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Module 16 · L05 of 16 ~40 min ⚡ +90 XP available

Simple Harmonic Motion — Derivation

A particle that accelerates back toward a fixed centre in proportion to its displacement traces out the most important oscillation in physics. In this lesson you'll meet the defining differential equation $\ddot{x} = -n^2 x$, verify the general solution $x = A\cos(nt) + B\sin(nt)$, and connect both to the mass-on-spring model. Master the derivation now and the applications in L06 fall into place.

Today's hook — A mass on a spring satisfies $m\ddot{x} = -kx$. Before reading on, rearrange to the standard form $\ddot{x} = -n^2 x$ and write down what $n$ must equal in terms of $m$ and $k$. Then try to verify by substitution that $x(t) = \cos(nt)$ is a solution. Compare your work after card 05.

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Recall — your gut answer first

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For a mass-on-spring with Hooke's law $F = -kx$, Newton's second law gives $m\ddot{x} = -kx$. Before checking — divide through by $m$ to get $\ddot{x} = -(k/m)x$, then identify what $n^2$ equals in the standard SHM form $\ddot{x} = -n^2 x$. Sketch your reasoning below.

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The two moves for setting up SHM

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Every SHM problem rewards two habits: measure displacement from the equilibrium point (where net force is zero), then write Newton II and match the form $\ddot{x} = -n^2 x$. Choose origin and sign convention correctly and the rest is calculus.

The setup-equation-identify flow: (1) place the origin at equilibrium and pick a positive direction, (2) write the net force in terms of displacement $x$, (3) compare $m\ddot{x} = F(x)$ with $\ddot{x} = -n^2 x$ and read off $n$.

Defining ODE: $\ddot{x} = -n^2 x$ · General solution: $x = A\cos(nt) + B\sin(nt)$

$\ddot{x} = -n^2 x \;\Longleftrightarrow\; x = A\cos(nt) + B\sin(nt)$

Restoring force, negative sign

The minus sign in $\ddot{x} = -n^2 x$ is non-negotiable. It says acceleration points back toward $x = 0$. Drop the sign and you get exponential growth, not oscillation.

$n$ is angular frequency

$n$ has units rad s$^{-1}$. Period $T = 2\pi/n$; frequency $f = n/(2\pi)$. For a spring, $n = \sqrt{k/m}$ — stiffer spring or lighter mass means faster oscillation.

Two equivalent forms

$x = A\cos(nt) + B\sin(nt)$ and $x = a\cos(nt + \alpha)$ describe the same motion. Use $a = \sqrt{A^2 + B^2}$ and $\tan\alpha = -B/A$ to convert.

What you'll master

Know

Key facts

SHM is defined by $\ddot{x} = -n^2 x$ where $n$ is the angular frequency
The general solution is $x = A\cos(nt) + B\sin(nt)$, equivalent to $a\cos(nt + \alpha)$
Period $T = 2\pi/n$; frequency $f = 1/T = n/(2\pi)$
For a spring, $n = \sqrt{k/m}$

Understand

Concepts

Why a linear restoring force produces sinusoidal motion
Why both $\cos(nt)$ and $\sin(nt)$ satisfy the ODE — and why their sum is the general solution
How Hooke's law plus Newton II gives the SHM equation

Can do

Skills

Verify by substitution that $x = A\cos(nt) + B\sin(nt)$ satisfies $\ddot{x} = -n^2 x$
Find $n$, $T$ and $f$ from a spring constant and mass
Identify SHM from a written or pictorial description of a system

Key terms

Simple harmonic motion (SHM)Oscillation in which acceleration is proportional to displacement from a fixed point and directed back toward it: $\ddot{x} = -n^2 x$.

Angular frequency ($n$)The constant appearing in the defining ODE. Units: rad s$^{-1}$. Determines how rapidly the system oscillates.

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

Frequency ($f$)Oscillations per second: $f = 1/T = n/(2\pi)$. Units: hertz (s$^{-1}$).

Amplitude ($a$)Maximum displacement from equilibrium: $a = \sqrt{A^2 + B^2}$ when $x = A\cos(nt) + B\sin(nt)$.

Phase ($\alpha$)The shift inside the cosine when written as $x = a\cos(nt + \alpha)$. Encodes initial position and velocity.

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

The defining equation and its general solution

core concept

A particle moves in simple harmonic motion if its displacement $x(t)$ from a fixed point satisfies

$$\ddot{x} = -n^2 x \qquad (n > 0)$$

The constant $n$ is the angular frequency. The general solution is

$$x(t) = A\cos(nt) + B\sin(nt)$$

where $A$ and $B$ are constants determined by initial conditions. The two terms are linearly independent solutions, and by the theory of second-order linear ODEs any solution is a combination of them.

Verification by substitution. Let $x = A\cos(nt) + B\sin(nt)$. Differentiate twice:

$\dot{x} = -An\sin(nt) + Bn\cos(nt)$
$\ddot{x} = -An^2\cos(nt) - Bn^2\sin(nt) = -n^2[A\cos(nt) + B\sin(nt)] = -n^2 x$ ✓

Equivalent single-cosine form. Using the auxiliary-angle identity, write $A\cos(nt) + B\sin(nt) = a\cos(nt + \alpha)$ where $a = \sqrt{A^2 + B^2}$ and $\tan\alpha = -B/A$. This form makes amplitude visible directly.

Why the form matters. The two-constant solution is what fits initial conditions; the single-cosine form is what shows the physical amplitude and phase. Most exam questions ask for both at some point.

Defining ODE: $\ddot{x} = -n^2 x$ (minus sign is essential) · General solution: $x = A\cos(nt) + B\sin(nt) = a\cos(nt + \alpha)$ · $a = \sqrt{A^2 + B^2}$, $\tan\alpha = -B/A$ · Period $T = 2\pi/n$, frequency $f = n/(2\pi)$

Pause — copy the SHM ODE $\ddot{x} = -n^2 x$, the general solution (both forms), $a = \sqrt{A^2+B^2}$, $\tan\alpha = -B/A$, and $T = 2\pi/n$ into your book.

Quick check: Which differential equation defines simple harmonic motion with angular frequency $n$?

Mass on a spring — where $n^2 = k/m$ comes from

core concept

We just saw that $\ddot{x} = -n^2 x$ has general solution $x = A\cos(nt)+B\sin(nt) = a\cos(nt+\alpha)$ with amplitude $a = \sqrt{A^2+B^2}$ and period $T = 2\pi/n$. That raises a question: where does the $-n^2 x$ restoring force come from physically? This card answers it → Hooke's law $F = -kx$ with Newton II gives $m\ddot{x} = -kx$, so $n^2 = k/m$, $T = 2\pi\sqrt{m/k}$, independent of amplitude.

A mass $m$ on a spring of stiffness $k$ satisfies Hooke's law: the restoring force when displaced by $x$ is $F = -kx$. Newton's second law $F = m\ddot{x}$ becomes

$$m\ddot{x} = -kx \;\Longrightarrow\; \ddot{x} = -\tfrac{k}{m} x$$

Comparing with $\ddot{x} = -n^2 x$ gives $n^2 = k/m$, so

$$n = \sqrt{\tfrac{k}{m}}, \qquad T = 2\pi\sqrt{\tfrac{m}{k}}, \qquad f = \tfrac{1}{2\pi}\sqrt{\tfrac{k}{m}}$$

Note that the period is independent of amplitude — a hallmark of SHM. Doubling the displacement doubles the maximum force, the maximum acceleration and the maximum speed, but leaves the period unchanged.

Common mistake. Don't confuse $n$ (the constant in the ODE, units rad s$^{-1}$) with $f$ (oscillations per second, units Hz). They differ by a factor of $2\pi$: $n = 2\pi f$.

Hooke's law $F = -kx$ + Newton II $\Rightarrow$ $m\ddot{x} = -kx$ · Standard form: $\ddot{x} = -(k/m)x$, so $n^2 = k/m$ · $n = \sqrt{k/m}$, $T = 2\pi\sqrt{m/k}$ · Period is independent of amplitude · Convert: $n = 2\pi f$

Pause — copy Hooke's law derivation of $n^2 = k/m$, $T = 2\pi\sqrt{m/k}$, the frequency $f = n/(2\pi)$, and the amplitude-independence of the period into your book.

Did you get this? True or false: for a mass on a spring undergoing SHM, doubling the initial displacement also doubles the period of oscillation.

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

PROBLEM 1 · VERIFY THE SOLUTION

Show by substitution that $x(t) = 3\cos(2t) + 4\sin(2t)$ satisfies the SHM differential equation $\ddot{x} = -n^2 x$ for some constant $n$, and state $n$.

Differentiate once: $\dot{x} = -6\sin(2t) + 8\cos(2t)$.

Apply the chain rule with inner function $u = 2t$. The coefficient of $t$ multiplies through each derivative — this is where $n$ appears.

PROBLEM 2 · SPRING CONSTANT TO $n$

A $0.5$ kg mass hangs on a spring of stiffness $k = 32$ N m$^{-1}$. Find the angular frequency $n$, period $T$ and frequency $f$ of the resulting SHM.

From Newton II with Hooke's law: $m\ddot{x} = -kx$, so $\ddot{x} = -(k/m)x$ and $n^2 = k/m = 32/0.5 = 64$.

Always start by writing the ODE in standard form. The constant $n^2$ is whatever multiplies $-x$.

PROBLEM 3 · IDENTIFY SHM FROM ODE

A particle moves so that $\ddot{x} + 9x = 0$. Show this is SHM, state $n$, write the general solution and find the period.

Rearrange: $\ddot{x} = -9x$, which matches $\ddot{x} = -n^2 x$ with $n^2 = 9$.

Any equation of the form $\ddot{x} + cx = 0$ with $c > 0$ is SHM. The coefficient of $x$ is $n^2$.

Fill the gap: A particle satisfies $\ddot{x} = -25x$. Then $n = $ rad s$^{-1}$, and the period is $T = 2\pi/$ seconds.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting the minus sign

$\ddot{x} = n^2 x$ (no minus) gives exponential solutions $e^{\pm nt}$, not oscillation. The negative sign in $\ddot{x} = -n^2 x$ encodes the restoring nature of the force. Always check the sign before reading off $n$.

Trap 02

Confusing $n$ with $n^2$

In $\ddot{x} = -9x$, the coefficient $9$ equals $n^2$, so $n = 3$ — not $n = 9$. Period $T = 2\pi/n = 2\pi/3$, not $2\pi/9$. Take the square root before computing $T$.

Trap 03

Period depends on amplitude (it doesn't)

SHM has the remarkable property that the period $T = 2\pi/n$ is independent of amplitude. Telling a student "release from further out, takes longer" is correct for a pendulum at large angles but not for ideal SHM.

Did you get this? True or false: if $\ddot{x} = -16x$, then the period of motion is $T = 2\pi/16$ seconds.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Verify that $x(t) = 5\sin(4t)$ satisfies $\ddot{x} = -n^2 x$. State $n$ and the period.

A mass of $2$ kg is attached to a spring with stiffness $k = 50$ N m$^{-1}$. Find the angular frequency $n$ and period $T$ of the SHM that results.

Show by direct differentiation that $x(t) = a\cos(nt + \alpha)$ satisfies $\ddot{x} = -n^2 x$ for any constants $a$, $\alpha$.

Rewrite $x = 3\cos(2t) + 4\sin(2t)$ in the form $a\cos(2t + \alpha)$, stating $a$ and $\tan\alpha$.

A spring is stiffened so $k$ is doubled while the mass is unchanged. By what factor does the period change? Justify using $T = 2\pi\sqrt{m/k}$.

Odd one out: Three of these are solutions of $\ddot{x} = -4x$. Which one is NOT?

Revisit your thinking

Earlier you took $m\ddot{x} = -kx$, divided by $m$ and tried to identify $n$ in the standard form.

The match gives $n^2 = k/m$, so $n = \sqrt{k/m}$ — a result you'll use whenever a mass-on-spring is described. The verification that $x = \cos(nt)$ solves $\ddot{x} = -n^2 x$ is one chain-rule calculation: $\ddot{x} = -n^2\cos(nt) = -n^2 x$. Together these two moves — set up Hooke + Newton, then read off $n$ — drive every SHM problem in L06.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. A particle satisfies $\ddot{x} = -36x$. State the angular frequency $n$ and period $T$ of its motion. (2 marks)

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ApplyBand 43 marks

Q2. Verify by direct substitution that $x(t) = 2\cos(3t) - 5\sin(3t)$ satisfies the differential equation $\ddot{x} = -n^2 x$ and state $n$. (3 marks)

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AnalyseBand 53 marks

Q3. A particle of mass $0.25$ kg is attached to a light spring of stiffness $k = 4$ N m$^{-1}$ on a smooth horizontal surface. Show that the particle executes SHM, find the angular frequency $n$ and the period $T$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\dot{x} = 20\cos(4t)$; $\ddot{x} = -80\sin(4t) = -16 \cdot 5\sin(4t) = -16x$. So $n^2 = 16$, $n = 4$ rad s$^{-1}$, $T = \pi/2$ s.

2. $n^2 = k/m = 50/2 = 25$, so $n = 5$ rad s$^{-1}$ and $T = 2\pi/5$ s $\approx 1.257$ s.

3. $\dot{x} = -an\sin(nt + \alpha)$; $\ddot{x} = -an^2\cos(nt + \alpha) = -n^2 \cdot a\cos(nt + \alpha) = -n^2 x$. ✓

4. $a = \sqrt{9 + 16} = 5$. Expanding $5\cos(2t + \alpha) = 5\cos\alpha\cos 2t - 5\sin\alpha\sin 2t$ matches $3\cos 2t + 4\sin 2t$ when $\cos\alpha = 3/5$, $\sin\alpha = -4/5$, so $\tan\alpha = -4/3$.

5. $T_{\text{new}}/T_{\text{old}} = \sqrt{m/(2k)}/\sqrt{m/k} = 1/\sqrt{2}$. New period is $1/\sqrt{2}$ of the old — about $70.7\%$.

Q1 (2 marks): $n^2 = 36$, so $n = 6$ rad s$^{-1}$ [1]. $T = 2\pi/n = \pi/3$ s [1].

Q2 (3 marks): $\dot{x} = -6\sin(3t) - 15\cos(3t)$ [1]; $\ddot{x} = -18\cos(3t) + 45\sin(3t)$ [1]. Factor: $\ddot{x} = -9[2\cos(3t) - 5\sin(3t)] = -9x$, so $n^2 = 9$, $n = 3$ [1].

Q3 (3 marks): By Hooke's law, $F = -kx$; by Newton II, $m\ddot{x} = F$, so $m\ddot{x} = -kx$ [1]. Hence $\ddot{x} = -(k/m)x = -16x$, matching $\ddot{x} = -n^2 x$ with $n^2 = 16$, so $n = 4$ rad s$^{-1}$ [1]. $T = 2\pi/4 = \pi/2$ s [1].

Boss battle · The Oscillation Architect

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Five timed questions on the SHM equation, verification, the spring connection and period calculations. 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