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Sinusoidal Wave Problem 
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#1
Apr1808, 04:05 PM

P: 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 [tex]\lambda[/tex] = 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[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! 


#2
Apr1808, 04:41 PM

Emeritus
Sci Advisor
PF Gold
P: 5,197

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
Apr1808, 04:57 PM

P: 3

Would x = 6.37 * [tex]\pi[/tex] because that would make it ([tex]\pi[/tex]/3 50.3t)



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