The ratio of the time periods of small oscillation of the insulated spring and mass system before and after charging the masses is

Options

A. `ge 1`
B. ` gt 1`
C. `le 1`
D. `=1`

Explanation

In equilibrium of the charged small bodies `(1)/(4piepsilon_(0))(q^(2))/((l_(0)+x_(0))^(2))=kx_(0)` where `x_(0)` is the elongation in the spring in equilibrium.
Let a further small elongation of `x` is given to the spring.
Then net restoring force on any of the charged particle is given by,
`F=-[k(x_(0)+x)-(1)/(4piepsilon_(0))(q^(2))/((l_(0)+x_(0)+x)^(2))]=-kx`. Since `x lt lt x_(0)`
`rArr a=-(2k)/(m)x` As `F=mua`, where reduced mass `mu=(mxxm)/(m+m)=(m)/(2)`
`rArr a=-omega^(2)x`,
Hence `omega=sqrt((2k)/(m))rArrT=2pisqrt((m)/(2k))`
In absence of charge, `T_(0)=2pisqrt((m)/(2k))`
Therefore `(T)/(T_(0))=1`
`:. (D)`