Phase Difference: Solving Homework Problems

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SUMMARY

The discussion focuses on calculating the phase difference between two sinusoidal waves defined by the equations y1 = (2.00 cm)sin(20x - 32t) and y2 = (2.00 cm)sin(25x - 40t). At the point x = 5 cm and t = 2 s, the calculated values are y1 = -1.9835 and y2 = 1.7018. The phase difference can be determined by evaluating the arguments of the sine functions at the given x and t values. Additionally, the position x where the two phases differ by +/- π at t = 2 s can be found by solving the equation 2sin(20x - 64) + 2sin(25x - 80) = 0.

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Homework Statement


Two sinusoidal waves in a string are defined by functions
y1= (2.00cm)sin(20x-32t)
y2=(2.00cm)sin(25x-40t)
where y and x are cm and t is sec
(a) What is the phase difference between these two waves at the point x = 5 cm and t = 2 s?
(b) What is the position x value closes to the orgin for which the two phases differ by +/- pi at t = 2 s? (This location is where the two waves add to zero)



Homework Equations


I use the equation given to me as y1 and y2
not sure if there is an equation for phase difference

The Attempt at a Solution



(a) I plugged in 5 cm and 2 s for y1 and y2
y1 = -1.9835 and y2 = 1.7018
but I am not sure what these answers mean and if they are relevant
b) (2sin(20x-32*2) + 2sin(25x-40*2) = 0 and solve for x? Is this correct?Do I do inverse sin and then move the (32*2 and 40*2) to the other side?

Thanks for any help!
 
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A few words on phase difference:

Phase is a way of telling "where" the graph of a function is.

For example, the graph of sin(x) is 0 at x=0 and 1 at x=pi/2. If I were to change the function to sin(x-pi/2), then the function would be -1 at x=0 and 0 at x=pi/2.

Effectively, I dragged the entire graph of the function to the right by pi/2, so we would say that this second function has a phase of pi/2 relative to the first (phase is typically only important when comparing two functions).
 

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