Paper 1 Oscillations and SHM Answers

Q	Ans	Brief reason
1	C	Period is time per cycle.
2	B	$T=1/f=0.40\mathrm{s}$.
3	B	$\omega =2\pi f$.
4	B	SHM requires restoring acceleration proportional to displacement.
5	D	$a=-{\omega }^{2}x$.
6	B	Turning point speed is zero.
7	A	$x=0$, so $a=0$.
8	B	$v_{\max }=\omega A$.
9	A	Standard spring-mass result.
10	C	$T=2\pi \sqrt{L/g}$ for small angles.
11	B	Energy is dissipated.
12	A	Resonance is identified from amplitude-frequency response.
13	A	Damping lowers and broadens the peak.
14	B	$E=\frac{1}{2}m{\omega }^2{A}^2$.
15	B	$\Delta \varphi =2\pi (T/4)/T=\pi /2$.
16	C	Antiphase means half a cycle apart.
17	C	Acceleration points toward equilibrium.
18	B	Speed is zero; energy is stored as potential energy.
19	A	Ideal SHM is sinusoidal.
20	B	Natural frequency is the free-oscillation frequency.