What is the length of a simple pendulum, which has a frequency of 0.5

What is the length of a simple pendulum, which has a frequency of 0.5 Hz (take g = 10 m/s²)?

[amp_mcq option1=”1 m” option2=”2 m” option3=”1.5 m” option4=”3 m” correct=”option1″]

This question was previously asked in

UPSC CISF-AC-EXE – 2024

The correct answer is A) 1 m.

The frequency (f) of a simple pendulum is related to its length (L) and the acceleration due to gravity (g) by the formula: f = 1 / (2π√(L/g)). Rearranging this formula to solve for L gives L = g / (4π²f²). Given f = 0.5 Hz and g = 10 m/s², and using the common approximation π² ≈ 10 when g ≈ 9.8 or 10 m/s², we get L = 10 / (4 * 10 * (0.5)²) = 10 / (40 * 0.25) = 10 / 10 = 1 meter.

The period (T) is the reciprocal of the frequency (T = 1/f = 1/0.5 = 2 seconds). The period formula is T = 2π√(L/g). Squaring both sides gives T² = 4π²(L/g), so L = gT² / (4π²). Using T=2s, g=10m/s², and π²≈10: L = 10 * (2)² / (4 * 10) = 10 * 4 / 40 = 40 / 40 = 1 meter. This confirms the result.

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