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Module 16 · L07 of 16 ~40 min ⚡ +90 XP available

SHM — Energy and Velocity

In simple harmonic motion, kinetic energy and potential energy trade off perfectly while their sum stays constant. From that single principle drops out the velocity formula $v^2 = n^2(a^2 - x^2)$ — the most useful equation in the SHM toolkit. With it you can read off speed at any displacement, locate where the particle is when given a speed, and check answers without integrating $\ddot{x} = -n^2 x$ a second time.

Today's hook — A particle moves in SHM with amplitude $a = 4$ m and $n = 3$. Before reading on, predict (a) its maximum speed and where it occurs, (b) its speed at $x = 2$. Sketch a quick graph of $v^2$ vs $x$ — what shape is it? Check after card 05.

0/5QUESTS

Recall — your gut answer first

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A particle undergoes SHM about $x = 0$ with $\ddot{x} = -n^2 x$. Before checking — where in the motion is the particle moving fastest, and where is it momentarily at rest? Try to give a one-line reason in terms of energy.

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The two moves for SHM energy problems

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Every SHM energy question rewards two habits: identify the amplitude $a$ and angular frequency $n$ from the equation of motion, then use $v^2 = n^2(a^2 - x^2)$ rather than re-integrating $\ddot{x} = -n^2 x$. Almost every velocity-at-displacement question collapses to a single substitution into this formula.

The read-identify-substitute workflow: (1) read off $n^2$ from $\ddot{x} = -n^2 x$, (2) identify amplitude $a$ from initial conditions or extremes, (3) substitute into $v^2 = n^2(a^2 - x^2)$.

$\ddot{x} = -n^2 x$ · $v^2 = n^2(a^2 - x^2)$ · $v_{\max} = na$ at $x = 0$

$v^2 = n^2(a^2 - x^2)$

Sign of $v$ is lost

$v^2 = n^2(a^2 - x^2)$ gives only the magnitude. The direction is determined by where the particle is in its cycle — use $x(t)$ or context to choose the sign.

Energy is constant

$\tfrac12 v^2 + \tfrac12 n^2 x^2 = \tfrac12 n^2 a^2$ (per unit mass) — the total energy never changes during SHM. Set up KE + PE = constant and read off any unknown.

$|x| \leq a$ always

If a question yields $x > a$ or $v^2 < 0$, you've made an error. The motion is confined to $-a \leq x \leq a$ and $v^2 \geq 0$ throughout.

What you'll master

Know

Key facts

$v^2 = n^2(a^2 - x^2)$ for SHM with $\ddot{x} = -n^2 x$ and amplitude $a$
KE per unit mass: $\tfrac12 v^2$; PE per unit mass: $\tfrac12 n^2 x^2$
Total energy $E = \tfrac12 n^2 a^2$ (per unit mass) is constant
$v_{\max} = na$ at $x = 0$; $v = 0$ at $x = \pm a$

Understand

Concepts

Why $v^2 = n^2(a^2 - x^2)$ falls out of energy conservation
Why the graph of $v^2$ vs $x$ is a downward parabola
Why KE and PE are out of phase but their sum is constant

Can do

Skills

Compute $v$ at any given $x$ given $n$ and $a$
Find $x$ where the particle has a given speed
Use KE + PE = constant to find amplitude from one (x, v) pair

Key terms

Simple harmonic motion (SHM)Motion satisfying $\ddot{x} = -n^2 x$, with restoring acceleration proportional to displacement from a fixed centre.

Amplitude $a$Maximum displacement from the centre. The motion is confined to $-a \leq x \leq a$.

Angular frequency $n$The constant in $\ddot{x} = -n^2 x$. Period $T = 2\pi/n$. Note: $n$ here is not an integer — it is a positive real.

Velocity formula$v^2 = n^2(a^2 - x^2)$. Gives speed at any displacement once $n$ and $a$ are known.

Kinetic energy (per unit mass)$\tfrac12 v^2$. Maximum at the centre $x = 0$; zero at the extremes $x = \pm a$.

Potential energy (per unit mass)$\tfrac12 n^2 x^2$. Zero at the centre; maximum at the extremes. Out of phase with KE.

MEX-M1NESA outcome (Applications of Calculus to Mechanics): solves problems involving the motion of a particle, including SHM, projectile motion, and resisted motion.

Deriving $v^2 = n^2(a^2 - x^2)$ from energy

core concept

Start from $\ddot{x} = -n^2 x$. Using the chain-rule identity $\ddot{x} = v\,\dfrac{dv}{dx}$:

$v\,\dfrac{dv}{dx} = -n^2 x$
Integrate: $\dfrac{v^2}{2} = -\dfrac{n^2 x^2}{2} + C$, so $v^2 = -n^2 x^2 + 2C$.
At the extreme $x = a$, $v = 0$, giving $2C = n^2 a^2$.
Therefore $v^2 = n^2 a^2 - n^2 x^2 = n^2(a^2 - x^2)$. $\quad\blacksquare$

Energy reading. Rearrange: $\tfrac12 v^2 + \tfrac12 n^2 x^2 = \tfrac12 n^2 a^2$. The left side is KE + PE (per unit mass); the right side is the total mechanical energy. Conservation of energy is exactly the statement that this sum is constant throughout the motion.

Worked through the hook: $a = 4$, $n = 3$.

(a) $v_{\max} = na = 3 \times 4 = 12$ m/s, at $x = 0$.
(b) At $x = 2$: $v^2 = 9(16 - 4) = 108$, so $v = \sqrt{108} = 6\sqrt{3} \approx 10.39$ m/s.
The graph of $v^2$ vs $x$ is a downward parabola $v^2 = 9(16 - x^2)$, peaking at $(0, 144)$ and meeting $v^2 = 0$ at $x = \pm 4$.

Why this beats integration. If a question only asks for $v$ at a particular $x$, you do not need $x(t) = a\cos(nt + \varphi)$ — the energy formula gives the answer in one line. Save the trigonometric form for time-dependent questions.

Derivation: $\ddot{x} = v\,\dfrac{dv}{dx} = -n^2 x \Rightarrow v^2 = n^2(a^2 - x^2)$ · Energy: $\tfrac12 v^2 + \tfrac12 n^2 x^2 = \tfrac12 n^2 a^2$ · $v_{\max} = na$ at $x = 0$; $v = 0$ at $x = \pm a$ · Graph of $v^2$ vs $x$: downward parabola, vertex $(0, n^2 a^2)$, roots $\pm a$

Pause — copy $v^2 = n^2(a^2-x^2)$, its derivation from $a = v\,dv/dx$, $v_{\max} = na$ at $x=0$, and the $v^2$-vs-$x$ parabola (vertex $(0,n^2a^2)$, roots $\pm a$) into your book.

Quick check: A particle moves in SHM with $\ddot{x} = -16x$ and amplitude $a = 5$. What is its speed at $x = 3$?

KE and PE — the balance that stays constant

core concept

We just saw that $v^2 = n^2(a^2-x^2)$ comes from integrating $\ddot{x} = -n^2 x$ via $a = v\,dv/dx$, with $v_{\max} = na$ at $x = 0$ and $v = 0$ at the amplitude $x = \pm a$. That raises a question: what is the energy interpretation of $v^2 = n^2(a^2-x^2)$? This card answers it → KE $= \tfrac{1}{2}v^2$ and PE $= \tfrac{1}{2}n^2x^2$ (per unit mass) sum to the constant total energy $\tfrac{1}{2}n^2a^2$; they are exactly out of phase.

Per unit mass, define

Kinetic energy $\mathrm{KE} = \tfrac12 v^2$.
Potential energy $\mathrm{PE} = \tfrac12 n^2 x^2$.
Total energy $E = \mathrm{KE} + \mathrm{PE} = \tfrac12 n^2 a^2$ (constant).

Trade-off through the cycle:

At $x = 0$: KE = $\tfrac12 n^2 a^2$ (maximum), PE = 0.
At $x = \pm a$: KE = 0, PE = $\tfrac12 n^2 a^2$ (maximum).
At general $x$: PE rises like $x^2$ and KE falls so that the sum is fixed.

$$\tfrac12 v^2 + \tfrac12 n^2 x^2 \;=\; \tfrac12 n^2 a^2$$

Finding amplitude from one $(x, v)$ pair. If you know $n$ and you are told that the particle has speed $v_1$ at $x_1$, you can recover the amplitude: $a^2 = x_1^2 + v_1^2/n^2$. No second equation needed.

KE = $\tfrac12 v^2$ (per unit mass); PE = $\tfrac12 n^2 x^2$ (per unit mass) · Total energy $E = \tfrac12 n^2 a^2$ — constant throughout the motion · KE max at centre; PE max at extremes; they are exactly out of phase · Amplitude from one snapshot: $a^2 = x_1^2 + v_1^2/n^2$

Pause — copy KE $= \tfrac{1}{2}v^2$, PE $= \tfrac{1}{2}n^2x^2$, total $E = \tfrac{1}{2}n^2a^2$ (constant), the KE-max-at-centre / PE-max-at-extremes rule, and the one-snapshot formula $a^2 = x_1^2 + v_1^2/n^2$ into your book.

Did you get this? True or false: at the centre of the motion ($x = 0$) the kinetic energy is at its maximum and the potential energy is zero.

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

PROBLEM 1 · SPEED AT A GIVEN DISPLACEMENT

A particle moves in SHM with $\ddot{x} = -25x$, amplitude $a = 6$ m. Find its speed when $x = 4$ m, and its maximum speed.

Read $n^2 = 25$, so $n = 5$. Amplitude $a = 6$. Use $v^2 = n^2(a^2 - x^2)$.

First step in any SHM energy question: pull $n^2$ off the equation of motion and confirm the amplitude.

PROBLEM 2 · DISPLACEMENT FROM A GIVEN SPEED

A particle in SHM with $n = 2$ and amplitude $a = 5$ has speed $v = 6$ m/s. Find the displacement $x$ at this instant.

Use $v^2 = n^2(a^2 - x^2)$ and solve for $x^2$: $\;x^2 = a^2 - \dfrac{v^2}{n^2}$.

Rearrange the formula before substituting — the algebra is cleaner that way.

PROBLEM 3 · AMPLITUDE FROM ENERGY BALANCE

A particle in SHM with $n = 4$ has speed $v = 8$ m/s when $x = 3$ m. Find the amplitude and the maximum speed.

Use $v^2 = n^2(a^2 - x^2) \Rightarrow a^2 = x^2 + \dfrac{v^2}{n^2}$.

When the unknown is $a$, rearrange to isolate $a^2$ — this is the energy-balance form.

Fill the gap: The SHM velocity formula is $v^2 = n^2(a^2 - $ $)$. The maximum speed is $v_{\max} =$ at $x = 0$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Dropping $n^2$ in the formula

Writing $v^2 = a^2 - x^2$ instead of $v^2 = n^2(a^2 - x^2)$ is the single most common slip. The $n^2$ multiplier sets the energy scale — without it your answers will be a factor of $n$ off.

Trap 02

Forgetting both signs of $x$

The formula is symmetric in $x$ — if a given speed gives $x^2 = 16$, both $x = 4$ and $x = -4$ are valid unless time-direction or initial conditions exclude one.

Trap 03

Reaching for $x(t) = a\cos(nt)$ unnecessarily

If the question asks only about speed and position (not time), use $v^2 = n^2(a^2 - x^2)$. Going via the trig form and differentiating wastes time and invites algebra errors.

Did you get this? True or false: for SHM with $n = 3$ and amplitude $a = 4$, the maximum speed is $v_{\max} = 12$ and it occurs at $x = 0$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A particle in SHM has $\ddot{x} = -9x$ and amplitude $a = 5$. Find its speed at $x = 3$ and its maximum speed.

A particle in SHM with $n = 4$ and amplitude $a = 3$ has what speed at $x = 2$?

For SHM with $n = 2$ and amplitude $a = 6$, find the displacements at which the particle has half its maximum speed.

A particle in SHM with $n = 5$ has speed $v = 20$ m/s at $x = 3$ m. Find the amplitude and the maximum speed.

For SHM with amplitude $a$ and frequency $n$, find the displacement at which KE equals PE.

Odd one out: Three of these statements about SHM are correct. Which one is NOT?

Revisit your thinking

Earlier you predicted the speed at $x = 2$ for SHM with $a = 4$ and $n = 3$, and sketched $v^2$ vs $x$.

The energy formula gave $v_{\max} = na = 12$ at $x = 0$, and $v = 6\sqrt{3} \approx 10.39$ at $x = 2$. The graph of $v^2$ vs $x$ is a downward parabola $v^2 = 9(16 - x^2)$, peaking at $(0, 144)$ and crossing the $x$-axis at $\pm 4$. The whole power of $v^2 = n^2(a^2 - x^2)$ is that it lets you skip $x(t)$ entirely whenever the question is about speed and position rather than time.

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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 undergoes SHM with $\ddot{x} = -36x$ and amplitude $a = 4$. Find its speed at $x = 2$ and its maximum speed. (2 marks)

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ApplyBand 43 marks

Q2. A particle in SHM with $n = 3$ has speed $v = 12$ m/s when $x = 3$ m. Find the amplitude and state the speed at $x = 0$. (3 marks)

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AnalyseBand 53 marks

Q3. A particle moves in SHM with amplitude $a$ and angular frequency $n$. (a) Show, by integrating $\ddot{x} = -n^2 x$, that $v^2 = n^2(a^2 - x^2)$. (b) Find the displacement at which the kinetic energy equals the potential energy. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $n = 3$, $a = 5$. At $x = 3$: $v^2 = 9(25 - 9) = 144$, $v = 12$. $v_{\max} = na = 15$ at $x = 0$.

2. $n = 4$, $a = 3$. At $x = 2$: $v^2 = 16(9 - 4) = 80$, $v = 4\sqrt{5} \approx 8.94$ m/s.

3. $v_{\max} = na = 12$, so half is $v = 6$. Then $36 = 4(36 - x^2)$, giving $x^2 = 27$, $x = \pm 3\sqrt{3}$.

4. $a^2 = x^2 + v^2/n^2 = 9 + 400/25 = 25$, so $a = 5$. $v_{\max} = na = 25$ m/s.

5. KE = PE means $\tfrac12 v^2 = \tfrac12 n^2 x^2$, i.e. $v^2 = n^2 x^2$. Combined with $v^2 = n^2(a^2 - x^2)$ this gives $a^2 = 2x^2$, so $x = \pm a/\sqrt{2}$.

Q1 (2 marks): $n = 6$, $a = 4$. $v^2 = 36(16 - 4) = 432$, so $v = 12\sqrt{3}$ m/s [1]. $v_{\max} = na = 24$ m/s at $x = 0$ [1].

Q2 (3 marks): Use $v^2 = n^2(a^2 - x^2)$: $144 = 9(a^2 - 9)$ [1], $a^2 - 9 = 16$, $a = 5$ m [1]. $v_{\max} = na = 15$ m/s at $x = 0$ [1].

Q3 (3 marks): (a) $v\,dv/dx = -n^2 x \Rightarrow v^2/2 = -n^2 x^2/2 + C$; at $x = a$, $v = 0$ gives $C = n^2 a^2/2$, so $v^2 = n^2(a^2 - x^2)$ [1]. (b) Setting $\tfrac12 v^2 = \tfrac12 n^2 x^2$ gives $v^2 = n^2 x^2$ [1]; combine with $v^2 = n^2(a^2 - x^2)$: $a^2 - x^2 = x^2$, so $x = \pm a/\sqrt{2}$ [1].

Boss battle · The Energy Conductor

earn bronze · silver · gold

Five timed questions on $v^2 = n^2(a^2 - x^2)$, KE/PE balance, and amplitude recovery. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

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

Mark lesson as complete

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