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

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SUMMARY

The discussion focuses on calculating the phase difference and minimizing sound between two speakers positioned 2.00 meters apart, with a listener 3.00 meters from one speaker. The frequency of the oscillator is set at 300Hz. The phase difference is calculated using the formula [psi] = (r2 - r1) / wavelength * 2[pi], where r1 and r2 are the distances to the listener. To minimize sound, the phase condition [psi] = [pi] is identified, indicating that adjustments in amplitude may be necessary for effective cancellation.

PREREQUISITES
  • Understanding of wave physics, specifically phase difference calculations
  • Knowledge of sound wave properties, including frequency and wavelength
  • Familiarity with the speed of sound in air, approximately 343 m/s
  • Ability to apply trigonometric principles in wave interference scenarios
NEXT STEPS
  • Learn how to calculate wavelength using the formula λ = v/f, specifically for sound waves
  • Study the principles of wave interference and conditions for constructive and destructive interference
  • Explore the effects of amplitude adjustments on sound cancellation between speakers
  • Investigate the impact of distance on sound intensity and perception in acoustics
USEFUL FOR

Acoustics engineers, audio technicians, and physics students interested in sound wave behavior and speaker placement optimization.

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.
 
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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].
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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