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`        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?`
one year ago

## Answers : (1)

```							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
Arun (askIITians forum expert)

```
one year ago
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