Find the correct frequency so two waves are in phase

[Jalo]

Homework Statement

A man is sitting between two speakers, 5 meters from the first and 6.5 meters away from the second.

Speaker 1~~~~~~(6.5m) ~~~~~~ Man ~~~~ (5m) ~~~~ Speaker 2

They both create sound waves in phase. What's the lowest frequency for which the man hears both sounds in phase?
The velocity of sound is 340m/s.

Homework Equations

The Attempt at a Solution

We have two speakers. Speaker 1 is positioned at x = 0, creating waves with a phase of
ϕ₁(x,t) = kx - wt
propagating in the positive direction.
Speaker 2 is positioned at x = 11.5, emiting waves with a phase
ϕ₂(x,t) = kx + wt
which propagate in the negative direction.

The man is sitting at x=6.5.
We know that both speakers emit in phase, therefore for any given instant t we have:
ϕ₁(0,t) = ϕ₂(11.5,t)

We also want the man to receive both waves in phase, therefore:
ϕ₁(6.5,t) = ϕ₂(6.5,t)

If we solve this equations we'll get:
ϕ₁(0,t) = ϕ₂(11.5,t) ⇔ 0*x - w*t = 11.5*k + wt ⇔ 2*wt = -11.5k
ϕ₁(6.5,t) = ϕ₂(6.5,t) ⇔ 6.5k - wt = 6.5k + wt ⇔ wt = 0

The result doesn't make any sense. I've probably assumed something wrongly, but I can't quite figure out what. If anyone could point my logic's holes I'd appreciate.

Thanks.
Daniel

 

[sankalpmittal]

Homework Statement

A man is sitting between two speakers, 5 meters from the first and 6.5 meters away from the second.

Speaker 1~~~~~~(6.5m) ~~~~~~ Man ~~~~ (5m) ~~~~ Speaker 2

They both create sound waves in phase. What's the lowest frequency for which the man hears both sounds in phase?
The velocity of sound is 340m/s.

Homework Equations

The Attempt at a Solution

Snip..

Jesus !
Such a long solution ?

Why do you not use,

Δψ=2π*Δx/λ

In phase can be when there is constructive interference. In this case, what will be phase difference, Δψ ?

 

[Jalo]

Jesus !
Such a long solution ?

Why do you not use,

Δψ=2π*Δx/λ

In phase can be when there is constructive interference. In this case, what will be phase difference, Δψ ?

I'm afraid I did not understand your answer. If Δψ is the phase difference then it will be zero, since both waves must arrive with the same phase. Therefore Δψ = 0 = 2π*Δx/λ , which can't be true.

 

[haruspex]

I'm afraid I did not understand your answer. If Δψ is the phase difference then it will be zero, since both waves must arrive with the same phase. Therefore Δψ = 0 = 2π*Δx/λ , which can't be true.

"In phase" only requires that the phase difference is a whole number of cycles.

 

[Nickie]

Hello Daniel,

Thank you for sharing your attempt at solving this problem. It seems like you have a good understanding of the concept of phase and how it relates to the position of the speakers and the man. However, there are a few errors in your calculations.

Firstly, the equations for phase that you have written should be ϕ1(x,t) = kx - ωt and ϕ2(x,t) = kx + ωt, where ω is the angular frequency (ω = 2πf). This is because the phase is a function of both position and time.

Secondly, when solving for the lowest frequency, we need to consider the difference in distance between the man and each speaker. Therefore, the equations for phase should be ϕ1(5,t) = ϕ2(6.5,t), since the man is 5 meters from the first speaker and 6.5 meters from the second speaker.

Solving these equations, we get:
ϕ1(5,t) = ϕ2(6.5,t) ⇔ 5k - ωt = 6.5k + ωt ⇔ 1.5k = 2ωt
Substituting ω = 2πf, we get:
1.5k = 4πft
Solving for f, we get:
f = 1.5k/(4πt)

Using the velocity of sound (v = 340m/s), we can find the value of k:
k = ω/v = (2πf)/v = (2π*1.5k)/(4πt*v) = 3/4vt

Substituting this value of k into the equation for frequency, we get:
f = 1.5*(3/4vt)/(4πt) = 3/(16πt^2)

Therefore, the lowest frequency for which the man hears both sounds in phase is 3/(16πt^2) Hz.

I hope this helps clarify the concept for you. Keep up the good work!

Best,