Phase Difference: Solving Homework Problems

[crazyog]

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!

 

[kreil]

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).

 

[Moetasim]

The phase difference between two waves is the difference in their starting phases at a specific time and location. To calculate the phase difference, we need to compare the phase angles of the two waves at the given point in space and time.

(a) To find the phase difference at x = 5 cm and t = 2 s, we can use the given equations for y1 and y2. Plugging in the values, we get y1 = -1.9835 cm and y2 = 1.7018 cm. These values represent the displacement of each wave at the given point in space and time.

To find the phase difference, we need to compare the phase angles of the two waves. The general equation for a sinusoidal wave is y = A*sin(ωt + φ), where A is the amplitude, ω is the angular frequency, t is time, and φ is the phase angle. Comparing the equations for y1 and y2, we can see that they both have the same amplitude (2.00 cm) and angular frequency (20x and 25x). However, the phase angles are different, with y1 having a phase angle of -32t and y2 having a phase angle of -40t.

To find the phase difference, we need to subtract the phase angle of y2 from the phase angle of y1. This gives us (-32t) - (-40t) = 8t. Plugging in the values of t = 2 s, we get a phase difference of 16 radians.

(b) To find the position x value closest to the origin where the two waves have a phase difference of +/- pi at t = 2 s, we can set the equation for y1 equal to the negative of the equation for y2 (since a phase difference of +/- pi indicates that the waves are completely out of phase and will cancel each other out).

2*sin(20x-32*2) = -2*sin(25x-40*2)

We can then use the inverse sine function to solve for x. However, note that this will give us the general solution for x, so we will need to use the given conditions (t = 2 s) to narrow it down to a specific value.

sin^-1(sin(20x-64)) = sin^-1(-sin(25x-80))

20x-64