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Find wavelength, Phase Difference, and amplitude from min and max |
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| Feb18-13, 06:25 PM | #18 |
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Find wavelength, Phase Difference, and amplitude from min and max
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? |
| Feb18-13, 08:06 PM | #19 |
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| Feb18-13, 08:11 PM | #20 |
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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?
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| Feb18-13, 08:30 PM | #21 |
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| Feb18-13, 09:52 PM | #22 |
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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).
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| Feb18-13, 10:19 PM | #23 |
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So to find the value of a, 3π/4 = a*cos(40k + 3[itex]\pi/4[/itex]) and solve for a, correct? |
| Feb18-13, 10:22 PM | #24 |
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![]() 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). |
| Feb18-13, 10:23 PM | #25 |
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And BTW -pi/4 is equally as good an answer here as +3pi/4 since it was not specified which speaker lagged which. |
| Feb18-13, 10:29 PM | #26 |
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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? |
| Feb18-13, 10:53 PM | #27 |
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I don't read it as asking for the amplitude only at x = 0.
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| Feb19-13, 10:24 AM | #28 |
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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). |
| Feb19-13, 10:31 AM | #29 |
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So in order to find the amplitude, I would isolate a, correct?
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| Feb19-13, 10:45 AM | #30 |
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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). |
| Feb19-13, 10:49 AM | #31 |
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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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