Transverse sinusoidal wave is travelling along a string

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Homework Statement


Prove that if a transverse sinusoidal wave is travelling along a string, then the slope at any point of the string is equal to the ratio of the particle speed to wave speed at that point.


The Attempt at a Solution


This is what i did isn't the equation for transverse = y=Asin(kx-wt)
So i differentiate it with respect to "t" therefore getting dy/dt = -wAcos(kx - wt)

cos(kx - wt) = 1

therefore dy/dt = -wA and w = 2pi(F)
So wouldn't it be equal to 2pi(F)A <------------------ just want to make sure but this is for the particle speed??

If so how am i to that the ratio of the particle speed to wave speed at that point is equal???
 

Answers and Replies

  • #2
tiny-tim
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HI SAGHTD! :smile:

(have an omega: ω and a pi: π :wink:)
Prove that if a transverse sinusoidal wave is travelling along a string, then the slope at any point of the string is equal to the ratio of the particle speed to wave speed at that point.

This is what i did isn't the equation for transverse = y=Asin(kx-wt)
So i differentiate it with respect to "t" therefore getting dy/dt = -wAcos(kx - wt)

cos(kx - wt) = 1

That's right … the particle speed at fixed position x is dy/dt.

But why are you putting cos(kx - ωt) = 1 ? :confused:

And where is your equation for the slope?
 
  • #3
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Oh....i remembered that before how our lecturer was telling us to sub cos(kx – ωt) as 1 to acquire Umax. Didn't really understand that much of it tho... :frown:
Equation for the slope is what i don’t really get . Didn’t the question say that they wanted the ratio of the particle speed to wave speed at that point? Now what little i understand the equation for wave speed would be v = λf so how am i to really give a ratio between these two equations?
 
  • #4
tiny-tim
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Hi SAGHTD! :smile:
Oh....i remembered that before how our lecturer was telling us to sub cos(kx – ωt) as 1 to acquire Umax.

He only meant that if you have y = Asin(kx – ωt) or dy/dt = Aωcos(kx – ωt), then the maximum values of y and dy/dt are by putting cos = 1, ie they're A and Aω. :wink:

(but this question doesn't ask you for that)
Equation for the slope is what i don’t really get .

Slope is distance/distance, so it here it must mean dy/dx …

they're asking for a formula to convert dy/dx to dy/dt. :smile:
 
  • #5
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OMG Thanks i got it out!!!! I understand :D
 

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