Oscillation frequency on the Moon versus on the Earth: For identical small-oscillation systems (e.g., a simple pendulum), how does the frequency of oscillation on the Moon compare with that on the Earth?

Introduction / Context:
Natural frequency of many gravity-controlled oscillators (like a simple pendulum) depends on the local gravitational acceleration g. Comparing environments with different g values, such as Earth and Moon, illustrates how gravity influences timing devices and dynamic response.

Given Data / Assumptions:

• For a simple pendulum of length L, small-angle frequency f = (1 / 2π) * sqrt(g / L).
• Approximate lunar gravity g_moon ≈ g_earth / 6.
• Length L is unchanged; small-angle approximation applies.

Concept / Approach:

Frequency scales with the square root of gravity: f ∝ sqrt(g). Therefore, the ratio f_moon / f_earth = sqrt(g_moon / g_earth) ≈ sqrt(1/6) ≈ 0.408. Thus, the lunar frequency is about 0.41 times the terrestrial value, meaning it is roughly 2.44 times less (the period is 2.44 times longer).

Step-by-Step Solution:

Verification / Alternative check:

Period scales inversely with frequency: T = 1/f. Therefore, T_moon / T_earth = 1 / 0.408 ≈ 2.45, matching the “2.44 times less” frequency statement.

Why Other Options Are Wrong:

(a) and (d) suggest frequency increases on the Moon, which contradicts f ∝ sqrt(g); (c) “3 times less” is an overestimate; (e) ignores the large difference in g.

Common Pitfalls:

Confusing frequency and period; using a linear instead of square-root dependence on g; rounding errors that change the qualitative conclusion.

Final Answer:

2.44 times less