# Two identical waves (except for direction of travel) oscillate through a springs and yield a superposition according to the equation y’ = (0.50 cm) sin What are the amplitude and speed of the two waves? What is the distance between nodes? What is the transverse speed of a particle of the string at the positon x = 1.5 cm when t =9/8 s?

Arun
25750 Points
6 years ago
Dear student
Yes, you do have to take the derivative of this. This is a partial differential equation because y' = (0.005 cm) sin[(π/3)x] cos[(45π)t] is a function of x and t. We need the speed, so that equation represents displacement along the y direction. Take dy'/dt, treating terms with x as constants.
First I'm going to convert the cm to m.
y' = (0.005) sin[(100π)/3)x] * cos[(45π)t]
Now derive using product rule and chain rule.
dy'/dt = 0.005 * sin[(100π)/3)x] * [-sin((45π)t)] * (45π)
= -0.225π * sin[(100π)/3)x] * sin[(45π)t]
Subbing in x = 0.015m and t = 9/8s gives v = -0.65m/s = -65cm/s.
You can check that this works by subbing the the values of x where there are nodes, and your speed should equal 0.
Regards