How Do You Calculate Phase Difference and Minimize Sound Between Two Speakers?

Click For Summary

Homework Help Overview

The discussion revolves around calculating the phase difference and minimizing sound between two speakers positioned 2.00 m apart, with a listener located 3.00 m from the wall in front of one speaker. The speakers are driven in phase at a frequency of 300 Hz.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of distances from the speakers to the listener and the subsequent phase difference. There is uncertainty regarding the need for wavelength and velocity in the calculations, with some questioning the appropriate formulas to use for both parts of the problem.

Discussion Status

Participants have made progress in calculating the distance difference and are exploring the relationship between wavelength, frequency, and phase difference. Some guidance has been provided regarding the formulas needed, but there remains uncertainty about how to apply the wavelength to find the answers.

Contextual Notes

There is a lack of information regarding the amplitudes of the sound waves, which may affect the analysis of sound cancellation between the speakers.

Jenn_Lane2001
Hi i posted here before and got great help, and I ran into another problem maybe you can help me this is the question.

Two loud speakers are placed on a wall 2.00 m apart. A listener stands 3.00m from the wall direclty infront of one of the speakers. A single ocillator is driving the speakers in phase at a frequency of 300Hz.
A. What is the phase difference between tehe two waves when they reach the observer.
B. What is the frequency closest to 300 hz. to which the oscillator may be adjusted such that the observer hears minimal sound?

This is what I did..

r1= square root of (3.00)^2 = 3
r2= square root of ((3.00)^2 + (2.00)^2) = 3.61

The difference is = r2-r1 = .61

Now this is where i got stuck I wasnt sure if i had to use .61/2pie to find the phase difference.
Also what forumla should I use for part B.
Thank you in advance.
 
Physics news on Phys.org
Hi Jenn_Lane2001,
I think you got the difference in distance correctly. Next you should find out what the wavelength is for a 300Hz sonic wave.
 
R u sure..?

I didnt think you need to know the wavelength. more the velocity but that is what i wasnt sure calculating. I am not sure of the equation to use. Thanks
 
[lamb] = v/f,
where
[lamb] is wavelength
v is velocity
f is frequency
I think, for sonic waves, v = 340 m/s or something.
 
Oh ok so sonic is the speed of sound

the speed of sound is 343v and then you will divide that by frequency which is 300 hz which equals wavelength= 1.143. now that i have the wavelenght where would i put taht into use as to determine the answers. Thanks a lot for the help.
 
Right. Since the progress in phase is 2[pi] for each wavelength travelled, your phase difference is
[psi] = .61m / 1.143m * 2[pi].
That's the answer to A.

Part B is more difficult. You know two waves cancel out each other when [psi] = [pi]. But that's only true if amplitudes are equal. You know the amplitude decreases (sound becomes softer) as you move away from a speaker. So the speaker that is more distant from the observer has to be turned up a bit if you want waves to cancel out. Anyway, as no info about amplitudes is given in the problem, it will probably do to solve for [psi] = [pi].
 

Similar threads

Replies
2
Views
6K
  • · Replies 9 ·
Replies
9
Views
12K
Replies
7
Views
5K
Replies
3
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
6K
Replies
14
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
3
Views
2K
Replies
3
Views
2K