Question

The displacement x of a simple harmonic motion is given by X = 24 cos (6.28t + 1.57) meters. What is the maximum speed of the oscillator.

Answer 1

`X(t)= 24 cos (6.28 t +1.57)`

And for this case we can write this expression like this:

`X(t) =A cos (wt +\phi)`

The velocity would be given by the first derivate and we got:

`V(t) = -wA sin (wt +\phi)`

And the maximum velocity would be:

`V_(max)= |6.28 *24| = 150.72 m/s`

Step-by-step explanation:

For this case we have the following function for the position:

`X(t)= 24 cos (6.28 t +1.57)`

And for this case we can write this expression like this:

`X(t) =A cos (wt +\phi)`

The velocity would be given by the first derivate and we got:

`V(t) = -wA sin (wt +\phi)`

And the maximum velocity would be:

`V_(max)= |6.28 *24 |= 150.72 m/s`