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.
0/5QUESTS
1
You are here
Recall, your gut answer first
+5 XP warm-up For a mass-on-spring with Hooke's law $F = -kx$, Newton's second law gives $m\ddot{x} = -kx$.
01
Recall, your gut answer first
+5 XP warm-up
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.
auto-saved
2
The two moves for setting up SHM
+5 XP to read Every SHM problem rewards two habits: measure displacement from the equilibrium point (where net force is zero),...
02
The two moves for setting up SHM
+5 XP to read
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 $n$.
$\ddot{x} = -n^2 x \;\Longleftrightarrow\; x = A\cos(nt) + B\sin(nt)$
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.
03
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
04
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.
3
The defining equation and its general solution
A particle moves in simple harmonic motion if its displacement from a fixed point satisfies
05
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:
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 $n$?
4
Mass on a spring, where $n^2 = k/m$ comes from
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 $T = 2\pi/n$.
06
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
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.
5
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
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 $n$.
Factor the right-hand side to expose the original $x$. Both cosine and sine pick up a factor of $-(2)^2 = -4$, leaving $-4x$.
3
So $\ddot{x} = -4x$, matching $\ddot{x} = -n^2 x$ with $n^2 = 4$, hence $n = 2$. The motion is SHM with angular frequency $2$ rad s$^{-1}$ and period $T = 2\pi/2 = \pi$ s.
Compare coefficient-by-coefficient. $n$ is read off the squared coefficient that emerged from the chain rule.
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.
1
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 $-x$.
2
$n = \sqrt{64} = 8$ rad s$^{-1}$.
$n$ is positive by convention. Units check: $\sqrt{\text{N m}^{-1}/\text{kg}} = \sqrt{\text{s}^{-2}} = \text{s}^{-1}$, as required for angular frequency.
3
$T = 2\pi/n = 2\pi/8 = \pi/4$ s $\approx 0.785$ s. $f = 1/T = 4/\pi$ Hz $\approx 1.27$ Hz.
Memorise both $T = 2\pi/n$ and $f = n/(2\pi)$. Exam answers usually want both an exact and a decimal.
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.
1
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 $n^2$.
2
$n = 3$ rad s$^{-1}$. General solution: $x(t) = A\cos(3t) + B\sin(3t)$ for constants $A$, $B$.
Write the two-constant form straight from $n$. Initial conditions (covered in L06) fix $A$ and $B$.
3
Period $T = 2\pi/n = 2\pi/3$ s. Equivalent form: $x(t) = a\cos(3t + \alpha)$ with $a = \sqrt{A^2 + B^2}$.
Always state both the period and the single-cosine form when asked to "describe" the motion, they make amplitude and timing explicit.
Fill the gap: A particle satisfies $\ddot{x} = -25x$. Then $n = $ rad s$^{-1}$, and the period is $T = 2\pi/$ seconds.
6
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
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.
7
Revisit your thinking
Activities · practice with the ideas
Activities · practice with the ideas
1
Verify that $x(t) = 5\sin(4t)$ satisfies $\ddot{x} = -n^2 x$. State $n$ and the period.
2
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.
3
Show by direct differentiation that $x(t) = a\cos(nt + \alpha)$ satisfies $\ddot{x} = -n^2 x$ for any constants $a$, $\alpha$.
4
Rewrite $x = 3\cos(2t) + 4\sin(2t)$ in the form $a\cos(2t + \alpha)$, stating $a$ and $\tan\alpha$.
5
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?
11
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.
auto-saved
1
Show what you have learned
Multiple choice, then short answer under exam conditions.
01
Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
02
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)
auto-saved
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)
auto-saved
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)
auto-saved
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.
