A particle of mass `2 kg` is moving of a straight line under the actin force `F = (8 - 2x)N`. It is released at rest from `x = 6m`.
a. Is the particle moving simple hormonically.
b. Find the equilibrium position of the particle.
c. Write the equation of motion of the particle.
d. Find the time period of SHM.

Explanation

`f = 8 - 2 x`
or `F = - 2 (x-4)`
For equilibrium position `F = 0`
`implies x = 4` is equilibrium position
Hence the motion of the partuical is SHM with force canstant 2 and equilibrium position`x = 4` .
a. Yes, motion is SHM.
b. Equilibrium position`x = 4` .

c. At `x = 6m` partical is at rest, i.e. it is one of the extreme position.
Hence amplitude is `A = 2 m` and initially partical is at the exterme position.
`:.` Equation of Shm can be written as
`x - 4 = 2 cos omega t`
where ` omega = sqrt((k)/(m))`
`= sqrt((2)/(2)) = 1`
i.e., `x = 4 + 2 cos t`
d. The time period
`T = (2 pi)/(omega) = 2 pi s`