Solving 2 Waves: Constructive Phase Difference

[goonking]

Homework Statement

Homework Equations

The Attempt at a Solution

so for 2 waves to be destructive, they need to be π , 3π, 5π... radians out of phase with each other.

to get constructive, they have to be 0, 2π, 4π... out of phase.

so to be constructive, the speaker needs to move 1π radian.

we solve for the wavelength: 343 m/s divided by 615 Hz = 0.557 meters

so the speaker just need to move 0.557 meters correct?

 

[gneill]

My interpretation of the problem is that the speakers are being fed by signals that are already out of phase by $\pi$ radians. You want to find the minimum distance between the speakers such that their outputs will be in phase and thus yield maximum constructive interference.

 

[goonking]

My interpretation of the problem is that the speakers are being fed by signals that are already out of phase by $\pi$ radians. You want to find the minimum distance between the speakers such that their outputs will be in phase and thus yield maximum constructive interference.

what is the relationship between the phase and distance between the 2 speakers?

 

[haruspex]

what is the relationship between the phase and distance between the 2 speakers?

Draw a diagram of a wave from one speaker, at some instant. Draw it with the wave at a maximum at the speaker. If you look at a point half a wavelength away, what is the phase there in relation to the phase at the speaker?

 

[goonking]

Draw a diagram of a wave from one speaker, at some instant. Draw it with the wave at a maximum at the speaker. If you look at a point half a wavelength away, what is the phase there in relation to the phase at the speaker?

I figured it out, having solved for wavelength (which is 0.557m), the phase difference had to move another π for it to be constructive so:

π = (2π/ 0.557m) (x₂ - x₁)

solving for the distance gives .278 meters