How Is Phase Difference Calculated Between Two Sound Waves?

[jrrodri7]

Homework Statement

Two Loud speakers are placed on a wall 3.79 meters apart. A listener stands directly in front of one of the speakers, 65.6 meters from the wall. The speakers are being driven by the same electric signal generated by a harmonic oscillator of frequency 4010 Hz The speed of sound in Air is 343 m/s. What is the phase difference $$\Delta\Phi$$ between the two waves? (answer in radians)

Homework Equations

y = A sin (kx (+/-) $$\omega$$t

The Attempt at a Solution

I tried finding the wavelength and using trig to find the distance between the observer and speakers...but I'm not sure if it has anything to do with the problem, and I haven't gotten an answer out of this yet...I'm lost. The only thing I know is superposition, but I know that if you just add sin to sin, you don't just get sin, and I don't know how to derive that to see how that works...

HELP!

 

[LowlyPion]

Homework Statement

Two Loud speakers are placed on a wall 3.79 meters apart. A listener stands directly in front of one of the speakers, 65.6 meters from the wall. The speakers are being driven by the same electric signal generated by a harmonic oscillator of frequency 4010 Hz The speed of sound in Air is 343 m/s. What is the phase difference $$\Delta\Phi$$ between the two waves? (answer in radians)

Homework Equations

y = A sin (kx (+/-) $$\omega$$t

The Attempt at a Solution

I tried finding the wavelength and using trig to find the distance between the observer and speakers...but I'm not sure if it has anything to do with the problem, and I haven't gotten an answer out of this yet...I'm lost. The only thing I know is superposition, but I know that if you just add sin to sin, you don't just get sin, and I don't know how to derive that to see how that works...

HELP!

Sure it does. The difference in distance is going to determine what lag there is between the 4010 hz sound waves.

 

[baba89]

Firstly, to find the wavelength of the sound waves, we can use the formula:

λ = v/f

Where λ is the wavelength, v is the speed of sound (343 m/s in this case), and f is the frequency (4010 Hz).

So, the wavelength is:

λ = 343/4010 = 0.0856 meters

We can then use trigonometry to find the distance between the observer and the speakers (let's call this distance d).

d = 3.79 + 65.6*cos(θ)

Where θ is the angle between the line connecting the two speakers and the line connecting the observer to one of the speakers.

To find this angle, we can use the inverse tangent function:

θ = arctan(65.6/3.79) = 86.2 degrees

Therefore, the distance between the observer and the speakers is:

d = 3.79 + 65.6*cos(86.2) = 65.9 meters

Now, to find the phase difference between the two waves, we can use the formula:

ΔΦ = 2π*(d/λ)

Where ΔΦ is the phase difference in radians, d is the distance between the observer and the speakers, and λ is the wavelength.

Substituting the values we found above, we get:

ΔΦ = 2π*(65.9/0.0856) = 4822 radians

Therefore, the phase difference between the two waves is approximately 4822 radians.