Compound pendulum — small-angle SHM relation for frequency: If a compound pendulum undergoes small oscillations with angular displacement θ and angular acceleration α proportional to θ (i.e., α = −(g/ℓ) θ), what is the frequency f of its motion expressed in terms of α and θ?

Introduction / Context:
The compound pendulum (physical pendulum) exhibits simple harmonic motion for small angular displacements. The relationship between the restoring angular acceleration and angular displacement determines the natural frequency. Expressing frequency directly in terms of the proportionality α/θ reinforces the SHM analogy to linear systems.

Given Data / Assumptions:

• Small-angle oscillations so that sin θ ≈ θ (in radians).
• Angular acceleration α is proportional to −θ: α = −(g/ℓ) θ.
• g is acceleration due to gravity; ℓ is the equivalent pendulum length.

Concept / Approach:

For SHM, the governing rotational equation is θ¨ + (g/ℓ) θ = 0. Angular frequency ω satisfies ω^2 = g/ℓ. Because α/θ = −g/ℓ (magnitude g/ℓ), we can write ω = sqrt(α/θ) using magnitudes. The cyclic frequency f relates to angular frequency by f = ω / (2π).

Step-by-Step Solution:

Verification / Alternative check:

From standard formula for a compound pendulum, f = (1 / 2π) * sqrt(g / ℓ). Substituting g/ℓ = α/θ reproduces the same expression.

Why Other Options Are Wrong:

(b) and (c) have incorrect dimensions (frequency cannot be linear in α/θ without a square root). (d) is not dimensionally consistent. (e) inverts the needed relationship and lacks the square root.

Common Pitfalls:

Forgetting the small-angle assumption; ignoring sign (use magnitudes for frequency); confusing angular frequency ω with cyclic frequency f.

Final Answer:

(1 / 2π) * sqrt(α / θ)