Question #c9010

Question

3804 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-27T13:59:14

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

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

$omega=(2pi)/T$

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