# Sinusoidal Wave Problem

1. ### jpond89

3
1. The problem statement, all variables and given/known data

Consider the sinusoidal wave in the figure, with the wave function below.

At a certain instant, let point A be at the origin and point B be the first point along the x axis where the wave is 60.0° out of phase with point A. What is the coordinate of point B?

y = (15.0 cm) cos(0.157x - 50.3t)

and I am not allowed to post the image but heres some helpful information:
its a cosine graph, and $$\lambda$$ = 40 cm, T = 0.125 s, f = 8.00 hz, k = 0.157 rads/cm

2. Relevant equations

Some equations that I have are:
k=2$$\pi$$/$$\lambda$$
v=$$\lambda$$*f
y=Asin(kx-$$\omega$$t)

3. The attempt at a solution
What I have tried so far is:
y=15*sin((.157)x-(50.3)t + $$\pi$$/3)
The example says " The vertical position of an element of the medium at t = 0 and x = 0 is also 15.0 cm "
The only problem is that I am really stuck and I cannot seem what to set the 60.0$$\circ$$ equal to. Help would be much appreciated! Thank you!

2. ### cepheid

5,189
Staff Emeritus
Everything in the argument of the cosine is the phase of the wave. So the phase is a function of both x and t. However, we're looking at this wave at a particular "instant", which means that t is constant. So, the only thing that would cause a phase change is x. So how far from the origin would the x coordinate of point B have to be in order for 0.157x - 50.3t to differ from 0.157*(0) - 50.3t by 60 degrees, given that t is constant?

3. ### jpond89

3
Would x = 6.37 * $$\pi$$ because that would make it ($$\pi$$/3 -50.3t)

4. ### cepheid

5,189
Staff Emeritus
I think you might be off by a factor of 3. Check your work again.