Wave problem It's been driving my crazy for last hour

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Wave problem! It's been driving my crazy for last hour!

Homework Statement


A)
Wave motion is characterized by two velocities: the velocity with which the wave moves in the medium (e.g., air or a string) and the velocity of the medium (the air or the string itself).

Consider a transverse wave traveling in a string. The mathematical form of the wave is y(x,t) = Acos2pi(kx-wt)

Find the velocity of propagation Vp of this wave.
Express the velocity of propagation in terms of some or all of the variables A, k, and W.

B)Two traveling waves move on a string that has a fixed end at . They are identical, except for opposite velocities. Each has an amplitude of 2.46 mm, a period of 3.65 ms, and a speed of 111 m/s.

The wave function of the resulting standing wave has the form y(x,t)=Asin(kx)sin(wt). Give the values of A, k, and W. Use meters (m) for A, inverse meters (1/m) for k, and inverse seconds (1/s) for W.




Homework Equations


A)
I believe I should use y(x,t) = Acos2pi(kx-wt)
B)
A is given
W=2(pi)f where f = 1/T
K = w/v

The Attempt at a Solution



A)
Ok so to find the velocity, we take the partial derivative and get
dy/dt = wAsin[kx-wt]

And I thought that was the answer but apparently it wasn't.

What is the velocity of propagation anyways? Maybe that's where I need to start.

B)
This one, I thought I had it right but apparently, Mastering Physics has been telling me its not.

Amplitude's given, so just convert 2.46mm to meters.

W(Omega) can be simply found by the equation w = 2pi/T

And since velocity is given, we can just plug that value in into k = w/v and find the wave number.

I got 2.46*10^-3 as Amplitude, 1720 for W and 15.5 for k and the program's telling me its wrong.

Am I tackling the problem wrong?

Any help would be deeply appriciated
 
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Can anyone help with this? I too have been trying to figure it out, only part B.
 


For the record, the answer turned out to be:

4.92×10−3, 15.5, 1720