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` or `v^2=9(12-x^2)`
We can compare the above expression with `v=omegasqrt(A^2+x^2)`, which is the expression of velocity of SHM.
From this we will get
`omega=3` and `A=sqrt(12)`
SHM is not a uniformly accelerated motion. Acceleration at a distance 3 cm from the mean position,
`a=omega^2(3cm)=27(cm)/(s^2)`
Maximum displacement from the mean position `=A=sqrt(12)cm`