The speed v of a particle moving along a straight line, when it is at a distance (x) from a fixed point of the line is given by `v^2=108-9x^2` (all equation are in CGS units):

Options

A. The motion is uniformly accelerated along the straight line
B. the magnitude of the acceleration at a distance 3 cm from the point is `27(cm)/(s^2)`
C. the motion is simple harmonic about the given fixed point
D. the maximum displacement from the fixed point is 4 cm

Answer

B, C

Explanation

`v^2=108-9x^2`
`(2vdx)/(dx)=-18ximpliesA=-9x` (non-uniform)
at `x=3cm`
`a=-27` or `|a|=27(cm)/(s^2)`
also `a=-9x` is a SHM equation so particle perform performs SHM about the give fixed point.
V is maximum at `x=0`
and V is zero at `x=sqrt(12)`
So amplitude `=2sqrt3`