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Module 16 · L06 of 16 ~40 min ⚡ +90 XP available

Simple Harmonic Motion — Solving and Applications

Knowing the general solution $x = A\cos(nt) + B\sin(nt)$ is only half the job. This lesson turns it into a working tool: use initial conditions to fix $A$ and $B$, compute amplitude $a = \sqrt{A^2 + B^2}$, find the period, and apply it all to pendulums and spring systems. By the end you'll predict where a real oscillator is at any time $t$.

Today's hook — A particle in SHM satisfies $\ddot{x} = -4x$ with $x(0) = 3$ and $\dot{x}(0) = 4$. Before reading on, write $x(t) = A\cos(2t) + B\sin(2t)$ and use the initial conditions to find $A$ and $B$. Then compute the amplitude. Compare your work after card 05.

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Recall — your gut answer first

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From L05 you have $x(t) = A\cos(nt) + B\sin(nt)$ as the general solution of $\ddot{x} = -n^2 x$. Before checking — if the particle starts at $x(0) = x_0$ with initial velocity $\dot{x}(0) = v_0$, find $A$ and $B$ in terms of $x_0$, $v_0$ and $n$. What does the amplitude become? Sketch your reasoning below.

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The two moves for solving an SHM problem

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Every applied SHM problem rewards two habits: substitute initial conditions into both $x(0)$ and $\dot{x}(0)$ to fix the constants, then read the amplitude as $\sqrt{A^2 + B^2}$ and the period as $2\pi/n$. Constants first, summary quantities second.

The constants-amplitude-period flow: (1) plug $t = 0$ into $x$ to get $A$, plug $t = 0$ into $\dot{x}$ to get $Bn$, (2) compute amplitude $a = \sqrt{A^2 + B^2}$, (3) state period $T = 2\pi/n$.

$A = x(0)$ · $B = \dot{x}(0)/n$ · $a = \sqrt{A^2 + B^2}$ · $T = 2\pi/n$

$a = \sqrt{A^2 + B^2}, \quad T = 2\pi/n$

Always use both ICs

One condition fixes $A$, the other fixes $B$. Omitting the velocity condition is the single most common error — without it, $B$ is undetermined and amplitude can't be found.

Pendulum: small angles only

For a simple pendulum, the equation $\ddot{\theta} = -(g/L)\sin\theta$ is approximated by $\ddot{\theta} = -(g/L)\theta$ when $\theta$ is small. Then $n^2 = g/L$ and $T = 2\pi\sqrt{L/g}$.

Maximum speed at $x = 0$

Energy conservation in SHM gives $\dot{x}^2 = n^2(a^2 - x^2)$. Maximum speed $v_{\max} = na$ occurs at the equilibrium point; the particle is momentarily at rest at the extremes $x = \pm a$.

What you'll master

Know

Key facts

$A = x(0)$ and $B = \dot{x}(0)/n$ for $x = A\cos(nt) + B\sin(nt)$
Amplitude $a = \sqrt{A^2 + B^2}$
Period $T = 2\pi/n$ (independent of amplitude)
Pendulum (small angle): $n = \sqrt{g/L}$, $T = 2\pi\sqrt{L/g}$
Velocity relation: $\dot{x}^2 = n^2(a^2 - x^2)$

Understand

Concepts

Why two initial conditions are needed (second-order ODE)
Why amplitude depends on both starting position and speed
Why small-angle approximation linearises the pendulum to SHM

Can do

Skills

Solve $\ddot{x} = -n^2 x$ given $x(0)$ and $\dot{x}(0)$
Compute amplitude, period and maximum speed for any SHM
Apply SHM to pendulums and mass-spring systems with numerical data

Key terms

Initial conditionsThe values of $x$ and $\dot{x}$ at $t = 0$. Two are needed to fully determine the solution of the second-order ODE $\ddot{x} = -n^2 x$.

Amplitude ($a$)Maximum displacement from equilibrium: $a = \sqrt{A^2 + B^2}$ where $x = A\cos(nt) + B\sin(nt)$. Equivalently $a^2 = x_0^2 + v_0^2/n^2$.

Period ($T$)Time for one full oscillation: $T = 2\pi/n$. Independent of amplitude for SHM.

Maximum speed$v_{\max} = na$, reached as the particle passes through equilibrium $x = 0$. Found from $\dot{x}^2 = n^2(a^2 - x^2)$.

Simple pendulumA mass on a light string of length $L$. For small angles $\theta$, $\ddot{\theta} \approx -(g/L)\theta$, giving SHM with $n = \sqrt{g/L}$ and $T = 2\pi\sqrt{L/g}$.

Energy form$\dot{x}^2 = n^2(a^2 - x^2)$ — a first integral of the SHM equation. Useful for finding speed at a given position without solving for $t$.

MEX-M1NESA outcome (Applications of Calculus to Mechanics): solves problems involving simple harmonic motion using a range of mathematical techniques.

Finding $A$ and $B$ from initial conditions

core concept

Given the general solution $x(t) = A\cos(nt) + B\sin(nt)$ and two initial conditions $x(0) = x_0$ and $\dot{x}(0) = v_0$, the constants follow directly.

Step 1. Substitute $t = 0$ into $x$:

$$x(0) = A\cos(0) + B\sin(0) = A = x_0$$

Step 2. Differentiate: $\dot{x}(t) = -An\sin(nt) + Bn\cos(nt)$. Substitute $t = 0$:

$$\dot{x}(0) = Bn = v_0 \;\Longrightarrow\; B = v_0/n$$

So the unique solution for given initial position $x_0$ and initial velocity $v_0$ is

$$x(t) = x_0\cos(nt) + \tfrac{v_0}{n}\sin(nt)$$

Amplitude and period. Using $a = \sqrt{A^2 + B^2}$:

$$a = \sqrt{x_0^2 + \tfrac{v_0^2}{n^2}}, \qquad T = \tfrac{2\pi}{n}$$

Special cases worth knowing. If $v_0 = 0$ (released from rest), then $B = 0$ and $a = |x_0|$ — the particle's starting position is the amplitude. If $x_0 = 0$ (launched from equilibrium), then $A = 0$ and $a = |v_0|/n$.

$A = x(0)$ (just plug in $t = 0$ to $x$) · $B = \dot{x}(0)/n$ (plug in $t = 0$ to $\dot{x}$, then divide by $n$) · $a = \sqrt{A^2 + B^2} = \sqrt{x_0^2 + v_0^2/n^2}$ · Released from rest: amplitude = starting displacement · Launched from equilibrium: amplitude = $|v_0|/n$

Pause — copy $A = x(0)$, $B = \dot{x}(0)/n$, the amplitude formula $a = \sqrt{A^2+B^2}$, and the two special cases (released from rest: amplitude $= x_0$; from equilibrium: amplitude $= v_0/n$) into your book.

Quick check: A particle satisfies $\ddot{x} = -9x$ with $x(0) = 4$ and $\dot{x}(0) = 0$. What is the amplitude of the motion?

Applications — pendulum and spring

core concept

We just saw that $A = x(0)$ and $B = \dot{x}(0)/n$ directly from the general solution, giving amplitude $a = \sqrt{x_0^2+v_0^2/n^2}$. That raises a question: what do these formulas look like for the two physically natural starting conditions — released from rest, or launched from equilibrium? This card answers it → released from rest at $x_0$: $x(t) = x_0\cos(nt)$, amplitude $= x_0$; launched from equilibrium at $v_0$: $x(t) = (v_0/n)\sin(nt)$, amplitude $= v_0/n$.

Simple pendulum (small angles). A mass on a light string of length $L$ swings under gravity. With $\theta$ measuring angle from vertical and $g$ the gravitational acceleration, the equation of motion is $\ddot{\theta} = -(g/L)\sin\theta$. For small $\theta$, $\sin\theta \approx \theta$, giving

$$\ddot{\theta} = -\tfrac{g}{L}\theta \;\Longrightarrow\; n = \sqrt{\tfrac{g}{L}}, \quad T = 2\pi\sqrt{\tfrac{L}{g}}$$

Two striking consequences: the period depends only on length and gravity (not on the mass or amplitude), and doubling the length increases $T$ by $\sqrt{2}$.

Mass on a spring. From L05, $n = \sqrt{k/m}$. If the mass is pulled aside a distance $x_0$ from equilibrium and released from rest ($v_0 = 0$), the motion is

$$x(t) = x_0\cos(nt), \qquad n = \sqrt{\tfrac{k}{m}}, \qquad v_{\max} = nx_0$$

If instead the mass is struck at equilibrium with initial speed $v_0$, the motion is $x(t) = (v_0/n)\sin(nt)$ with amplitude $v_0/n$.

Energy shortcut. Without solving the full ODE, the relation $\dot{x}^2 = n^2(a^2 - x^2)$ gives the speed at any position. Setting $x = 0$ recovers $v_{\max} = na$; setting $\dot{x} = 0$ recovers $x = \pm a$ (the turning points).

Pendulum (small angle): $\ddot{\theta} = -(g/L)\theta$, $T = 2\pi\sqrt{L/g}$ · Spring: $n = \sqrt{k/m}$, $T = 2\pi\sqrt{m/k}$ · Released from rest at $x_0$: $x(t) = x_0\cos(nt)$, $v_{\max} = nx_0$ · Launched from equilibrium at speed $v_0$: $x(t) = (v_0/n)\sin(nt)$, amplitude $= v_0/n$ · Energy form: $\dot{x}^2 = n^2(a^2 - x^2)$

Pause — copy the pendulum period $T = 2\pi\sqrt{L/g}$, the spring period $T = 2\pi\sqrt{m/k}$, and the energy velocity formula $\dot{x}^2 = n^2(a^2-x^2)$ into your book.

Did you get this? True or false: the period of a simple pendulum (small swings) depends on the mass of the bob.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · INITIAL CONDITIONS → CONSTANTS

A particle satisfies $\ddot{x} = -4x$ with $x(0) = 3$ and $\dot{x}(0) = 4$. Find $A$, $B$, the amplitude and the period.

From $\ddot{x} = -4x$: $n^2 = 4$, so $n = 2$. General solution: $x(t) = A\cos(2t) + B\sin(2t)$.

Read $n$ off the ODE first — every subsequent step uses it. Then write the general solution with $n$ inside the trig functions.

PROBLEM 2 · PENDULUM PERIOD

A simple pendulum of length $L = 1.2$ m swings through small angles. Taking $g = 9.8$ m s$^{-2}$, find the period $T$ and frequency $f$ of the resulting SHM.

For small angles, $\ddot{\theta} = -(g/L)\theta$. So $n^2 = g/L = 9.8/1.2 \approx 8.167$.

Always state the small-angle approximation before quoting $n^2 = g/L$ — examiners look for this justification.

PROBLEM 3 · SPRING WITH GIVEN AMPLITUDE

A $0.5$ kg mass on a spring of stiffness $k = 8$ N m$^{-1}$ is pulled $0.1$ m from equilibrium and released from rest. Find $x(t)$, the period and the maximum speed of the mass.

$n^2 = k/m = 8/0.5 = 16$, so $n = 4$ rad s$^{-1}$. Released from rest: $x(0) = 0.1$, $\dot{x}(0) = 0$.

List the ICs before applying the formulas. "Released from rest" always means $\dot{x}(0) = 0$.

Fill the gap: A particle in SHM has $x(0) = 0$ and $\dot{x}(0) = 6$ with $n = 3$. Then $A = $ and $B = $ , giving amplitude $a = $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Reading $B = \dot{x}(0)$ instead of $\dot{x}(0)/n$

Because $\dot{x}(t) = -An\sin(nt) + Bn\cos(nt)$, evaluating at $t = 0$ gives $\dot{x}(0) = Bn$, NOT $B$. Always divide the initial velocity by $n$ to get $B$. Skipping this step is the single most common SHM error.

Trap 02

Amplitude equals initial displacement (only sometimes)

Amplitude $a = \sqrt{A^2 + B^2}$, not $|A|$ alone. If the particle starts at $x_0$ with non-zero initial speed, the amplitude is strictly larger than $|x_0|$. Only when $v_0 = 0$ does $a = |x_0|$.

Trap 03

Pendulum formula at large angles

$T = 2\pi\sqrt{L/g}$ only holds for small angles, where $\sin\theta \approx \theta$. At large amplitudes, the exact equation $\ddot{\theta} = -(g/L)\sin\theta$ is non-linear and the period grows with amplitude. State the small-angle assumption when applying the formula.

Did you get this? True or false: a particle in SHM with $\ddot{x} = -4x$ starting at $x(0) = 3$ with $\dot{x}(0) = 8$ has amplitude $a = 3$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A particle satisfies $\ddot{x} = -25x$ with $x(0) = 2$ and $\dot{x}(0) = 0$. Find $x(t)$, amplitude and period.

A particle satisfies $\ddot{x} = -16x$ with $x(0) = 0$ and $\dot{x}(0) = 12$. Find $x(t)$ and the amplitude.

A pendulum of length $0.8$ m swings through small angles. Using $g = 9.8$ m s$^{-2}$, find the period.

A mass on a spring undergoes SHM with $n = 5$ and amplitude $a = 0.2$ m. Find the maximum speed and the speed when $x = 0.1$ m. (Use $\dot{x}^2 = n^2(a^2 - x^2)$.)

A particle satisfies $\ddot{x} = -4x$ with $x(0) = 3$ and $\dot{x}(0) = -8$. Find the amplitude and write $x(t)$ in the form $a\cos(2t + \alpha)$.

Odd one out: Three of these statements about SHM with $\ddot{x} = -n^2 x$ are true. Which one is NOT?

Revisit your thinking

Earlier you took $\ddot{x} = -4x$ with $x(0) = 3$ and $\dot{x}(0) = 4$ and tried to find $A$, $B$ and the amplitude.

With $n = 2$, the position IC gives $A = 3$ and the velocity IC gives $2B = 4$, so $B = 2$. The amplitude is $a = \sqrt{9 + 4} = \sqrt{13}$ — strictly larger than the initial displacement because the particle was also moving when timing began. The lesson here generalises: once you've fixed $A$ and $B$ from the two ICs, every other quantity (amplitude, period, maximum speed, phase) follows from the formulas in card 02. That recipe handles every standard MEX-M1 SHM application.

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Multiple choice

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Short answer

ApplyBand 32 marks

Q1. A particle satisfies $\ddot{x} = -9x$ with $x(0) = 5$ and $\dot{x}(0) = 0$. Write down $x(t)$ and state the amplitude and period. (2 marks)

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ApplyBand 43 marks

Q2. A particle moves so that $\ddot{x} = -16x$. At $t = 0$, $x = 2$ and $\dot{x} = 12$. Find $x(t)$, the amplitude of the motion and the maximum speed. (3 marks)

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AnalyseBand 53 marks

Q3. A simple pendulum of length $L$ has period $T = 2$ s on Earth where $g = 9.8$ m s$^{-2}$. (a) Find $L$. (b) If the same pendulum is taken to the Moon where $g_{\text{moon}} = 1.62$ m s$^{-2}$, find the new period. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $n = 5$. $A = 2$, $B = 0$. $x(t) = 2\cos(5t)$. Amplitude $= 2$, period $T = 2\pi/5$ s.

2. $n = 4$. $A = 0$, $B = 12/4 = 3$. $x(t) = 3\sin(4t)$. Amplitude $= 3$.

3. $T = 2\pi\sqrt{L/g} = 2\pi\sqrt{0.8/9.8} \approx 1.796$ s.

4. $v_{\max} = na = 5 \times 0.2 = 1$ m s$^{-1}$. At $x = 0.1$: $\dot{x}^2 = 25(0.04 - 0.01) = 0.75$, so $|\dot{x}| = \sqrt{0.75} \approx 0.866$ m s$^{-1}$.

5. $n = 2$. $A = 3$, $B = -8/2 = -4$. $a = \sqrt{9 + 16} = 5$. In single-cosine form: $5\cos(2t + \alpha)$ with $5\cos\alpha = 3$, $5\sin\alpha = 4$, so $\cos\alpha = 3/5$, $\sin\alpha = 4/5$, $\tan\alpha = 4/3$.

Q1 (2 marks): $n^2 = 9$, $n = 3$. $A = x(0) = 5$, $B = \dot{x}(0)/n = 0$ [1]. $x(t) = 5\cos(3t)$, amplitude $= 5$, $T = 2\pi/3$ s [1].

Q2 (3 marks): $n = 4$. $A = 2$, $B = 12/4 = 3$ [1]. $x(t) = 2\cos(4t) + 3\sin(4t)$. Amplitude $a = \sqrt{4 + 9} = \sqrt{13}$ [1]. Maximum speed $v_{\max} = na = 4\sqrt{13}$ m s$^{-1}$ [1].

Q3 (3 marks): (a) From $T = 2\pi\sqrt{L/g}$: $L = g(T/(2\pi))^2 = 9.8 \times (1/\pi)^2 = 9.8/\pi^2 \approx 0.993$ m [1]. (b) $T_{\text{moon}} = 2\pi\sqrt{L/g_{\text{moon}}} = 2\pi\sqrt{0.993/1.62}$ [1] $\approx 4.92$ s — about $2.46\times$ the Earth period [1].

Boss battle · The Pendulum Whisperer

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Five timed questions on initial conditions, amplitude, period and pendulum/spring applications. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering quick SHM questions. Lighter alternative to the boss.

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Module overview · Extension 2