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Find wavelength, Phase Difference, and amplitude from min and max 
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#20
Feb1813, 08:11 PM

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?



#21
Feb1813, 08:30 PM

HW Helper
Thanks
PF Gold
P: 4,761




#22
Feb1813, 09:52 PM

P: 616

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).



#23
Feb1813, 10:19 PM

P: 95

So to find the value of a, 3π/4 = a*cos(40k + 3[itex]\pi/4[/itex]) and solve for a, correct? 


#24
Feb1813, 10:22 PM

HW Helper
Thanks
PF Gold
P: 4,761

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). 


#25
Feb1813, 10:23 PM

HW Helper
Thanks
PF Gold
P: 4,761

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


#26
Feb1813, 10:29 PM

P: 95

I'm sorry, I was trying to find amplitude when they are placed sidebyside.
So to find amplitude when they are placed sidebyside, 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? 


#27
Feb1813, 10:53 PM

P: 616

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



#28
Feb1913, 10:24 AM

HW Helper
Thanks
PF Gold
P: 4,761

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). 


#29
Feb1913, 10:31 AM

P: 95

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



#30
Feb1913, 10:45 AM

HW Helper
Thanks
PF Gold
P: 4,761

I still don't know what you mean.
a is given to you. It's the amplitude of each speaker by itself. Putting them sidebyside gets you the net amplitude of the two outputs. So if that amplitude is called a', then a' = a + a cos(ψ). 


#31
Feb1913, 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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