Question

A spring of spring constant k is cut into two equal half AB & CD and are attached horizontally at the two ends of a uniform horizontal rod AC of length l and mass m as shown. The rod is pivoted at its centre ‘O’ and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is:

• A. $\frac{1}{2\pi}\sqrt{\frac{k}{m}}$
• B. $\frac{1}{2\pi}\sqrt{\frac{2k}{m}}$
• C. $\frac{1}{2\pi}\sqrt{\frac{3k}{m}}$
• D. $\frac{1}{2\pi}\sqrt{\frac{6k}{m}}$

Answer 1

$\frac{1}{2\pi} \sqrt{\frac{3k}{m}}$

Answer 2

Explanation:

1. Spring Constant of Halves: When a spring of spring constant k is cut into two equal halves, the spring constant of each half doubles. Therefore, each of the two springs attached to the rod has a spring constant k' = 2k.

2. Restoring Torque:

   • When the rod rotates by a small angle θ from its equilibrium position, each end (A and C) moves a distance x = (l/2)θ perpendicular to the rod.
   • The restoring force exerted by each spring is F = k'x = (2k)(l/2)θ = klθ.
   • The torque due to each spring about the pivot O (center of the rod) is τ_spring = F × (l/2) = (klθ)(l/2) = (kl^2/2)θ.
   • Since both springs exert a restoring torque in the same direction, the total restoring torque is τ_total = 2 × τ_spring = 2 × (kl^2/2)θ = kl^2θ.

3. Moment of Inertia: The moment of inertia of a uniform rod of mass m and length l about its center is I = (1/12)ml^2.

4. Equation of Rotational SHM:

   • The equation of motion for rotational SHM is Iα = τ_total, where α = d^2θ/dt^2.
   • Substituting the values: (1/12)ml^2 (d^2θ/dt^2) = -kl^2θ.
   • Rearranging to the standard SHM form: d^2θ/dt^2 = - (kl^2 / ((1/12)ml^2)) θ
   • d^2θ/dt^2 = - (12k/m) θ.

5. Frequency of Oscillation:

   • Comparing the equation with d^2θ/dt^2 = -ω^2θ, we find the angular frequency squared ω^2 = 12k/m.
   • The angular frequency is ω = \sqrt{\frac{12k}{m}}.
   • The frequency of oscillation f is related to the angular frequency by f = ω/(2π).
   • Therefore, f = \frac{1}{2\pi} \sqrt{\frac{12k}{m}}.

   Simplifying the expression inside the square root:

   $$f = \frac{1}{2\pi} \sqrt{\frac{12k}{m}} = \frac{1}{2\pi} \sqrt{\frac{4 \cdot 3k}{m}} = \frac{1}{2\pi} 2\sqrt{\frac{3k}{m}}$$

   However, there seems to be an error in the final simplification in the original explanation. It should be:

   $$f = \frac{\sqrt{3k/m}}{\pi}$$