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.

 

[Moetasim]

Hello! It seems like you are working on a problem related to waves and vibrations. Part c asks for the transverse speed at a specific point (x = 1) and time (t = 1). Transverse speed refers to the speed of the wave as it moves perpendicular to the direction of propagation. In this case, it would be the speed of the wave as it moves up and down at x = 1 and t = 1. To find this, you can use the formula v = A\omega\cos(kx - \omega t), where A is the amplitude, \omega is the angular frequency, k is the wave number, x is the position, and t is the time. Plug in the values you have already found for \omega and k, and the given values for x and t, and you should get the correct answer of -0.022 m/s. I hope this helps!