 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2015/10/b/solution

θ

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

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

For

ω \neq = ω_{0}

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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