Difference in Wave Phase Question

In summary, one speaker remains at the origin while the other can slide along the track. When the adjustable speaker is at positions 0.6m and 1.2m, there are interference maxima. The frequency of the sound from the speakers is 571.67Hz. In part B, the speakers are allowed to have different phase constants and are adjusted so that there are interference maxima at positions 0.6m and 1.05m. The difference in phase constant between the two speakers is 2.093 radians.
  • #1
thatguy4000
8
0

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?
 
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  • #2
thatguy4000 said:
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?
 
  • #3
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?
 
  • #4
thatguy4000 said:
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.
 
  • #5
thatguy4000 said:

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
 
  • #6
thatguy4000 said:

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 said:
5)=0.15m.

haruspex said:
What is the wavelength in part b? Is it the same as in part a?
Wavelength (Lamda)=1.05-0.6=0.45m
 
  • #7
alanbo11 said:
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.
 
  • #8
alanbo11 said:
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.
alanbo11 said:
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.
 
  • #9
haruspex said:
That's why I asked. So suppose the wavelength has changed. You can still get an answer.

kuruman said:
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.
 

Related to Difference in Wave Phase Question

1. What is wave phase?

Wave phase refers to the position of a wave at a specific point in time. It is often measured in degrees or radians and can be used to describe the relationship between two waves or the position of a single wave.

2. How is wave phase different from wave frequency?

Wave frequency refers to the number of complete cycles a wave completes in a given amount of time, while wave phase describes the position of a wave at a specific point in time. Frequency is a measure of how often a wave occurs, while phase is a measure of where a wave is in its cycle.

3. Can wave phase be negative?

Yes, wave phase can be negative. It is measured in degrees or radians and can have values ranging from -180 degrees to 180 degrees. A negative phase indicates that the wave is in the opposite direction of a positive phase.

4. How does wave phase affect interference?

Wave phase plays a crucial role in interference, which is the interaction of two or more waves. When waves with the same frequency and amplitude are in phase (their peaks and troughs line up), they will reinforce each other and create a larger amplitude. However, when they are out of phase (their peaks and troughs do not line up), they will cancel each other out and create a smaller amplitude.

5. How does the difference in wave phase affect the Doppler effect?

The Doppler effect is the change in frequency of a wave due to the relative motion between the source of the wave and the observer. The difference in wave phase can affect the perceived frequency of the wave, as waves that are in phase will have a higher frequency and waves that are out of phase will have a lower frequency. This is why the sound of a passing siren seems to change as it moves towards and away from an observer.

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