Two springs, each of unstretched length 20 cm but having different spring constant `k_1=1000(N)/(m)` and `k_2=3000(N)/(m)` are attached to two opposite faces of a small block of mass `m=100g` kept on a smooth horizontal surface as shown in fig. The outer ends of the two springs are now attached to two pins `P_1` and `P_2` whose locations are shown in the figure. as a result of this, the block acquires a new equilibrium position. The block has been displaced by small amount from its equilibrium position and released to perform simple harmonic motion: then

Options

A. new equilibrium position is at 35 cm from `P_1` and time period of simple harmonic motion is `(pi)/(100)`s.
B. New equilibrium position is at 20 cm from `P_1` and time period of simple harmonic motion is `(pi)/(100)s`
C. new equilibrium position is at 35 cm from `P_1` and time period of simple harmonic motion is `(pi)/(25)s`
D. new equilibrium position is at 30 cm from `P_1` and time period of simple harmonic motion is `(pi)/(26)`s

Answer

A

Explanation

In equilibrium position, net force acting on the object (block of mass m) is zero. Let spring of spring constant `k_1` is stretched by `x_1` and spring of spring constant `k_2` is streched by `x_2`, then free body diagram of the block is as shown in Fig.
Now, `x_1+x_2=20cm` and `k_1x_1=k_2x_2`
`x_1=15cm` and `x_5cm`
So, new equilibrium position is at `x=(20+15)cm` from `P_1`, and time period of oscillation of the block is given by
`T=2pisqrt((m)/(k_1+k_2))=2pisqrt((0.1)/(4000))=(pi)/(100)s`