Compound Pendulum – Frequency of Oscillation Given a compound pendulum with radius of gyration k_G about its centroidal axis and distance h between the point of suspension and the center of gravity (C.G.), what is the frequency of small oscillations f?

Introduction / Context: A compound (physical) pendulum is any rigid body oscillating about a horizontal axis not passing through its center of mass. Its period and frequency depend on geometry and mass distribution via the radius of gyration and the suspension distance.

Given Data / Assumptions:

• Small angular oscillations (small-angle approximation).
• Radius of gyration about C.G.: k_G.
• Distance between suspension point and C.G.: h.
• Acceleration due to gravity: g.

Concept / Approach: The time period T of a compound pendulum is T = 2π * sqrt( (k_G^2 + h^2) / (g * h) ). Frequency is the reciprocal: f = 1 / T. Substituting gives f = (1 / 2π) * sqrt( g * h / (k_G^2 + h^2) ).

Step-by-Step Solution: Start with T = 2π * sqrt( (k_G^2 + h^2) / (g * h) ). Invert to get f = 1 / T. Thus f = (1 / 2π) * sqrt( g * h / (k_G^2 + h^2) ).

Verification / Alternative check: Special case: if k_G → 0 (point mass at distance h), f reduces to (1 / 2π) * sqrt( g / h ), the simple pendulum result with length L = h. This matches known theory.

Why Other Options Are Wrong: Option B swaps numerator and denominator, giving period instead of frequency form. Options C and D ignore k_G and only apply to a simple pendulum. Option E has incorrect dependence on k_G and h.

Common Pitfalls: Confusing radius of gyration with geometric length. Forgetting that frequency is 1 / T, not 2π / T.

Final Answer: f = (1 / 2π) * sqrt( g * h / (k_G^2 + h^2) )