Question #c9010

1 Answer
Mar 27, 2017

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#