hbase/sound/db.html#c1

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

~H

 

[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?

 

[pizzasky]

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?

 

[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?

 

[pizzasky]

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.

 

[FlipStyle1308]

So I solve for:

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

 

[pizzasky]

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

 

[FlipStyle1308]

I got it correct, thanks for the help!

 

[pizzasky]

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.