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

by rocapp
Tags: amplitude, difference, phase, wavelength
HW Helper
Thanks
PF Gold
P: 4,406
 Quote by 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?
Not what I got. tms?
 P: 95 I think my problem is that delta phi, the distance they are apart needs to be in radians. How could I find that value?
HW Helper
Thanks
PF Gold
P: 4,406
 Quote by 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?
37.5 deg = 37.5/180 * pi radians. (But I didn't get that. Waiting for tms to reply ...).
 P: 501 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).
P: 95
 Quote by 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).
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?
HW Helper
Thanks
PF Gold
P: 4,406
 Quote by 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).
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).
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Thanks
PF Gold
P: 4,406
 Quote by rocapp 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.
 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?
 P: 501 I don't read it as asking for the amplitude only at x = 0.
HW Helper
Thanks
PF Gold
P: 4,406
 Quote by 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?
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).
 P: 95 So in order to find the amplitude, I would isolate a, correct?
 HW Helper Thanks PF Gold P: 4,406 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).
 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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