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 partical moving simple harmonically.
b.Find the equilibrium position of the particle.
c. Write the equation of motion of the partical.
d. Find the time period of SHM.

Answer

a. Yes
b. `x = 4m`
c.`x = 4 + 2 cos t`
d.`2 pi sec`

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. Equilibrum position is `x = 4`
c. At `x = 6m`. Partical is rest ia at rest , i. e. it is one of the the extreme position.
Hence amplitude A is 2 m and initial partical is at the exterme position.
Therefore, 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 + cos t`
d. Time period `T = (2 pi)/(omega) = 2 pi sec`.