Question 7 of 8: Simple pendulum striking a wall (impact & free vibration)

Question 7 — Simple pendulum striking a wall (impact & free vibration) (20 marks — Part B)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. Natural period; pre/post‑impact speeds and rebound angles for cycles 1–2; average impact force.

Approach. Energy conservation on the swing gives the bottom speed; the coefficient of restitution reverses and scales it ($v'=ev$); repeat for successive cycles. The average impact force follows from the impulse–momentum theorem over $\Delta t$.

1. Natural period (free vibration). For small oscillations the simple‑pendulum period is $$T=2\pi\sqrt{\frac{L}{g}}=2\pi\sqrt{\frac{1}{9.81}}=\boxed{2.006\ \text{s}}\quad(f=0.499\ \text{Hz}).$$ (For a large first swing of $\theta_0=90^\circ$ the amplitude‑corrected period is $\approx2.35$ s; the small‑angle value is the nominal natural period.)
2. Speed at the wall (cycle 1). Energy conservation from $\theta_0$ to the bottom: $$v_1=\sqrt{2gL(1-\cos\theta_0)}=\sqrt{2(9.81)(1)(1-\cos90^\circ)}=\boxed{4.43\ \text{m/s}}.$$
3. Rebound (cycle 1). Restitution reverses and scales the speed: $v_1'=e\,v_1=0.9(4.43)=\boxed{3.99\ \text{m/s}}$. It climbs to $$\cos\theta_1=1-\frac{v_1'^2}{2gL}=1-e^2(1-\cos\theta_0)=1-0.81=0.19\ \Rightarrow\ \theta_1=\boxed{79.1^\circ}.$$
4. Impact force (cycle 1). The momentum reverses from $-mv_1$ to $+mv_1'$, so the impulse is $J_1=m(v_1+v_1')=4(4.43+3.99)=33.7\ \text{N}\cdot\text{s}$, and over $\Delta t=0.010$ s the average force is $$F_{\text{avg},1}=\frac{J_1}{\Delta t}=\frac{33.7}{0.010}=\boxed{3366\ \text{N}}.$$ Kinetic energy lost this impact: $\tfrac12 m v_1^2(1-e^2)=\tfrac12(4)(4.43^2)(0.19)=\boxed{7.46\ \text{J}}.$
5. Cycle 2. The bob returns from $\theta_1$ to the bottom with the same speed $v_2=v_1'=3.99$ m/s (energy conserved on the free swing), strikes again to $v_2'=e\,v_2=e^2 v_1=3.59$ m/s, rebounding to $\cos\theta_2=1-e^4(1-\cos\theta_0)=0.344\Rightarrow\theta_2=\boxed{69.9^\circ}$, with $F_{\text{avg},2}=m(v_2+v_2')/\Delta t=\boxed{3030\ \text{N}}$. Each cycle the amplitude decays geometrically ($\cos\theta$ climbs as $1-e^{2k}(1-\cos\theta_0)$) and the impact force scales by $e$.

Computed for the assumed release $\theta_0=90^\circ$; results scale with $(1-\cos\theta_0)$ — substitute the actual $\theta_0$ if it is known.