Question

Question #c9010

Answer 1

The amplitude is `=32m`
The period is `=5.03s`

Explanation:

For a SHM, we have the following

displacement, `x=asin(omega t+phi)`

velocity, `v=dx/dt=aomegacos(omega t+phi)`

acceleration, `a=(d^2x)/dt^2=-a omega^2sin(omega t+phi)`

The maximum velocity is

`aomega=40`

The maximum acceleration is

`aomega^2=50`

From, those 2 equations, we deduce

`omega=50/40=5/4=1.25`

and

amplitude, `a=40/1.25=32m`

To find the period, we use

`omega=(2pi)/T`

`T=(2pi)/omega=(2pi)/1.25=5.03s`