Find wavelength, Phase Difference, and amplitude from min and max

[rocapp]

Homework Statement

See attached image

Homework Equations

Δphi = 2pi(Δx/λ) + Δphi₀ = 2pi(m) -- at the maximum
Δphi = (m + .5) (2pi) -- at the minimum

The Attempt at a Solution

The first step should be to solve for lambda, the wavelength.

So I plug in the knowns:

At 30 cm the amplitude is a maximum, so:

Δphi = 2pi(Δx/λ) + Δphi₀ = 2pi(m)

Δphi = 2pi(30/λ) + Δphi₀ = 2pi(m)

I am not sure what to do next. I tried too many times, so it now displays the answer as λ = 80 cm, but I'm not sure why that is true. I need help on the rest.

I'm just not sure what to do since I am missing more than one variable.

 

[tms]

You can figure out the wavelength by inspection: while the speaker moves 40 cm the wave goes from a minimum to a maximum. That tells you that 40 cm is what fraction of the wavelength?

 

[rude man]

Spkr 1 output = a*cos(kx - wt), k = 2π/λ and w = 2πf.
Put spkr 1 at x = 0 and let t = 0 arbitrarily:
Then spkr 1 = a.

Similarly, Spkr 2 output is a*cos(kx - wt - ψ), allowing for a phase difference ψ between the two speakers when they're side-by-side (as you might get if spkr 2 had a much longer hookup cable than spkr 1).

Now, what is the equation for the output of spkr 2 heard at x = 0 when placed a distance x0 in front of spkr 1? And what is the combined output at x = 0?

From this you get 2 equations and 2 unknowns: k and ψ.

 

[rocapp]

Spkr 1 output = a*cos(kx - wt), k = 2π/λ and w = 2πf.
Put spkr 1 at x = 0 and let t = 0 arbitrarily:
Then spkr 1 = a.

Similarly, Spkr 2 output is a*cos(kx - wt - ψ), allowing for a phase difference ψ between the two speakers when they're side-by-side (as you might get if spkr 2 had a much longer hookup cable than spkr 1).

Now, what is the equation for the output of spkr 2 heard at x = 0 when placed a distance x0 in front of spkr 1? And what is the combined output at x = 0?

From this you get 2 equations and 2 unknowns: k and ψ.

You can figure out the wavelength by inspection: while the speaker moves 40 cm the wave goes from a minimum to a maximum. That tells you that 40 cm is what fraction of the wavelength?

Thanks!

λ = 80 cm, since 40 cm = (1/2)λ.

The equation for the output of speaker 2 at x = 0 would be:

S2 = a*cos(-ψ)


The equation for the the combined output would be:

S1 + S2 = a*cos(-ψ₂ + ψ₁)

k = 2π/80 = 0.0785

So at x = -10 cm,

S2 = a*cos(0.0785*(-10) + ψ2).


I'm not sure what to do from here, though.

 

[tms]

The equation for the output of speaker 2 at x = 0 would be:

S2 = a*cos(-ψ)

Aren't you leaving out something in the argument to the cosine?

The equation for the the combined output would be:

S1 + S2 = a*cos(-ψ₂ + ψ₁)

That's not how you add cosines.

 

[rocapp]

If t=0 and x=0 then both terms besides the psi should be 0, no?

 

[rocapp]

And the correct equation for the combined outputs is:

S1 + S2 = 2a*cos(0.5*(ψ2+ψ1))*cos(0.5*(ψ2-ψ1))

Yes?

 

[rude man]

Thanks!

λ = 80 cm, since 40 cm = (1/2)λ.

The equation for the output of speaker 2 at x = 0 would be:

S2 = a*cos(-ψ)

No, that would be the output of spkr 2 at its own position, not at x = 0. Call its position x' = 0. Along the x axis, x' = 0 is a distance x0 to the right of x = 0 if spkr 2 were located x0 to the right of spkr 1.
So x' = x - x0.

OK, remember, we are arbitrarily fixing t = 0 everywhere.

So spkr 2 output = cos(kx' - ψ). But then x' = x - x2 if spkr 2 is in front of spkr 1 by x2 meters. And, if we put spkr 2 behind spkr 1 by x1 meters, spkr 2 output = cos(kx'' - ψ) where x'' = x + x1. Draw these three positions along an x-axis with x = 0 at spkr 1 and x = -x1 and then x = + x2 for spkr 2 and this should be pretty clear.

Now for each of the two position of spkr 2, add the signals from both spkrs and deduce what the cosine arguments must be in each case for the spkr signals to add up to 0 and 2a respectively. Remember we still can let x = 0 in these equations without loss of generality.

 

[rude man]

And the correct equation for the combined outputs is:

S1 + S2 = 2a*cos(0.5*(ψ2+ψ1))*cos(0.5*(ψ2-ψ1))

Yes?

Ugh! Nope!
Why do you want to multiply? We are adding the signals from the two speakers.
Also, there is no need for two phases ψ1 and ψ2. There is only one phase, namely the phase between the two speakers. As I said, think of this phase as being produced by a difference in cable lengths between the two speakers.

 

[tms]

Also, there is no need for two phases ψ1 and ψ2. There is only one phase, namely the phase between the two speakers.

Just to clarify for the OP, the most general solution would have a phase for both waves, but one can adjust when time = 0 in order to eliminate one of the phases.

 

[tms]

If t=0 and x=0 then both terms besides the psi should be 0, no?

At that one time, at that one place, yes, but you want a more general solution, for part C at least.

 

[tms]

Ugh! Nope!

Actually, yes, assuming x = t = 0. The formula for adding two cosines is:
\cos\theta + \cos\phi = 2 \cos\left(\frac{\theta + \phi}{2}\right)<br /> \cos\left(\frac{\theta - \phi}{2}\right).

 

[rocapp]

How would I find Psi from this?

 

[tms]

How would I find Psi from this?

From what?

 

[rocapp]

How would I find it from the equations I gave?

 

[tms]

Think about the information you are given. You now know the wavelength, and you know about how far apart the speakers are when the total output is at a maximum (that is, when the speakers are in phase). From that you can figure out the phase shift.

 

[rude man]

Or you can use the equation cos(kx - wt - psi) for spkr 2 in terms of the two speaker locations to find psi (and lamba).

In other words, cos(kx' - wt - psi) for spkr 2 in front of spkr 1, and cos(kx'' - wt - psi) for spkr 2 behind spkr 1.

With x' and x'' as I previously explained.

 

[rocapp]

Just to clarify, I have diagnosed myself Physics-retarded. I'm not bad at many other things, but Physics sometimes baffles me.

So from this, I see that I might be able to use the equation I provided originally:



Distance they are apart at a maximum = ΔPhi = 40

= 2π*(30 cm/80 cm) + ψ

40 = 2.35 + ψ

ψ = 37.65

Correct?

 

[rude man]

Just to clarify, I have diagnosed myself Physics-retarded. I'm not bad at many other things, but Physics sometimes baffles me.

So from this, I see that I might be able to use the equation I provided originally:



Distance they are apart at a maximum = ΔPhi = 40

= 2π*(30 cm/80 cm) + ψ

40 = 2.35 + ψ

ψ = 37.65

Correct?

Not what I got. tms?

 

[rocapp]

I think my problem is that delta phi, the distance they are apart needs to be in radians. How could I find that value?

 

[rude man]

I think my problem is that delta phi, the distance they are apart needs to be in radians. How could I find that value?

37.5 deg = 37.5/180 * pi radians. (But I didn't get that. Waiting for tms to reply ...).

 

[tms]

My thinking is that the speakers are in phase when they are separated by 3/8 of a cycle, so the phase difference is 3\pi/4 (or maybe -\pi/4; I always mix up the signs).

 

[rocapp]

My thinking is that the speakers are in phase when they are separated by 3/8 of a cycle, so the phase difference is 3\pi/4 (or maybe -\pi/4; I always mix up the signs).

Thanks, it was 3\pi/4.

So to find the value of a,

3π/4 = a*cos(40k + 3\pi/4)

and solve for a, correct?

 

[rude man]

My thinking is that the speakers are in phase when they are separated by 3/8 of a cycle, so the phase difference is 3\pi/4 (or maybe -\pi/4; I always mix up the signs).

That was my thinking too.

I agree, keeping track of the phase sign is a chore. Here I defined spkr 2 to lag spkr 1 by having the cable for spkr 2 longer than that of spkr 1. So the argument in my expression for spkr 2 was cos(kx - wt - ψ) compared to spkr 1 = cos(kx - wt).

 

[rude man]

Thanks, it was 3\pi/4.

So to find the value of a,

40 = 2a*cos(kx'' + 3\pi/4)

and solve for a, correct?

a is given!
And BTW -pi/4 is equally as good an answer here as +3pi/4 since it was not specified which speaker lagged which.

 

[rocapp]

I'm sorry, I was trying to find amplitude when they are placed side-by-side.

So to find amplitude when they are placed side-by-side, I would have to find the equation for both at x=0:

S1+S2 = a*cos(kx''' + PSI + 3π/4)


And solve for a,
Correct?

 

[tms]

I don't read it as asking for the amplitude only at x = 0.

 

[rude man]

I'm sorry, I was trying to find amplitude when they are placed side-by-side.

So to find amplitude when they are placed side-by-side, I would have to find the equation for both at x=0:

S1+S2 = a*cos(kx''' + PSI + 3π/4)


And solve for a,
Correct?

If you placed them side-by-side, the two signals would add up to

a cos(kx - wt) + a cos(kx - wt - ψ)
where you can arbitrarily pick x = t = 0 so the addition reduces to
a + a cos(-ψ).

 

[rocapp]

So in order to find the amplitude, I would isolate a, correct?

 

[rude man]

I still don't know what you mean.

a is given to you. It's the amplitude of each speaker by itself.

Putting them side-by-side gets you the net amplitude of the two outputs. So if that amplitude is called a', then a' = a + a cos(-ψ).