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


by rocapp
Tags: amplitude, difference, phase, wavelength
rude man
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#19
Feb18-13, 08:06 PM
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Quote Quote by rocapp View Post
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
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#20
Feb18-13, 08:11 PM
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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
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#21
Feb18-13, 08:30 PM
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Quote Quote by rocapp View Post
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
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#22
Feb18-13, 09:52 PM
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My thinking is that the speakers are in phase when they are separated by 3/8 of a cycle, so the phase difference is [itex]3\pi/4[/itex] (or maybe [itex]-\pi/4[/itex]; I always mix up the signs).
rocapp
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#23
Feb18-13, 10:19 PM
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Quote Quote by tms View Post
My thinking is that the speakers are in phase when they are separated by 3/8 of a cycle, so the phase difference is [itex]3\pi/4[/itex] (or maybe [itex]-\pi/4[/itex]; I always mix up the signs).
Thanks, it was 3[itex]\pi[/itex]/4.

So to find the value of a,

3π/4 = a*cos(40k + 3[itex]\pi/4[/itex])

and solve for a, correct?
rude man
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#24
Feb18-13, 10:22 PM
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Quote Quote by tms View Post
My thinking is that the speakers are in phase when they are separated by 3/8 of a cycle, so the phase difference is [itex]3\pi/4[/itex] (or maybe [itex]-\pi/4[/itex]; 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
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#25
Feb18-13, 10:23 PM
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Quote Quote by rocapp View Post
Thanks, it was 3[itex]\pi[/itex]/4.

So to find the value of a,

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

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
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#26
Feb18-13, 10:29 PM
P: 95
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
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#27
Feb18-13, 10:53 PM
P: 501
I don't read it as asking for the amplitude only at x = 0.
rude man
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#28
Feb19-13, 10:24 AM
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Quote Quote by rocapp View Post
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
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#29
Feb19-13, 10:31 AM
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So in order to find the amplitude, I would isolate a, correct?
rude man
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#30
Feb19-13, 10:45 AM
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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).
rocapp
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#31
Feb19-13, 10:49 AM
P: 95
Just had a breakthrough moment. My textbook must be awful because I just understood what these cosine functions are actually representing from your comment and then found an equation that would help.

Amplitude of both = 2a*cos(Δψ/2)

= 2a*cos(1.1775)


Amplitude in terms of a = 0.767a

Submitted, and that was correct.

Thank you both tms and rudeman for your help! I really appreciate you guys and hope one day to give back to the community here at Physics Forums.


Thanks again!


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