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 Incoming links: Harmonic forcing of a horizontally driven pendulum

Past exam of the physics 1 paper of the University of Oxford / 2015 / 10 / b / Solution 2026-10-25

The Euler-Lagrange equation for

θ

is

l \ddot{θ} + \overset{x}{¨} cos θ + g sin θ = 0.

For small angles, set

cos θ ≃ 1

and

sin θ ≃ θ

. With

x = A cos ω t

and

ω_{0}^{2} = g / l

, the harmonic forcing of a horizontally driven pendulum obeys

\ddot{θ} + ω_{0}^{2} θ = \frac{A ω ^{2}}{l} cos ω t .

For

ω \neq = ω_{0}

, its general solution is

θ (t) = C cos ω_{0} t + D sin ω_{0} t + \frac{A ω ^{2}}{l ( ω _{0}^{2} - ω ^{2} )} cos ω t .

C, D

set the initial angle and angular velocity. If the bob starts vertical with zero angular velocity at

t = 0

, this reduces to

θ (t) = \frac{A ω ^{2}}{l ( ω _{0}^{2} - ω ^{2} )} (cos ω t - cos ω_{0} t) .

The solution remains an approximation only while the resulting angle is small.

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