Question

In a free space, a thin rod carrying uniformly distributed negative charge `-q` is placed symmetrically along the axis of a tin ring of radius R varrying uniformly dostributed charge Q. The mass of the rod is m and length is `l=2R`. The ring is fixed and the rod is free to move. The rod is displaced slightly along the axis of the ring and then released. Find the period T to the small amplitude oscillations of the rod.

Answer 1

If we displace the rod by length upward the unbalaced force will be due to length 2 dy. Hence the unbalanced force on the element is
`dF=(dq)E`
`=(lambda 2dy)(1)/(4 pi epsilon_(0))(Qq)/((R^(2)+R^(2))^(3//2))`
`a = (1)/(4pi epsilon_(0)) (Qq)/(2sqrt(2)R^(3)m)dy`
Hence,
`omega^(2)=(1)/(4pi epsilon_(0)) (Qq)/(2sqrt(2)R^(3)m)`
or `(2 pi)/(T) =((1)/(4 pi epsilon_(0))(Qq)/(2 sqrt(2)R^(3)m))^(1//2)`
or `T=2 pi((4 pi epsilon_(0)2 sqrt(2)R^(3)m)/(Qq))^(1//2)=4 pi R((2 sqrt(2)pi epsilon_(0)Rm)/(Qq))^(1//2)`