Time difference between two sine waves

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SUMMARY

The discussion focuses on calculating the time difference between two sine waves with a phase difference of 20 degrees and a frequency of 60 Hz. The first wave reaches its maximum at time t=0 and position x=0. The second wave's maximum is determined using the equation A*sin(-ωt + φ) = A, where φ is converted to radians (20 degrees = π/9). The final calculation reveals that the time difference until the second wave reaches its maximum is 7/2160 seconds.

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"There are two sine waves having a phase difference of 20 degrees. After one reaches its maximum value, how much time will pass until the other reaches its maximum, assuming a frequency of 60 Hz."

Should I go about this by assuming...
sin(120pi*t) = sin(120pi(t + x) - 20)

Any hints appreciated
 
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You have two sines which have are functions of position and time. (actually, considering them function of time alone is sufficient)
They both have the form:
A\sin(kx-\omega t + \phi)

Assume the first wave reaches its maximum A at time t=0 and position x=0.
Then you have to find t when the second wave reaches its max A:

A\sin(-\omega t + \phi)=A
Where \phi is 20 degrees expressed in radians.
 
Last edited:
I get it. Since 20 degrees = pi/9, moving LHS A to RHS gives
sin (-120pi*t + pi/9) = 1
so, -120pi*t + pi/9 = pi/2
and finally t = 7/2160 seconds

Many thanks
 

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