A particle executing simple harmonic motion has angular frequence `6.28s^-1` and amplityude 10 cm. Find a the time period, b the maximum speed c. the maximum acceleration d. the speed when thedisplacement is 6 cm from the mean position e. the speed t `t=1/6s` assuming that the motion starts from rest t t=0.

Explanation

`omega=6.28=2pirad//s, A=10cm`
(a) `T=(2pi)/(omega)=(2pi)/(2pi)=1s`
(b) `v_(max)=Aomega=10xx2pi=20pi cm//s`
(c) `a_(max)=Aomega^2=10xx(2pi)^2=40pi^2 cm//s^2`
(d) `v=omegasqrt(A^2-x^2)`
`=2pisqrt((10)^2-(6)^2)=16picm//s`
(e) Let `x=Asin(omegat+phi)`
`v=Aomegacos(omegat+phi)`
At `t=0`, `v=0`
`cos(omegat+phi)=0`
`cosphi=cospi/2impliesphi=pi//2`
`v=Aomegacos(omegat+pi//2)=-Aomegasinomegat`
At `t=0`, `x=Asin(omegaxx0+pi//2)=A`

At `t=1/6s`
`v=-Aomegasin(omegaxx1//6)=-10xx2pisin(2pi//6)`
`=-10sqrt3picm//s`
At `t=1//6s, x=Asin(2pixx1//6+pi//2)`
`=Acospi//3=A//2=5cm`