The displacement time relation for a particle can be expressed as `y=0.5[cos^2(npit)-sin^2(npit)]` This relation shows that

Options

A. the particle executing a SHM with amplitude `0.5`m
B. the particle is executing a SHM with frequency `n` time that of a second's pedulum
C. the particle is executing a SHM and the velocity in its mean position is `(npi)(m)/(s)`
D. the paritcle is not executing a SHM at all

Answer

A, C

Explanation

`y=0.5[cos^2(npit)-sin^2(npit)]=0.5cos2npit`
`(dy)/(dt)=-0.5xx2npixxsin2npit`
`(d^2y)/(dt^2)=-(0.5)(2npi)^2cos2npit=-4n^2pi^2t` .(i)
i.e, `(d^2y)/(dt^2)prop-y`
i.e., particle is executing SHM with amplitude `0.5`m i.e., choice (a) is correct
As standard equation of SHM is
`(d^2y)/(dt^2)=-omega^2y`
Hence `omega=2npi`
Force second's pendulum `T=2s`
Hence `omega'=(2pi)/(T)=pi=2n`
i.e, Choice (b) is not correct .
`((dy)/(dt))_(max)=npi`
i.e., Choice (c ) is not correct.