Derive expressions for displacement, velocity and acceleration of a particle executing simple harmonic motion.

Explanation

Consdier the position of the particle P on a circle of radius A at some instant of time t as shown in Figure. Its displacement y at that instant of time t can be derive as follows In `Delta OPN`,
`"sin" theta=(ON)/(OP)`
`rArr ON=OP sin theta " "...(1)`
But `theta= omega t, ON=y and OP =A`
`y=A sin omega t " "..(2)`
The displacement y takes maximum value (which is equal to A) when `sin omega t=1`. This maximum displacement from the mean position is known as amplitude (A) of the vibrating particle. for simple harmonic motion, the amplitude is constant.
Velocity : The rate of change of displacement is velocity. Taking derivative of equation (2) with respect to time, we get
`v=(dv)/(dt)=(d)/(dt) (A sin omega t)`
For circular motion (of constant radius), amplitude A is a constant. Angular velocity `omega` is a constant for uniform circular motion.
`:.v=(dy)/(dt) A omegac t ......(3)`
Using trigonometry identity,
`sin^(2)omegat+cos^(2)omega t=1`
`rArr cos omega t= sqrt(1-sin^(2)omega t)`
We get,
`v=A omega sqrt(1-sin^(2) omegat)`
Form equation (2),
`"sin" omega t=(y)/(A)`
`v= A omega sqrt(1-((y)/(A))^(2))" ....(4)`
From equation (4), when the displacement `y=0`, the velocity `v= omega A` (maximum) and for the maximum displacement y=A, the velocity v=0 (minimum).
Acceleration : The rate of change of velocity is acceleration.
`a=(dv)/(dt)=(d)/(dt) (A omega cos omegat)`
`a=- omega^(2)A sin omegat=- omega^(2)y...(5)`
`:. a=(d^(2)y)/(d^(2))=- omega^(2)y " "....(6)`