Difference in Wave Phase Question

[thatguy4000]

Homework Statement

A. Two identical speakers, with the same phase constant, are arranged along a 1D track. One speaker remains at the origin. The other speaker can slide along the track to any position x. You are on the track at x=10 m. You hear interference maxima when the adjutable speaker's position is 0.6 m and 1.2 m and at no points in between. What is the frequency of the sound from the speakers in Hz?
B. The speakers are now allowed to have different phase constants. They are adjusted so that you hear interference maxima when the adjustable speaker is at x = 0.6 m and again when it is at x = 1.05 m. What is the difference in the phase constant between the two speakers in rad?

Homework Equations

velocity = wavelength * frequency
delta phi = phi2 - phi1 = (2pi * delta x)/wavelength

The Attempt at a Solution

The wavelength in part A would be 0.6m because that's the distance between the maximum interference, giving me frequency = 343m/s / 0.6m = 571.67Hz (correct).
I could not get B correctly though. Using the difference in phase constant equation, delta phi (phase difference) = (2pi * 0.45m) / 0.6m = 4.71 radians (wrong). What did I do wrong?

 

[haruspex]

Using the difference in phase constant equation, delta phi (phase difference) = (2pi * 0.45m) / 0.6m = 4.71 radians

What is the wavelength in part b? Is it the same as in part a?

 

[thatguy4000]

In my attempt at a solution, I just assumed that the frequency from part A carried over (am I wrong to assume this?), giving me the same wavelength of 0.6m. In light of your question, I'm confused as to why the distance between interference maxima is only 0.45m instead of 0.60m, because shouldn't the distance between maxima always be a multiple of the wavelength?

 

[haruspex]

I'm confused as to why the distance between interference maxima is only 0.45m instead of 0.60m,

That's why I asked. So suppose the wavelength has changed. You can still get an answer.

 

[alanbo11]

Homework Statement

A. Two identical speakers, with the same phase constant, are arranged along a 1D track. One speaker remains at the origin. The other speaker can slide along the track to any position x. You are on the track at x=10 m. You hear interference maxima when the adjutable speaker's position is 0.6 m and 1.2 m and at no points in between. What is the frequency of the sound from the speakers in Hz?
B. The speakers are now allowed to have different phase constants. They are adjusted so that you hear interference maxima when the adjustable speaker is at x = 0.6 m and again when it is at x = 1.05 m. What is the difference in the phase constant between the two speakers in rad?

Homework Equations

velocity = wavelength * frequency
delta phi = phi2 - phi1 = (2pi * delta x)/wavelength

The Attempt at a Solution

The wavelength in part A would be 0.6m because that's the distance between the maximum interference, giving me frequency = 343m/s / 0.6m = 571.67Hz (correct).
I could not get B correctly though. Using the difference in phase constant equation, delta phi (phase difference) = (2pi * 0.45m) / 0.6m = 4.71 radians (wrong). What did I do wrong?

Your mistake: delta x not equal (1.05-0.6)=0.45m, it should be (1.2-1.05)=0.15m. The new wavelength (Lamda)=(1.05-0.6)=0.45m. Hence delta phi=(2pi * 0.15)/0.45=2.093.

HTH

 

[alanbo11]

Homework Statement

A. Two identical speakers, with the same phase constant, are arranged along a 1D track. One speaker remains at the origin. The other speaker can slide along the track to any position x. You are on the track at x=10 m. You hear interference maxima when the adjutable speaker's position is 0.6 m and 1.2 m and at no points in between. What is the frequency of the sound from the speakers in Hz? B. The speakers are now allowed to have different phase constants. They are adjusted so that you hear interference maxima when the adjustable speaker is at x = 0.6 m and again when it is at x = 1.05 m. What is the difference in the phase constant between the two speakers in rad?

Homework Equations

velocity = wavelength * frequency delta phi = phi2 - phi1 = (2pi * delta x)/wavelength

The Attempt at a Solution

The wavelength in part A would be 0.6m because that's the distance between the maximum interference, giving me frequency = 343m/s / 0.6m = 571.67Hz (correct). I could not get B correctly though. Using the difference in phase constant equation, delta phi (phase difference) = (2pi * 0.45m) / 0.6m = 4.71 radians (wrong). What did I do wrong?

Your mistake: delta x not equal (1.05-0.6)=0.45m, it should be (1.2-1.05)=0.15m. The new wavelength (Lamda)=(1.05-0.6)=0.45m. Hence delta phi=(2pi * 0.15)/0.45=2.093. HTH

5)=0.15m.

What is the wavelength in part b? Is it the same as in part a?

Wavelength (Lamda)=1.05-0.6=0.45m

 

[kuruman]

Your mistake: delta x not equal (1.05-0.6)=0.45m, it should be (1.2-1.05)=0.15m. The new wavelength (Lamda)=(1.05-0.6)=0.45m. Hence delta phi=(2pi * 0.15)/0.45=2.093. HTH
Wavelength (Lamda)=1.05-0.6=0.45m

Thank you for posting. However, please note that the original post is 4 years old and the original poster is unlikely to profit from your contribution now.

 

[haruspex]

delta x not equal (1.05-0.6)=0.45m, it should be (1.2-1.05)=0.15m

It works either way. Just have to reduce the phase difference modulo $2\pi$ later.

Wavelength (Lamda)=1.05-0.6=0.45m

Yes, I know that, but clearly the OP had missed it. Since he did not come back, my guess is he got the point and the right answer but didn't have the courtesy to acknowledge.

 

[alanbo11]

That's why I asked. So suppose the wavelength has changed. You can still get an answer.

Thank you for posting. However, please note that the original post is 4 years old and the original poster is unlikely to profit from your contribution now.

Ha, I hear you but is may be useful to some other folks.