The equation of a transverse wave travelling in a rope is given by: a) find the

Question

The equation of a transverse wave travelling in a rope is given by:

a) find the amplitude frequency, velocity and wavelength of the wave.

Ok do not need help with a) i got 0.1sin[2Pi(x/2-t)], A=0.1m,=2.0m,f=1,v=2.0 (from f)

* it is b) that I need help with

b) find the maximum transverse speed of a particle (by particale we mean any small element of the rope which will be set in motion as the wave moves past a chosen position) in the rope.

Ok so I did dy/dx, and I got dy/dx=0.1cos(x-2t) (let cos(pheta)=1)

0.1pi

Now the soloution goes as follows dy/dx=2.00*0.1cos(x-2t), they then make cos(pheta =1) and finish the solution as I have. What I don't understand is how they got dy/dx=2.00*0.1cos(x-2t)... can someone please explain steps for part b) as thorough as possible.

Explanation / Answer

Speed is given by dy/dt , not dy/dx

So you must differentiate y w.r.t t

dy/dt = 0.1 cos (x-2t) x -2

dy/dt = -0.2 cos (x-2t)

Since this has a negative sign, we must take the minimum value of cos to get the maximum value of dy/dt

Minimum value of cos  is -1.

Hence maximum velocity = -0.2 x -1 = 0.2 = 0.628

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