A block of mass 2 kg is attached with two identical springs of spring constant 20 N m$$^{-1}$$ each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is $$\frac{\pi}{\sqrt{X}}$$ in SI unit. The value of $$X$$ is _____.

Solution & Explanation

1. Understand the Spring Configuration

A block of mass $$m = 2 \,\, \text{kg}$$ is connected between two identical horizontal springs, each having a spring constant $$k = 20 \,\, \text{N/m}$$. The outer ends of the springs are anchored to rigid vertical walls.

When the mass is displaced to one side by a distance $$x$$:

• One spring is compressed by a distance $$x$$, exerting a restoring force of $$F_1 = -k \cdot x$$ directed back toward the equilibrium center.
• The other spring is stretched by the exact same distance $$x$$, pulling the block with a restoring force of $$F_2 = -k \cdot x$$ in the same direction.

Because both restoring forces pull the mass in the same direction to restore equilibrium, the total net force is:

$$F_{\text{net}} = F_1 + F_2 = -kx - kx = -2kx$$

This additive restoring behavior means the two springs are structurally operating in a parallel combination.

2. Calculate the Effective Spring Constant ($$k_{\text{eq}}$$)

For springs connected in parallel, their individual stiffness constants add together directly to create a stiffer system:

$$k_{\text{eq}} = k + k$$

$$k_{\text{eq}} = 20 \,\, \text{N/m} + 20 \,\, \text{N/m} = 40 \,\, \text{N/m}$$

3. Determine the Time Period of Oscillation ($$T$$)

The standard formula for the time period of a spring-mass system executing Simple Harmonic Motion (SHM) is:

$$T = 2\pi \sqrt{\frac{m}{k_{\text{eq}}}}$$

Substitute the mass ($$m = 2 \,\, \text{kg}$$) and the effective spring constant ($$k_{\text{eq}} = 40 \,\, \text{N/m}$$) into the formula:

$$T = 2\pi \sqrt{\frac{2}{40}} = 2\pi \sqrt{\frac{1}{20}}$$

Simplify the expression under the radical sign:

$$T = 2\pi \cdot \frac{1}{\sqrt{4 \times 5}} = 2\pi \cdot \frac{1}{2\sqrt{5}}$$

Canceling out the factor of 2 from both the numerator and denominator yields:

$$T = \frac{\pi}{\sqrt{5}}$$

4. Find the Value of $$X$$

We are given that the time period of oscillation is expressionally equal to $$\frac{\pi}{\sqrt{X}}$$ in SI units:

$$\frac{\pi}{\sqrt{X}} = \frac{\pi}{\sqrt{5}} \implies X = 5$$

Correct Numerical Answer: 5