A light pulley is suspended at the lower end of a spring of constant `k_(1)`, as shown in figure. An inextensible string passes over the pulley. At one end of string a mass `m` is suspended, the other end of the string is attached to another spring of constant `k_(2)`. The other ends of both the springs are attached to rigid supports, as shown. Neglecting masses of springs and any friction, find the time period of small oscillations of mass `m` about equilibrium position.

Answer

A, B, D

Explanation

Let `F` be the extra tension in the string, when the block is displaced `x` from its mean position.
Extention in spring `- 2` is
`x_(2) = (F)/(k_(2))`
Extension is spring `- 1` is
`x_(1) = (2F)/(k_(1))`
`x = 2x_(1) + x_(2) = (4F)/(k_(1) + (F)/(k_(2))`
Extra tension `F` will become restoring force for the block. Therefore, above equation can be written as,
`F = - ((1)/((4)/(k_(1)) + (1)/(k_(2)))) x = ((k_(1)k_(2))/(4k_(2) + k_(1)))x`
or `k_(e) = (k_(1)k_(2))/(4k_(2) + k_(1))`
`T = 2pi sqrt((m)/(k_(e)))`
`= 2pi sqrt((m(4k_(2) + k_(1)))/(k_(1)k_(2))`