If the displacement of a moving point at any time is givenby an equation of the form `y (t) = a cos omega t + b sin omega t`, shown that the motion is simple harmonic . If `a = 3 m, b = 4m and omega = 2`: determine the period , amplitude, maximum velocity and maximum acceleration.

Explanation

`y=acos omegat+bsin omegat`
Let, `a=Rsin theta and b=Rcos theta`
Then, `y=Rsintheta cos omega t +R cos theta sin omega t`
`y=Rsin (omega t + theta)`
Velocity, `V=(dy)/(dt)=Romega cos (omegat+theta)`
Acceleration, `A=(dV)/(dt)=-omega^(2)Rsin(omegat+theta)`
`=omega^(2)y`
It means `A prop y` and `-ve` sign shows that A is directed towards the mean position, hence the motion is SHM.
Amplitude, `R=sqrt(a^(2)+b^(2))=sqrt(3^(2)+4^(2))`
`=5` units
Time period, `T=(2pi)/(omega)=(2pi)/(2)=3.14s`
Max. velocity, `V_(max)=Romega=5xx2=10unit`
Max. acceleration `=-omega^(2)R`
`=-(2)^(2)xx5=-20units`