Transverse sinusoidal wave is travelling along a string

AI Thread Summary
A transverse sinusoidal wave traveling along a string can be represented by the equation y = A sin(kx - ωt). The particle speed at a fixed position is derived as dy/dt = -ωA cos(kx - ωt), with maximum values obtained by setting cos(kx - ωt) to 1. The slope of the wave, represented as dy/dx, is crucial for establishing the relationship between particle speed and wave speed. The discussion clarifies that the ratio of particle speed to wave speed can be expressed through the differentiation of the wave equation, leading to a better understanding of the underlying physics. The participants ultimately grasp the connection between these concepts, confirming their understanding of the problem.
SAGHTD
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Homework Statement


Prove that if a transverse sinusoidal wave is traveling 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?
 
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HI SAGHTD! :smile:

(have an omega: ω and a pi: π :wink:)
SAGHTD said:
Prove that if a transverse sinusoidal wave is traveling 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?
 
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?
 
Hi SAGHTD! :smile:
SAGHTD said:
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:
 
OMG Thanks i got it out! I understand :D
 
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