Paper 3 Oscillations and SHM Answers

Q1 Spring-mass oscillator

1. Spring force is $F=-kx$ relative to equilibrium. Newton’s second law gives $ma=-kx$, so $a=-(k/m)x$. Acceleration is proportional to displacement and towards equilibrium, so the motion is SHM. [3]
2. $\omega =\sqrt{k/m}=\sqrt{64.0/0.250}=16.0\mathrm{rad}{\mathrm{s}}^{-1}$. $T=2\pi /\omega =0.393\mathrm{s}$. [3]
3. $v_{\max }=\omega A=16.0(0.045)=0.720\mathrm{m}{\mathrm{s}}^{-1}$. [2]
4. $E=\frac{1}{2}k{A}^2=\frac{1}{2}(64.0)(0.045{)}^2=0.0648\mathrm{J}$. [2]
5. Elastic potential energy decreases as the spring returns towards equilibrium. Kinetic energy increases, becoming maximum at equilibrium, while total mechanical energy remains constant if losses are negligible. [2]

Q2 Pendulum and approximation

1. $T=2\pi \sqrt{L/g}=2\pi \sqrt{0.720/9.81}=1.70\mathrm{s}$. [2]
2. In the ideal small-angle derivation, gravitational force and inertia are both proportional to bob mass, so mass cancels from the equation of motion. [2]
3. Small angular displacement; light inextensible string; point-like bob; negligible air resistance/friction; uniform $g$. Any two. [2]
4. The restoring component is proportional to $\sin \theta$, not exactly $\theta$. For larger angles, $\sin \theta \approx \theta$ is no longer accurate, so the motion is not exactly SHM and the period changes. [2]
5. Time many complete oscillations and divide by the number; use a fiducial marker; repeat and average; start/stop timing at the same phase point. [2]