Simple harmonic motion — definition, restoring force F = −kx: Question bank

Reminders you may lean on: $F = - k x$ (force opposes displacement), $a = - ω^{2} x$ , $ω = k / m$ , $T = 2 π m / k$ , $v = ω A^{2} - x^{2}$ , $E = \frac{1}{2} k A^{2}$ .

True or false — justify

Every claim below is either subtly right or subtly wrong. The reason matters more than the verdict.

All SHM is oscillation, and all oscillation is SHM.

False. SHM is a special oscillation requiring

F \propto - x

; a bouncing ball or large-angle pendulum oscillates but its force is not linear in

x

, so no constant period.

Doubling the amplitude doubles the period.

False.

T = 2 π m / k

has no amplitude in it — a wider pull gives both a larger force and a larger top speed that cancel exactly (isochronism). See Energy in SHM.

Doubling the amplitude doubles the maximum speed.

True.

v_{m a x} = A ω

is linear in

A

; with

ω

fixed by

k / m

, twice the amplitude means twice the peak speed.

Doubling the amplitude quadruples the total energy.

True.

E = \frac{1}{2} k A^{2} \propto A^{2}

, so

A \to 2 A

multiplies energy by

4

The minus sign in

F = - k x

means the force is a negative number.

False. It is a direction rule: force points opposite to displacement. At

x < 0

F = - k (neg) > 0

, i.e. positive, still pointing home.

At the turning points

x = \pm A

the acceleration is zero because the mass is momentarily at rest.

False. Velocity is zero there, but

a = - ω^{2} x

is maximum at

x = \pm A

— that peak force is exactly what reverses the motion.

x = 0

the mass has zero force but is not in equilibrium in the "nothing happening" sense.

True-ish.

F = 0

x = 0

so no instantaneous push, yet the mass streaks through at top speed — it does not stay because it carries momentum.

A heavier mass on the same spring always oscillates slower.

True.

T = 2 π m / k

grows with

m

; more inertia responds more sluggishly to the same restoring force.

A stiffer spring (larger

k

) on the same mass oscillates faster.

True.

ω = k / m

rises with

k

; a stronger pull-back for the same displacement means quicker cycles.

If you know

ω

, you know the period regardless of amplitude or phase.

True.

T = 2 π / ω

depends only on

ω = k / m

; amplitude

A

and phase

ϕ

set how big and when, never how long.

Spot the error

Each snippet contains one flawed step. Name it and fix it.

F = - k x^{3}

is SHM because the force still opposes displacement."

Error: SHM needs force proportional (linear) to

x

F \propto - x^{3}

opposes displacement and oscillates, but

T

depends on amplitude, so it is not SHM.

"Since

F = ma

and

F = - k x

, then

a = - k x

directly."

Error: dropped the mass.

ma = - k x \Rightarrow a = - (k / m) x = - ω^{2} x

; forgetting

m

gives wrong units and wrong

ω

U = \frac{1}{2} k x^{2}

so the potential energy is largest at

x = 0

Error:

\frac{1}{2} k x^{2}

is smallest (zero) at

x = 0

and largest at

x = \pm A

. The equilibrium is the bottom of the energy valley, not the top.

"Speed is greatest at

x = \pm A

because that's the biggest displacement."

Error:

v = ω A^{2} - x^{2}

vanishes at

x = \pm A

and peaks at

x = 0

. Big displacement means big potential, not big speed.

"Because

ω = k / m

, a bigger amplitude gives a bigger

ω

Error:

ω

contains only

k

and

m

, no

A

. Amplitude has zero effect on angular frequency.

x = A cos (ω t + ϕ)

can't be SHM because it starts at

A

, not

0

Error: the starting point is set by

ϕ

, not by whether it's SHM. Any

ϕ

still satisfies

a = - ω^{2} x

; cosine and sine are the same motion shifted in phase.

"Adding a constant gravitational pull to

F = - k x

(vertical spring) destroys SHM."

Error: gravity only shifts the equilibrium to a new point

x_{0} = m g / k

. About that new centre

F = - k (x - x_{0})

— still linear, still SHM, same

ω

Why questions

Explain the mechanism, don't just restate the rule.

Why does the same motion (a sine wave) show up for springs, pendulums, and atoms in a crystal?

Near any stable equilibrium the potential is a parabola

U \approx \frac{1}{2} k x^{2}

, so

F = - d U / d x = - k x

for all of them — the shared linear restoring force forces the shared motion. See Taylor Series.

Why must the SHM solution be a sine or cosine and nothing else?

The equation

a = - ω^{2} x

asks "which function returns to minus-itself after two derivatives?" — only

sin

and

cos

(and their combinations) do, so they are forced, not guessed.

Why does energy conservation let us find speed without solving for time?

E = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2}

links

v

directly to position

x

; rearranging gives

v = ω A^{2} - x^{2}

with no clock involved. Full derivation in Energy in SHM.

Why does a bigger initial pull not make the oscillation "run out of steam" sooner?

SHM is frictionless; energy

\frac{1}{2} k A^{2}

is conserved, so amplitude never decays. Only added dissipation (a

- b v

term) shrinks it — see Damped Oscillations.

Why is

x = 0

the fastest point even though nothing is pushing there?

All the spring's potential energy has converted to kinetic energy at

x = 0

; zero force means no change of speed at that instant, but the mass already carries its maximum speed.

Why does the projection of uniform circular motion trace out SHM?

A point going round a circle at constant

ω

has its shadow on one axis obey

x = A cos (ω t + ϕ)

— the same equation as SHM. See Uniform Circular Motion.

Why do we insist

k > 0

F = - k x

With

k > 0

the force pulls back toward equilibrium (stable, oscillates). If

k < 0

the force pushes away — the point is unstable and the object runs off, no oscillation at all.

Edge cases

Boundary and degenerate situations the standard formulas quietly assume.

What happens if the amplitude

A = 0

The mass sits at equilibrium forever:

x (t) = 0

v = 0

F = 0

E = 0

. It's the trivial (degenerate) solution — technically SHM with zero motion.

What if

k \to 0

(spring becomes floppy)?

ω = k / m \to 0

and

T \to \infty

: no restoring force, so no oscillation — the mass just drifts freely. SHM breaks down at the boundary.

What if

m \to 0

ω = k / m \to \infty

and

T \to 0

: an infinitely light mass would oscillate infinitely fast — physically a signal that inertia is what sets the timescale.

For a real spring pulled very far, why might

F = - k x

(and hence SHM) fail?

Real springs obey Hooke's Law

F = - k x

only within an elastic range; stretch too far and the force is no longer linear (it stiffens or permanently deforms), so the motion is no longer SHM. See Hooke's Law.

For a pendulum, at what point does the "SHM" description quietly break?

Only for small angles does

F \approx - (m g / L) x

(linear). At large swings the true force

\propto sin θ

is nonlinear, the period grows with amplitude, and it stops being SHM. See Simple Pendulum.

If the mass starts exactly at rest at equilibrium (

x = 0

v = 0

), what is the subsequent motion?

Nothing moves: with zero displacement there is zero force and zero speed, so

E = 0

and it stays put. Motion needs either an initial displacement or an initial velocity.

Is the restoring force linear in displacement, $F = - k x$ with $k > 0$ ? ::: If yes it's SHM (constant period, no amplitude dependence); if the force is nonlinear (like $- k x^{3}$ or large-angle $sin θ$ ) it may still oscillate but is not SHM.

Connections

Parent: SHM definition
Energy in SHM — energy reasoning behind the speed and amplitude traps
Hooke's Law — the elastic-limit edge case
Simple Pendulum — where the small-angle approximation dies
Damped Oscillations — why real amplitudes decay
Uniform Circular Motion — the projection view
Taylor Series — why every valley bottom is a parabola

Question bank — Simple harmonic motion — definition, restoring force F = −kx

True or false — justify

Spot the error

Why questions

Edge cases

Quick Recall

Connections