A force `F = - 10x + 2` acts on a particle of mass `0.1 kg` where x is in m and F in newton. If is released from rest at `x = 0`, find:
a. Amplitude:
b. Time period:
c. Equation of motion.

Explanation

a. given , `F = - 10x + 2 implies Delta F = - 10 Delta x`
But `Delta F = ma`
`ma = - 10 Delta x`
`:. a = 0 ((10)/(0.1)) Delta x = - 100 Delta x b`
or `omega ^(2) = 100 or omega = 10 rad//s`
b. `:. T = (2 pi)/(omega) = (2 pi)/(10) = (pi)/(5) sec`.
Let at `x = x_(0)`, mean positionbe situated . At mean position `F = 0`
`:. F = - 10x + 2 implies 0 = - 10 x_(0) + 2`
This gives `x_(0) = (2)/(10) = 0.2 m`
Let us write equation of SHM about `x = x_(0)`. here we are measuring displacement `(Deltax)` from mean position.
Let `Delta x = A sin (omega t +phi)`
`= A sin (10 t + phi)` (i)
At `t = 0` , the partical is released from rest, i.e., amplitudeposition.
Hence, amplitude of oscillation `|A| = (0.2 - 0) = 0.2m` hence, from eq, (i)
`- 0.2 = 0.2 sin phi impliesphi = - 1 which gives phi = - (pi)/(2)`
New Eq. (i) becomes
`Delta x = 0.2 sin (10 t - (pi)/(2)) = - 0.2 cos 10 t (m)`
The displacement of the partical from origin will be
`x = 0.2 + Delta x = 0.2 - 0.2 cos 10 t (m)`