Question

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

Answer 1

`F = 8 - 2x `
or `F = -2(x - 4)`
at equilibrium position `F = 0`
Hence the motion of particle is `SHM` with force constant `2` and equilibrium position `x = 4`
(a) Yes, motion is `SHM`.
(b) Equilibrium position is `x = 4`
(c) At `x = 6 m`, particle is at rest i.e., it is one of the extreme position Hence amlitude is `A = 2 m` and initially particle is at the extreme position.

`:.` Equation of `SHM` can be written as
`x - 4 = 2 omegat`, where `omega = sqrt((k)/(m)) = sqrt((2)/(2)) = 1`
i.e., `x = 4 + 2 cos t`
(d) Time period, `T = (2pi)/(omega) = 2pisec`.