Simple harmonic motion (free, damped, forced)

At a Glance

• Frequency: 8 sub-parts across 8 of 13 years (2013, 2014, 2015, 2018, 2019, 2023, 2024, 2025)
• Priority tier: T1
• Marks (count): 10 (4), 12 (1), 15 (2), 20 (1)
• Average solve time: ~11 min
• Difficulty mix: medium 3, easy 3, hard 2
• Section: B | Dominant type: derivation

Why This Chapter Matters

SHM has appeared in 8 of the last 13 years with a consistently high mark ceiling — half the questions are 12 marks or above. The questions decompose into four templates: extract amplitude/period/frequency from given velocity-position or velocity-acceleration data; find the time to return to a given point; analyse two-phase motion when an elastic string alternates between taut (SHM) and slack (free flight); or bound the period of a nonlinear oscillator. The first two templates are quick and reliable marks; the two-phase and period-bounds templates are the hard 15/20-mark questions and require careful phase-matching. Mastering the velocity formula $v^2 = \omega^2(A^2-x^2)$ and the parametric form $x=A\sin(\omega t+\phi)$ unlocks all four.

Minimum Theory

Equation and solutions. SHM about a centre $O$ is defined by $\ddot x = -\omega^2 x$ (acceleration proportional to displacement, directed toward $O$). The general solution is $x(t)=A\sin(\omega t+\phi)$, where $A$ is the amplitude and $\omega=2\pi/T$ the angular frequency. The velocity formula obtained by first integration (energy integral) is $v^2=\omega^2(A^2-x^2).$ At the extremes $x=\pm A$: $v=0$ (momentarily at rest). At the centre $x=0$: $v=A\omega$ (maximum speed). Acceleration: $a=-\omega^2x$, so $|a|$ is maximum at the extremes ($|a|_{\max}=\omega^2 A$) and zero at the centre.

Extracting the period. Two velocity-position pairs $(v_1,x_1)$ and $(v_2,x_2)$ immediately give $\omega$ by subtracting the velocity formula: $v_1^2-v_2^2=\omega^2(x_2^2-x_1^2)\;\Longrightarrow\;\omega^2=\frac{v_1^2-v_2^2}{x_2^2-x_1^2},\quad T=2\pi\sqrt{\frac{x_2^2-x_1^2}{v_1^2-v_2^2}}.$ One velocity-acceleration pair $(v,a)$ at displacement $x$ gives: $\omega^2=-a/x$, $A^2=x^2+v^2/\omega^2$.

Time to return to a point. In the parametric form, $x(t)=A\sin(\omega t+\phi_0)$ with $\phi_0$ determined by initial position and direction. The next time the particle passes through $x=b$ (while moving outward from $O$: $\phi_0=\arcsin(b/A)$ with $\cos\phi_0>0$) is when $\omega t + \phi_0 = \pi-\phi_0$, i.e. $t = (\pi-2\phi_0)/\omega$. Using $\pi/2-\arcsin(b/A)=\arccos(b/A)$: $t=(T/\pi)\arccos(b/A)$. Similarly from $\phi_0=\arctan(p\omega/v)$: $t=(T/\pi)\arctan(v/(p\omega))$.

Elastic string SHM. For a particle on an elastic string (natural length $\ell$, modulus $\lambda$, mass $m$), when the extension is positive the equation of motion gives SHM about the equilibrium length $\ell_0 = \ell + mg\ell/\lambda$, with $\omega^2=\lambda/(m\ell)$. When the string goes slack, the motion switches to free projectile under gravity alone.

Question Archetypes

Four patterns cover every SHM question in the corpus.

shm-parameters (4 question(s); 2013, 2015, 2018, 2024)

Recognition Cues

• “Find the time of one complete oscillation” from data about velocity at specific displacements.
• “Find the new amplitude” when velocity is instantaneously changed but position is unchanged.
• Two $(v_i, x_i)$ pairs are given — subtract the velocity formula to eliminate $A$ and find $\omega$.
• A velocity-acceleration pair is given — use $\omega^2 = |a|/|x|$ and $v^2+a^2/\omega^2 = A^2\omega^2$.

Solution Template

From two extreme positions and mid-velocity:

1. Identify the extremes (zero-velocity points) $x_1,x_2$; amplitude $A=|x_2-x_1|/2$, centre at the midpoint.
2. At the midpoint (centre), velocity is maximised: $v_{\max}=A\omega$.
3. $\omega=v_{\max}/A$, $T=2\pi/\omega$.

From two velocity-position pairs:

1. Write $v_1^2=\omega^2(A^2-x_1^2)$ and $v_2^2=\omega^2(A^2-x_2^2)$.
2. Subtract: $v_1^2-v_2^2=\omega^2(x_2^2-x_1^2)$; solve for $\omega^2$ (then $T$).
3. If $A$ is needed: add back to find $A^2$.

From one velocity-acceleration pair:

1. From $a=-\omega^2 x$: $\omega^2=|a|/|x|$ (but need sign — $a$ and $x$ must have opposite signs).
2. From two states: subtract $u^2-v^2=(f_2^2-f_1^2)/\omega^2$ to find $\omega^2$.

Worked Example(s)

2013 Paper 1, 2013-P1-Q5c (10 marks)

SHM in line $OPQ$; velocity zero at $P$ (distance $x$ from $O$) and $Q$ (distance $y$); velocity $v$ at midpoint. Find period.

$P,Q$ are the extremes; midpoint $M$ is the centre. Amplitude $A=|y-x|/2$. At $M$: $v=A\omega=(|y-x|/2)\omega$. Hence $\omega=2v/|y-x|$ and: $\boxed{\;T=\frac{\pi|y-x|}{v}.\;}$

Note: the distances $x,y$ from $O$ fix $P,Q$; only their difference matters. The origin $O$ may or may not be the centre of oscillation.

2015 Paper 1, 2015-P1-Q5c (10 marks)

Amplitude $a$, period $T$. Velocity trebled at displacement $2a/3$ from mean; period unchanged. Find new amplitude.

Original velocity at $x=2a/3$: $v_1^2=\omega^2(a^2-4a^2/9)=5\omega^2a^2/9$.

After trebling velocity (same $x$, same $\omega$), new amplitude $A'$ satisfies $(3v_1)^2=\omega^2(A'^2-4a^2/9)$: $9\cdot\frac{5\omega^2a^2}{9}=\omega^2\!\left(A'^2-\frac{4a^2}{9}\right)\;\Longrightarrow\;A'^2=5a^2+\frac{4a^2}{9}=\frac{49a^2}{9}.$ $\boxed{\;A'=\dfrac{7a}{3}.\;}$

2018 Paper 1, 2018-P1-Q6b (12 marks)

Velocities $v_1,v_2$ at distances $x_1,x_2$ from centre. Find period.

Subtract: $v_1^2-v_2^2=\omega^2(x_2^2-x_1^2)$, so $\omega=\sqrt{(v_1^2-v_2^2)/(x_2^2-x_1^2)}$: $\boxed{\;T=2\pi\sqrt{\dfrac{x_2^2-x_1^2}{v_1^2-v_2^2}}.\;}$

Amplitude $A$ cancels and is not required.

2024 Paper 1, 2024-P1-Q6b (15 marks)

Velocities $u,v$ and accelerations $f_1,f_2$ at two positions. Find $k$ such that distance $=k(v^2-u^2)$; show amplitude formula.

From $a=-\omega^2x$: positions are $x_i=-f_i/\omega^2$. The velocity identity becomes $u^2=\omega^2A^2-f_1^2/\omega^2$ and $v^2=\omega^2A^2-f_2^2/\omega^2$. Subtract: $u^2-v^2=\frac{f_2^2-f_1^2}{\omega^2}\;\Longrightarrow\;\omega^2=\frac{f_2^2-f_1^2}{u^2-v^2}.$

Distance: $|x_2-x_1|=|f_2-f_1|/\omega^2=\dfrac{|f_2-f_1|(u^2-v^2)}{f_2^2-f_1^2}=\dfrac{u^2-v^2}{f_1+f_2}$ (for $f_2>f_1>0$).

This equals $k(v^2-u^2)=-k(u^2-v^2)$, so $\boxed{k=-1/(f_1+f_2)}$.

Amplitude: $\omega^2A^2=u^2+f_1^2/\omega^2$. Substituting $\omega^2$: $A^2=\frac{(u^2f_2^2-v^2f_1^2)(u^2-v^2)}{(f_2^2-f_1^2)^2}\;\Longrightarrow\;\boxed{\;A=\frac{1}{f_2^2-f_1^2}\sqrt{(u^2-v^2)(u^2f_2^2-v^2f_1^2)}.\;}$

Common Traps

• The amplitude is half the distance between the extreme positions ($P$ and $Q$), not the full distance. The midpoint of $PQ$ is the centre — at that point velocity is maximum, not zero.
• When velocity is changed by an impulse (2015 problem), the position is unchanged at the instant of impulse. Only the velocity changes; substitute the same $x=2a/3$ into the new energy equation with the new velocity.
• In the two-pair subtraction (2018), the numerator carries the $x^2$ terms ($x_2^2-x_1^2$) and the denominator the $v^2$ terms ($v_1^2-v_2^2$). Swapping produces a negative radicand.

shm-time (2 question(s); 2014, 2023)

Recognition Cues

• A particle passes through a specific point $P$ at a given position $b$ (or $p$) from the centre, moving outward. “Find the time before it returns to $P$.”
• Or: given the result is $T/2$, find the corresponding position.

Solution Template

1. Write $x(t)=A\sin(\omega t+\phi_0)$ with $\phi_0$ chosen so that $A\sin\phi_0=b$ and $\dot x(0)=A\omega\cos\phi_0>0$ (outward). Hence $\phi_0=\arcsin(b/A)\in(0,\pi/2)$.
2. The next passage through $x=b$ satisfies $\omega t+\phi_0=\pi-\phi_0$, giving $t=(\pi-2\phi_0)/\omega$.
3. Substitute and simplify using $\pi/2-\arcsin(b/A)=\arccos(b/A)$: $t=(T/\pi)\arccos(b/A)$.

If amplitude $A$ is not given but velocity $v$ at position $p$ is:

• $A^2=p^2+v^2/\omega^2$ and $\phi_0=\arctan(p\omega/v)$; time $=(T/\pi)\arctan(v/(p\omega))$.

Worked Example(s)

2014 Paper 1, 2014-P1-Q5c (10 marks)

SHM period $T$, amplitude $a$; particle at $x=b$ moving outward. Prove time to return is $(T/\pi)\cos^{-1}(b/a)$.

Set $x=a\sin(\omega t+\phi_0)$; initial conditions give $\phi_0=\arcsin(b/a)$ with $\cos\phi_0>0$. Next return when $\omega t+\phi_0=\pi-\phi_0$: $t=\frac{\pi-2\arcsin(b/a)}{\omega}=\frac{2(\pi/2-\arcsin(b/a))}{\omega}=\frac{2\arccos(b/a)}{\omega}.$ With $\omega=2\pi/T$: $\boxed{\;t=\dfrac{T}{\pi}\cos^{-1}\!\left(\dfrac{b}{a}\right).\;}\qquad\blacksquare$

Boundary checks: $b=a$ (extreme) → $t=0$ ✓; $b=0$ (centre) → $t=(T/\pi)(\pi/2)=T/2$ ✓.

2023 Paper 1, 2023-P1-Q5d (10 marks)

Period $T$; particle at $P$ ($OP=p$) with velocity $v$ outward. Find time to return. Find $p$ when return time is $T/2$.

Since amplitude $A$ is unknown, use: $A^2=p^2+v^2/\omega^2$ and $\phi_0=\arctan(p\omega/v)$. The return time is $t=(\pi-2\phi_0)/\omega$; using $\pi/2-\arctan(x)=\arctan(1/x)$: $\boxed{\;t=\dfrac{T}{\pi}\arctan\!\left(\dfrac{v}{p\omega}\right),\qquad\omega=\dfrac{2\pi}{T}.\;}$

For $t=T/2$: $\arctan(v/(p\omega))=\pi/2$, requiring $v/(p\omega)\to\infty$, hence $\boxed{p=0}$. Geometrically: from the centre, the particle travels to the extreme and back in exactly $T/2$.

Common Traps

• At the extreme ($b=A$), the formula gives $t=0$, which means the “next return” is immediate — the particle is at the turning point and will return through $b=A$ after a full period. This is a limiting case of the formula.
• The identity $\pi/2-\arctan x=\arctan(1/x)$ (valid for $x>0$) is the key simplification from the $2\phi_0$ form to the $\arctan$ form. Derive it, don’t assume it.
• “Along $OP$” means outward (velocity $>0$ in the direction away from $O$). If the particle is moving inward, the phase is in $(\pi/2,\pi)$ and the return time is different.

shm-composite (1 question(s); 2025)

Recognition Cues

• A heavy particle attached to an elastic string fixed at one end; the particle is released from a stretched position.
• While the string is taut (extended beyond natural length), the motion is SHM about the equilibrium length.
• Once the particle rises past the natural length of the string, the string goes slack and motion is free flight under gravity alone.
• The question asks for total time to return to the release point.

Solution Template

1. Find the equilibrium length $\ell_0$ (where tension = weight): $\lambda(\ell_0-\ell)/\ell = mg$, so $\ell_0 = \ell(1+mg/\lambda)$.
2. Write the SHM centre (at $\ell_0$) and amplitude $A=|y_0-\ell_0|$ from the release point.
3. Use $x(t)=\ell_0+A\cos(\omega t)$ (starts at rest). Find the time $t_1$ when $x(t_1)=\ell$ (string goes slack) by solving $\cos(\omega t_1)=({\ell-\ell_0})/A$.
4. Find the speed $v_0 = |\dot x(t_1)| = A\omega|\sin(\omega t_1)|$ at the transition.
5. Free flight: projectile upward with speed $v_0$; time to return to $\ell$ is $t_2=2v_0/g$.
6. By time-reversal symmetry, the return through the SHM region mirrors phase 1. Total $T_{\text{return}}=2t_1+t_2$.

Worked Example(s)

2025 Paper 1, 2025-P1-Q6b (15 marks)

Elastic string, natural length $a$, fixed at $O$; particle at lower end, released from $C$ where string has length $4a$; modulus $=mg$. Show particle returns to $C$ in time $\sqrt{a/g}(2\sqrt3+4\pi/3)$.

Setup. With downward as positive and $y$ = distance below $O$: equilibrium at $y_0=2a$ (Hooke + weight balance), $\omega^2=g/a$, amplitude $A=4a-2a=2a$. Position: $y(t)=2a+2a\cos(\omega t)$.

Phase 1 — SHM until string goes slack ($y=a$): $2a+2a\cos(\omega t_1)=a\;\Rightarrow\;\cos(\omega t_1)=-\tfrac12\;\Rightarrow\;\omega t_1=\tfrac{2\pi}{3}\;\Rightarrow\;t_1=\tfrac{2\pi}{3\omega}=\dfrac{2\pi}{3}\sqrt{\dfrac{a}{g}}.$ Speed at $y=a$: $|\dot y|=2a\omega\sin(2\pi/3)=2a\omega\cdot\tfrac{\sqrt3}{2}=\sqrt{3ag}$ (directed upward).

Phase 2 — Free flight (upward speed $\sqrt{3ag}$): $t_2=\dfrac{2\sqrt{3ag}}{g}=2\sqrt{\dfrac{3a}{g}}=2\sqrt3\sqrt{\dfrac{a}{g}}.$

Phase 3 — SHM back to $C$ (symmetric to Phase 1): $t_3=t_1$.

Total: $T_{\text{return}}=2t_1+t_2=\dfrac{4\pi}{3}\sqrt{\dfrac{a}{g}}+2\sqrt3\sqrt{\dfrac{a}{g}}=\sqrt{\dfrac{a}{g}}\!\left(2\sqrt3+\dfrac{4\pi}{3}\right).\qquad\blacksquare$

Common Traps

• The SHM centre is the equilibrium length $2a$ (where tension equals weight), not the natural length $a$ and not the midpoint of the total path.
• At $y=a$, $\cos(\omega t_1)=-1/2$ comes from $(a-2a)/2a=-1/2$; this gives $\omega t_1=2\pi/3$ (not $\pi/3$). Using the wrong branch gives the wrong time.
• The free-flight time uses $v_0=\sqrt{3ag}$ (from the SHM velocity formula at the transition point). Using $v_0=2a\omega$ (maximum SHM speed) gives the wrong speed.

shm-period-bounds (1 question(s); 2019)

Recognition Cues

• The acceleration is $Fy$ with $F>0$ a positive even function of $y$ — a generalised SHM where the “spring constant” varies with position.
• The question asks to bound the period $T$ by $2\pi/\sqrt{F_{\max}}<T<2\pi/\sqrt{F_{\min}}$.
• Often applied to the nonlinear pendulum oscillating at large angle.

Solution Template

1. Energy integral. From $\ddot y=-F(y)y$ and $v=0$ at $y=a$: multiply by $\dot y$ and integrate to get $\dot y^2=2\int_y^a F(s)s\,ds$.
2. Period. By symmetry: $T=4\int_0^a\frac{dy}{\sqrt{2\int_y^a F(s)s\,ds}}$.
3. Sandwich. For $F_2\le F(s)\le F_1$: $F_2\cdot\tfrac12(a^2-y^2)\le\int_y^a Fs\,ds\le F_1\cdot\tfrac12(a^2-y^2)$. Inverting the inequality in the period integral and integrating $\int_0^a dy/\sqrt{a^2-y^2}=\pi/2$ gives the bound.
4. Pendulum. $F=(g/l)\sin\theta/\theta$ (write $\ddot y=-(g/l)(\sin\theta/\theta)y$ with $y=l\theta$). On $[0,\pi/6]$, $\sin\theta/\theta$ decreases from 1 to $3/\pi$, so $F_1=g/l$ and $F_2=(3g)/(l\pi)$.

Worked Example(s)

2019 Paper 1, 2019-P1-Q7c (20 marks)

Acceleration $Fy$ toward origin, $F$ positive and even; particle vibrates between $\pm a$. Show $2\pi/\sqrt{F_1}<T<2\pi/\sqrt{F_2}$; apply to pendulum at $30°$.

Step 1 — Energy. $\dot y^2=2\displaystyle\int_y^a F(s)s\,ds$.

Step 2 — Bounds. $F_2\le F\le F_1$ gives $F_2\cdot\tfrac12(a^2-y^2)\le\int_y^a Fs\,ds\le F_1\cdot\tfrac12(a^2-y^2)$. In the period integral, larger denominator → smaller $T$: $4\int_0^a\!\frac{dy}{\sqrt{F_1(a^2-y^2)}} < T < 4\int_0^a\!\frac{dy}{\sqrt{F_2(a^2-y^2)}}.$ Both integrals equal $\pi/2$ after substituting $y=a\sin u$: $\boxed{\;\dfrac{2\pi}{\sqrt{F_1}} < T < \dfrac{2\pi}{\sqrt{F_2}}.\;}\qquad\blacksquare$

Pendulum ($\theta_{\max}=\pi/6$). Write $l\ddot\theta=-g\sin\theta$; set $y=l\theta$ so $\ddot y=-(g/l)(\sin\theta/\theta)\cdot y$ with $F=(g/l)\sin\theta/\theta$. Since $\sin\theta/\theta$ is decreasing on $[0,\pi/6]$: $F_1=\frac{g}{l}\quad\text{(at }\theta=0\text{)},\qquad F_2=\frac{g}{l}\cdot\frac{\sin(\pi/6)}{\pi/6}=\frac{g}{l}\cdot\frac{3}{\pi}.$

Substituting: $2\pi\sqrt{\frac{l}{g}} < T < 2\pi\sqrt{\frac{l\pi}{3g}} = 2\pi\sqrt{\frac{l}{g}}\sqrt{\frac{\pi}{3}}.\ \blacksquare$

Common Traps

• The bounding is on $\int_y^a F(s)s\,ds$ — the integrand is the moment $F(s)\cdot s$, not $F(s)$ alone. Bounding $F$ by its extremes pulls the constant out but leaves $\int_y^a s\,ds=\tfrac12(a^2-y^2)$.
• Larger $F$ → larger speed → smaller period: so $F_1$ (max) gives the lower bound $2\pi/\sqrt{F_1}$ and $F_2$ (min) gives the upper bound. This direction is easy to reverse.
• Pendulum: $\sin(\pi/6)=1/2$, so $\sin(\pi/6)/(\pi/6)=(1/2)/(\pi/6)=3/\pi$. The resulting $\sqrt{\pi/3}$ factor ≈ 1.023 means the true period barely exceeds the small-angle approximation.

Marks-Aware Writing

10-mark questions (2013-Q5c, 2014-Q5c, 2015-Q5c, 2023-Q5d): Two to three steps. Write the energy relation or parametric form; derive the answer algebraically; box the result. For time-return questions, state the phase $\phi_0$ and the symmetry argument $\sin(\pi-\phi)=\sin\phi$ explicitly — these are the method marks.

12-mark questions (2018-Q6b): Show the two energy equations, perform the subtraction, solve for $\omega^2$. The examiner marks the subtraction step and the period formula separately.

15-mark questions (2024-Q6b, 2025-Q6b): For the velocity-acceleration problem, there are two sub-results ($k$ and the amplitude formula); treat them as two separate derivations. For the elastic-string problem, label the three phases clearly and verify the $\cos(\omega t_1)=-1/2$ step in detail.

20-mark questions (2019-Q7c): Full two-part structure. Part 1 (the bound): set up the integral, sandwich, integrate. Part 2 (pendulum): identify $F=(g/l)\sin\theta/\theta$, state monotonicity, compute $F_1$ and $F_2$, substitute. Each sub-result carries method marks — present both explicitly.

Practice Set

No additional practice items beyond the worked examples above.