Determine the period of small longitudinal oscillations of a body with mass `m` in the system shown in figure. The stiffness values of the springs are `x_(1)` and `x_(2)`. The friction and the masses of the springs are negligible.

Explanation

The net unbalanced force on the block `m` when it is at a small horiozontal distance `x` from the equilibrium position becomes `(k_(1)+k_(2))x`
From `F_(x)=mw_(x)` for the block `:`
`-(k_(1)+k_(2))x=m ddot(x)`
Thus` ddot (x)=-((k_(1)+k_(2))/(m))x`
Hence the sought time period `T=2pi sqrt((m)/(k_(1)+k_(2)))`
Alternate `:` Let us set the block `m` in motion to perform small oscillation . Let us locate the block when it is at a distance `x` from its equilibrium position.
As the spring force is restoring conservative force and deformation of both the springs are same, so from the conservation of mechanical energy of the spring `-` block system `:` `(1)/(2)m ((dx)/(dt))^(2)+(1)/(2) k_(1)x^(2)+(1)/(2)k_(2)x^(2)=` Constant
Differentiating with respect to time
`(1)/(2) m 2 ddot(x)ddot(x)+(1)/(2)(k_(1)+k_(2))2xdot(x)=0`
`ddot(x)=-((k_(1)+k_(2)))/(m)x`
Hence the sought time period `T=2pisqrt((m)/(k_(1)+k_(2)))`