For a simple pendulum of length L in a uniform gravitational field g, which expression gives its frequency?

For a simple pendulum of length L in a uniform gravitational field g, which expression gives its frequency?

• A. f = (1/(2π)) * sqrt(g/L)
• B. T = 2π sqrt(L/g)
• C. f = sqrt(g/L) / (2π)
• D. T = (2π) * sqrt(L/g)

Answer: (A)

A simple pendulum behaves like simple harmonic motion when the swing angle is small. The equation of motion for the angle θ becomes θ'' + (g/L) θ ≈ 0, which gives an angular frequency ω = sqrt(g/L). The frequency is f = ω/(2π), so f = (1/2π) sqrt(g/L). This shows that opening up the motion to SHM directly leads to the familiar result that frequency grows with the square root of gravity and decreases with the square root of length. For reference, the period is T = 2π sqrt(L/g), since f = 1/T.

A simple pendulum behaves like simple harmonic motion when the swing angle is small. The equation of motion for the angle θ becomes θ'' + (g/L) θ ≈ 0, which gives an angular frequency ω = sqrt(g/L). The frequency is f = ω/(2π), so f = (1/2π) sqrt(g/L). This shows that opening up the motion to SHM directly leads to the familiar result that frequency grows with the square root of gravity and decreases with the square root of length. For reference, the period is T = 2π sqrt(L/g), since f = 1/T.