Sound waves problem

  • #1
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.
 

Answers and Replies

  • #2
172
2
For the 1st part of the question, it may be helpful to note that [tex]Intensity\propto (Amplitude)^2 [/tex]
 
  • #3
Just to double check, does that symbol mean indirectly proportional?
 
  • #4
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
7
No, it mean directly proportional to. If you wish to form an equality you must add in a constant [itex]I = kAmp^2[/itex]

~H
 
  • #5
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
nrqed
Science Advisor
Homework Helper
Gold Member
3,764
294
FlipStyle1308 said:
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
[itex] I= {1 \over 2} \rho v \omega^2 s_{max}^2 [/itex] in terms of the displacement amplitude or [itex] I = {1\over 2} {\Delta P_{max}^2 \over \rho v} [/itex] ( 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 :smile: ).

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
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
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
7
Decibels is given in terms of logarithms,

[tex]I(dB) = 10\log \left( \frac{I}{I_0} \right)[/tex]

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
 
  • #9
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
172
2
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:
  • #11
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
172
2
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 [tex]I_{0}[/tex] in the equation given is a constant.
 
  • #13
So I solve for:

10log(10,000)? How do I plug these intensities in?
 
  • #14
172
2
I don't really get what you mean by "plugging the intensities in". Perhaps you would like to show some working?
 
  • #15
I got it correct, thanks for the help!
 
Last edited:
  • #16
172
2
Good! But take note of the fact that [tex]I_{0}[/tex] is a constant value (the threshold of hearing, if I understand it correctly). The [tex]log_{10}(I_{0})[/tex] term should cancel off when you are doing your working.
 

Related Threads on Sound waves problem

  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
2
Views
941
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
2
Views
973
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
8K
Top