A particle of mass 2 kg moving along x-axis has potential energy given by, `U=16x^(2)-32x` ( in joule), where x is in metre. Its speed when passing through `x=1 is 2ms^(-1)`

Options

A. the motion of particle is a uniformly acceleration motion
B. the motion of particle is an oscillatory from `x=1.5m to 3.0m`
C. the motion of particle is not a simple harmonic motion
D. the motion of oscillatory motion is `pi//2 ` s

Answer

D

Explanation

Given, `U=16x^(2)-32x,`
`F=(dU)/(dx)=-(d)/(dx)(16x^(2)-32x)=-32x+32`
`:. k=32N//m,`
If means `x=1m` is the mean position . At mean position, velcoity is maximum. So `romega =2`
or `rsqrt((k)/(m))=2`
or `r=2sqrt((m)/(k))=2sqrt((2)/(32))=(1)/(2)m =0.5m`
Time period, `T=2pisqrt((m)/(k))=2pisqrt((2)/(32))=(pi)/(2)s`
Since the mean position is at `x=1m` and amplitude is 0.5m, hence, particle will oscillate betweent `(1+0.5)m` and `(1-0.5)m` `[i.e., 1.5m` and `0.5m]`