engineering physics MCQ #4605

The frequency of a simple pendulum on the surface of the earth and at height h above it are related as

engineering physics MCQ #4605

  1. Question 1

    Q1. The frequency of a simple pendulum on the surface of the earth and at height h above it are related as

    • A) f_h = f₀ √(R / (R + h))
    • B) f_h = f₀ √(R + h) / R
    • C) f_h = f₀ (R / (R + h))
    • D) f_h = f₀ (R + h) / R

    Answer: f_h = f₀ √(R / (R + h))

    Explanation: Frequency f ∝ √g, and g at height h is g_h = g₀ (R / (R + h))², so f_h = f₀ √(R / (R + h)).