A particle move along y-axis according to equation `y =3+4 cos omega t`. The motion of the particle is

Explanation

The given force does not correspond to the standard equation of the SHM, that is `F= -kx`. But still the relation between force and displacement is linear. The equation can be written in a standard format by just shifting the origin.
Calculation: We can write the given force as
`F= - k (y-(C )/(k))`
Let us define
`s=y -(C )/(k)`
where s is the displacement from mean position because at y= C/k, F is zero.
Thus, the mean position is at y= C/k.
Using Eq. 15.28, force can be written as
`F= -ks`
which is the equation of SHM about y= C/k.
Differntiating Eq. 15.28 with respect to time twice, we get
`(d^(2)y)/(dt^(2)) = (d^(2)s)/(dt^(2))`
Thus, force `F= m(d^(2)y//dt^(2))` can also be written as `F= m(d^(2)s//dt^(2))`. Substituting in Eq.15.29
`m(d^(2)s)/(dt^(2)) = - ks`
Comparing with standard equation we know that the solution of the above SHM equation is
`s= y_(m) sin (omega t+ phi)`
where `omega= sqrt((k)/(m))`
Replacing with value of s from Eq.15.28, we get
`y= y_(m) sin (omega t+ phi) + (C )/(k)`