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Sinusoidal Wave Problem

  • Thread starter jpond89
  • Start date
3
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1. Homework Statement

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 [tex]\lambda[/tex] = 40 cm, T = 0.125 s, f = 8.00 hz, k = 0.157 rads/cm

2. Homework Equations

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

3. The Attempt at a Solution
What I have tried so far is:
y=15*sin((.157)x-(50.3)t + [tex]\pi[/tex]/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[tex]\circ[/tex] equal to. Help would be much appreciated! Thank you!
 

Answers and Replies

cepheid
Staff Emeritus
Science Advisor
Gold Member
5,183
35
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
0
Would x = 6.37 * [tex]\pi[/tex] because that would make it ([tex]\pi[/tex]/3 -50.3t)
 
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,183
35
I think you might be off by a factor of 3. Check your work again.
 

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