Solve Transverse Speed Q: Part C & D

[roam]

Hello!
The following is my problem (a question from an old exam paper);
http://img341.imageshack.us/img341/5142/69855901wv3.gif




Well, I'm stuck on part c and d.
I already got the correct answers for a & b;
v = \frac{\omega}{k} => \frac{4}{3} = 1.3 m/s

K = \frac{2\pi}{\lambda} => \lambda = 2.09 m

v = f\lambda => f = 0.65 Hz

v = \sqrt{\frac{T}{\mu}} => 1.3 = \sqrt{\frac{T}{4}} => T = 7.1


I checked the answers and it is correct up to this point. Now I don't know how to do question c. What is it meant by "transverse speed". What formula should I use?

(The right answer is -0.022 m/s is the transverse speed at x = 1 & t = 1)

I don't know which formula to use...

Thanks.

 

[dynamicsolo]

The transverse velocity is the speed at which a point on the string is moving perpendicularly to the length of the string. It is the time derivative of the displacement y. Keep in mind, though, that here y is a function of two variables, so you will want the partial derivative of y with respect to t. You would then evaluate this derivative function at
x = 1, t = 1.