Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in the figure. The time period of oscillation is

Explanation

Let mass m gets displaced by a small distance x to the left side of initial position. As a result, the left spring get compressed by length x and right spring gets elongated by the same length x.

`therefore` Force exerted by the left spring
`F_(1) = Kx`, towards right
Force exerted by the right spring.
`F_(2) = -kx`, towards right
`therefore` total force of mass m, `F = F_(1) + F_(2) = -2Kx`
Since the force acting on the mass m is proportional to displacement x and is directed towards its mean position (negative sign), the motion of mass m is S.H.M.
Force constant of combination, K = 2K
`therefore` Time period of oscillation
`therefore T = 2pi sqrt((m)/(k.)) = 2pisqrt((m)/(2K))`