Transverse sinusoidal wave is travelling along a string

[SAGHTD]

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?

 

[tiny-tim]

HI SAGHTD!

(have an omega: ω and a pi: π )

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 ?

And where is your equation for the slope?

 

[SAGHTD]

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...
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?

 

[tiny-tim]

Hi SAGHTD!

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.

(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.

 

[SAGHTD]

OMG Thanks i got it out! I understand :D