Question

Two identical springs of spring constant '2k' are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is: 

 

• A. $2\pi\sqrt{\frac{m}{2k}}$
• B. $2\pi\sqrt{\frac{m}{k}}$
• C. $\pi\sqrt{\frac{m}{k}}$
• D. $\pi\sqrt{\frac{m}{2k}}$

Hint:

When two springs are connected such that a displacement causes one to stretch and the other to compress while both exert forces in the same direction, they are considered to be in parallel. The effective spring constant for parallel springs is $k_{eff} = k_1 + k_2$.

Answer 1

The Correct Option is C

When the mass 'm' is displaced by a distance 'x' to the right from its equilibrium position, the left spring is stretched by 'x' and the right spring is compressed by 'x'.
The restoring force from the left spring is $F_1 = -(2k)x$ (pointing left).
The restoring force from the right spring is $F_2 = -(2k)x$ (also pointing left).
The total restoring force on the mass is the sum of these two forces:
$F_{total} = F_1 + F_2 = -2kx - 2kx = -4kx$.
The total force is of the form $F = -k_{eff}x$, where the effective spring constant is $k_{eff} = 4k$.
This arrangement is equivalent to two springs connected in parallel.
The time period 'T' of a simple harmonic oscillator is given by the formula $T = 2\pi\sqrt{\frac{m}{k_{eff}}}$.
Substituting the value of $k_{eff}$:
$T = 2\pi\sqrt{\frac{m}{4k}} = 2\pi\frac{\sqrt{m}}{2\sqrt{k}} = \pi\sqrt{\frac{m}{k}}$.