A particle of mass 5 kg is placed in a field of gravitational potential
`v=(7x^(2)-21x)J//kg` . Then its motion

Options

A. is SHM with angular frequency `1.67 rad//s`
B. is SHM with angular frequency `3.74 rad//s`
C. Is oscillarotry but no SHM
D. is SHM with amplitude 5.5m

Answer

B

Explanation

Potential energy, `U=mV=5(7x^(2)-21x)`
`=(35x^(2)-105x)`
`F=(dU)/(dx)=-(70x-105x)`
`=-70x+105` …(i)
As `F prop x` and this force `F` is directed towards mean position of particle, hence if the particle is left free, it will execute linear SHM.
`:. ` Spring factor, `k=70N//m`
Angular frequency,
`omega=sqrt((k)/(m))=sqrt((70)/(5))=3.74rad//s`
Without knowing the energy, we cannot find its amplitude.