Pendulum striking a wall — restitution and period: 22-Mec-A2 Kinematics and Dynamics of Machines, Undated paper, Q6

Question 6: Pendulum striking a wall — restitution and period (20 marks)

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. Simple pendulum $L=2$ m, $m=1$ kg, released from rest at $\theta_0=10^\circ$; a vertical wall at the equilibrium (bottom) position; COR $e=0.85$; contact time $\tau=5$ ms; $g=9.81$ m/s$^2$.

Find. The oscillation period for the 1st and 2nd cycles (and the supporting impact/amplitude quantities).

Approach. Use the small-angle isochronous period for the free swing; the wall (at the vertical) turns the motion into a sequence of one-sided swings whose period is unchanged by amplitude but includes the brief contact; energy methods give the impact speed, restitution the rebound and amplitude decay.

Fig. 6. The bob is released at $10^\circ$, swings down to the wall at the vertical, rebounds with speed $ev$, and climbs to a slightly smaller angle each cycle.

Free natural period (isochronous). For small amplitude, $$T=2\pi\sqrt{\frac{L}{g}}=2\pi\sqrt{\frac{2}{9.81}}=\boxed{2.837\ \text{s}}.$$ This is independent of amplitude, so the natural period is the same for the 1st and 2nd cycles — the wall reduces the swing amplitude, not the period.
Effect of the wall on the cycle. With the wall at the vertical, the bob swings only on one side. The time between successive impacts is a half free-period plus the contact time: $$T_{\text{cycle}}=\tfrac{1}{2}T+\tau=1.418+0.005=\boxed{1.423\ \text{s}}\quad(\text{1st}= \text{2nd}).$$
Impact speed and rebound. Energy from release to the bottom: $v_0=\sqrt{2gL(1-\cos\theta_0)}=\sqrt{2(9.81)(2)(1-\cos10^\circ)}=0.772$ m/s. Restitution: $v_0'=e\,v_0=0.85(0.772)=0.656$ m/s.
Amplitude decay. Each rebound climbs to $\cos\theta_{k}=1-\dfrac{v_k'^2}{2gL}$, giving $\theta_0=10^\circ\to\theta_1=8.50^\circ\to\theta_2=7.22^\circ$ (amplitude scales roughly as $e$ per impact).
Impulsive wall force. The momentum reverses, so incoming and rebound speeds add: $$\bar F=\frac{m(v_0+v_0')}{\tau}=\frac{1(0.772+0.656)}{0.005}=\boxed{285.7\ \text{N}}.$$

Quantity	Value
Free natural period $T$	2.837 s (1st = 2nd, isochronous)
Constrained cycle period (wall + contact)	1.423 s
Impact speed / rebound	0.772 / 0.656 m/s
Amplitude: $\theta_0\to\theta_1\to\theta_2$	10° → 8.50° → 7.22°
Average impulsive force	285.7 N

Topic. Isochronism of the simple pendulum, coefficient of restitution, and impulse–momentum at an impact.

Key relations. $T=2\pi\sqrt{L/g}$ (amplitude-independent, small angle); energy $\tfrac12mv^2=mgL(1-\cos\theta)$; restitution $v'=ev$; impulse $\bar F\,\tau=m(v+v')$.

Why this works. For small angles the restoring torque is linear in $\theta$, so the motion is simple-harmonic with a period set only by $L$ and $g$ — the amplitude (and hence the wall’s energy removal) does not change it. Restitution removes a fixed fraction of speed at each contact, so amplitude decays geometrically while the period holds.

Common pitfall. Using $mv'/\tau$ for the wall force (forgetting the momentum reverses, so $v$ and $v'$ add), and expecting the period to shrink with amplitude (it does not, to first order).

Source. Rao, Mechanical Vibrations, §1.10 (pendulum, isochronism) and Hibbeler, Dynamics, §15.4 (coefficient of restitution).

Question 6 of 7: Pendulum striking a wall — restitution and period

Question 6: Pendulum striking a wall — restitution and period (20 marks)