Transverse wave: altitude at x=0.25 m, t=0.15 s

[Gil-H]

Homework Statement

Monochromatic transverse wave is propagating
in the positive x direction on a string (neither end is fixed).
The wave velocity is 12 m/s, amplitude 0.05 m, and wavelength is 0.4 m.
At t=0 and x=0 the string altitude is 0, and moving in
the positive y direction [i.e. (dy/dt)|(t=0) > 0]

Find the altitude y at x=0.25 m and t=0.15 sec.

The Attempt at a Solution

I figured that since y is 0 when x and t are 0, it must sine:
y(x,t) = Asin(kx-ωt)

I found the parameters I need:
k = (2π/λ)=(2π)/(0.4) = 5π
ω = vk = 12*5π = 60π

and I get
y(x,t) = 0.05sin(5πx-60πt)

I plug in the values x=0.25 and t=0.15 and get
y(0.25,0.15) = -0.02

But this is not the answer. Why?
What did I do wrong?

 

[tiny-tim]

Welcome to PF!

Hi Gil-H! Welcome to PF!

y(x,t) = 0.05sin(5πx-60πt)

I plug in the values x=0.25 and t=0.15 and get
y(0.25,0.15) = -0.02

The wave equation looks ok,

but how did you get .02? ​

 

[Gil-H]

I got -.02 because I was unawarely working in degrees.
I should be .0354
But after advice I got, I changed the wavefunction to be:
y(x,t) = -0.05sin(5πx-60πt) (I added a minus sign)
That's because of the condition that (dy/dt)|(t=0) > 0.

Anyway, thanks for trying to help (-: