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Calculating resultant amplitude, given 2 ampli's and phase diff 
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#1
Jul2211, 09:30 AM

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1. The problem statement, all variables and given/known data
Two waves of the same frequency have amplitudes 1.02 and 2.27. They interfere at a point where their phase difference is 59.5°. What is the resultant amplitude? 2. Relevant equations Resultant Amp = (Amp1 + Amp2)cos(theta/2) 3. The attempt at a solution Okay this seemed like a simple plug and chug problem. I ended up getting an answer of 2.86, but this is marked wrong. I then did the "practice another version" thing on webassign and worked another version of the same exact problem (just with different values of the two amps and a different theta), but used the exact same formula and guess what... it was correct! I've spent the last 15 minutes trying to figure out why the formula works for one version of the problem but not the one that matters. Can anyone help? 


#2
Jul2211, 10:16 AM

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ehild 


#3
Jul2211, 10:44 AM

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#4
Jul2211, 12:50 PM

P: 33

Calculating resultant amplitude, given 2 ampli's and phase diff
Can someone help, please? I don't expect anyone to just solve the problem and give me the answer like they do on cramster, but statements like the two above (with all due respect) don't really help me one single iota. No, I'm not sure if that equation works 100% of the time (obviously, it doesn't). But, what formula should I try instead?



#5
Jul2211, 12:57 PM

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1. There are some units that should have been specified 2. There is a typo in the question (no way to fix that!) 3. The marking program is incorrectly programmed for this question (call your instructor) 


#6
Jul2211, 03:08 PM

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Try the following formula for the resultant amplitude :
A=sqrt[A_{1}^{1}+A_{2}^{2}2A_{1}A_{2}cos(θ)]. ehild 


#7
Jul2211, 03:37 PM

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#8
Jul2211, 03:43 PM

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The interfering waves are A_{1}sin(wtkx) and A_{2}sin(wtkx+θ). Their resultant is a single wave with amplitude A. Denoting wtkx by α, we have the sum
A_{1}sin(α)+A_{2}sin(α+θ)=Asin(α+φ)*. sin(α+θ)=sin(α)cos(θ)+cos(α)sin(θ), and sin(α+φ)=sin(α)cos(φ)+cos(α)sin(φ). Replacing into eq. * : sin(α)[A_{1}+A_{2}cos(θ)]+cos(α)A_{2}sin(θ)=A[sin(α)cos(φ)+cos(α)sin(φ)] The equation is valid for any angles, that is for any values of sin(α) and cos(α) between 1 and 1. Therefore A1+A_{2}cos(θ)=Acos(φ) and A_{2}sin(θ)=Asin(φ) Square both equation and add up: A^{2}=A_{1}^{2}+A_{2}^{2}+2A_{1}A_{2}cos(θ). ehild 


#9
Jul2211, 03:46 PM

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#10
Jul2211, 04:15 PM

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#11
Jul2211, 04:26 PM

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I've been looking more closely at the OP's formula. It would appear to be a good approximation for small phase angles, say less than 50 degrees or so. I'm currently investigating a derivation that will provide an error term.



#12
Jul2211, 04:28 PM

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#13
Jul2211, 04:32 PM

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I wonder why not the exact formula is used? It is derived in every books on General Physics.
ehild 


#14
Jul2211, 04:40 PM

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Makes the math simpler. I've derived an expression for the signal sum that looks as follows: [tex] (A + B) cos\left(\frac{\phi}{2}\right) cos(\omega t) + (A  B)sin\left(\frac{\phi}{2}\right) sin(\omega t) [/tex] The amplitude of the left term is clearly the OP's formula. The right term represents a deviation from that formula. Apparently the OP's formula works best for similar amplitudes and small angles. 


#15
Jul2211, 04:52 PM

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#16
Jul2211, 05:16 PM

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#17
Jul2211, 11:05 PM

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What if students try to use the resultant for A1=3 A2=1 and phase difference of 180°? They see only a formula to plug in data. Applying the correct formula is not much more work than using the wrong one.Why do they teach the wrong one then?
ehild 


#18
Jul2311, 07:24 AM

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