How does the period T of a simple pendulum depend on length L, assuming small oscillations?

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Multiple Choice

How does the period T of a simple pendulum depend on length L, assuming small oscillations?

Explanation:
The key idea is that for small swings a pendulum behaves like a simple harmonic oscillator, so its period depends on length through a square root. Using sin θ ≈ θ for small angles, the equation of motion becomes θ'' + (g/L) θ = 0, giving the angular frequency ω = sqrt(g/L). The period is T = 2π/ω = 2π sqrt(L/g). This shows T grows with the square root of the length: if you lengthen the pendulum, the period increases, but only as the square root of that length. Options that suggest T decreases with length, or grows linearly with length, or is independent of length, don’t match the physics. The correct relationship is T ∝ √L.

The key idea is that for small swings a pendulum behaves like a simple harmonic oscillator, so its period depends on length through a square root. Using sin θ ≈ θ for small angles, the equation of motion becomes θ'' + (g/L) θ = 0, giving the angular frequency ω = sqrt(g/L). The period is T = 2π/ω = 2π sqrt(L/g). This shows T grows with the square root of the length: if you lengthen the pendulum, the period increases, but only as the square root of that length. Options that suggest T decreases with length, or grows linearly with length, or is independent of length, don’t match the physics. The correct relationship is T ∝ √L.

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