SHM differential equation — solution - x = A cos(ωt + φ): Question bank

Before we start, one reminder of the vocabulary so nothing below sneaks a symbol past you:

$x$ = displacement from the centre (metres), measured $+$ one way, $-$ the other.
$ω$ = angular frequency (rad/s), a fixed number baked in by the physics ( $k / m$ for a spring).
$A$ = amplitude, the farthest $x$ ever reaches.
$ϕ$ = phase, the "head start" that tells you where in the wiggle the clock started.
$\overset{x}{¨}$ = acceleration (the second time-derivative of $x$ ); the defining law is $\overset{x}{¨} = - ω^{2} x$ .

True or false — justify

Every extra oscillation cycle takes exactly as long as the previous one.

True — the period

T = 2 π / ω

depends only on

ω

, and

ω

never changes during ideal SHM, so cycles are identical clockwork.

Doubling the amplitude halves the period.

False —

T = 2 π / ω

contains no

A

at all, so period is untouched. The bigger swing is faster (larger

v_{m a x} = A ω

), covering the extra distance in the same time.

At the exact centre (

x = 0

) the acceleration is zero.

True —

a = - ω^{2} x

, and

x = 0

makes

a = 0

. The restoring pull is zero at home because you have nowhere to be pulled back from.

At the exact centre the velocity is zero.

False — the centre is where speed is maximum (

v_{m a x} = A ω

). Zero speed happens only at the turning points

x = \pm A

x = A sin (ω t + ϕ)

describes a fundamentally different motion than

x = A cos (ω t + ϕ)

False — sine is just cosine shifted by a phase (

sin θ = cos (θ - π /2)

), so it is the same SHM started at a different point in the cycle; absorb the shift into

ϕ

If two identical springs have masses

m

and

2 m

, the heavier one oscillates faster.

False —

ω = k / m

, so more mass means smaller

ω

and a longer period; the heavier mass is more sluggish.

The graph of acceleration versus displacement for SHM is a straight line through the origin.

True —

a = - ω^{2} x

is linear in

x

with slope

- ω^{2}

; a straight line through the origin with negative slope is the signature of SHM.

Velocity and displacement reach their maxima at the same instant.

False — they are a quarter-cycle apart:

v

peaks at the centre (

x = 0

) while

x

peaks at the ends where

v = 0

. They are

9 0^{\circ}

out of phase — see Phase and phase difference.

The total energy of an SHM oscillator changes as it moves between centre and extreme.

False — energy just trades between kinetic and potential; the total is constant, which is exactly the conserved quantity Step 1 of the derivation uncovered. See Energy in SHM.

Spot the error

"Since

v_{0} = 0

at release, the phase must be

ϕ = 0

Incomplete —

v_{0} = 0

only tells you the start is a turning point (

cos ϕ = \pm 1

). You still must check the sign of

x_{0}

x_{0} > 0

gives

ϕ = 0

, but

x_{0} < 0

gives

ϕ = π

"I plugged

f = 2

Hz straight into

cos (f t)

Wrong variable — the argument uses angular frequency

ω = 2 π f

, not

f

. Using

f

undercounts the rotation by a factor of

2 π

, so the motion would appear far too slow.

A = x_{0}^{2} + v_{0}^{2}

from the initial conditions."

Units clash — you can't add a length

^{2}

to a velocity

^{2}

. Velocity must be divided by

ω

first:

A = x_{0}^{2} + v_{0}^{2} / ω^{2}

, so both terms are lengths squared.

"The solution is

x = A cos (ω t)

, full stop — one equation, one answer."

Missing a constant —

\overset{x}{¨} = - ω^{2} x

is second-order and demands two arbitrary constants. Dropping

ϕ

throws away the information about where in the cycle you began.

"Max acceleration and max speed happen at the same place."

No —

a_{m a x} = A ω^{2}

occurs at the ends (

x = \pm A

, strongest pull), while

v_{m a x} = A ω

occurs at the centre (

x = 0

, zero pull). They are at opposite locations.

"Because

tan ϕ = - v_{0} / (ω x_{0})

, I just take

arctan

and I'm done."

Not quite —

tan

repeats every

18 0^{\circ}

, so

arctan

alone can't tell

ϕ

from

ϕ + π

. You must use the sign of

x_{0}

(fixes

cos ϕ

) or

v_{0}

(fixes

sin ϕ

) to pick the correct quadrant.

"A pendulum swung to

6 0^{\circ}

obeys

x = A cos (ω t + ϕ)

with

ω = g / L

Only at small angles — the neat formula assumes the restoring force is linear in displacement, which is a small-angle approximation. At

6 0^{\circ}

the motion is no longer true SHM. See Simple pendulum.

Why questions

Why does the SHM solution have to be a sinusoid and not, say, a parabola?

Because the law

\overset{x}{¨} = - ω^{2} x

says the second derivative is a flipped, scaled copy of the function itself — and cosine/sine are the only functions with that self-referential property; a parabola's second derivative is a constant, not proportional to itself.

Why do we divide

v_{0}

ω

when computing the amplitude?

To convert a speed into a length so it can be compared with position: since

v_{m a x} = A ω

, dividing any velocity by

ω

yields the equivalent displacement (metres), letting you add it under one square root.

Why does the minus sign in

\overset{x}{¨} = - ω^{2} x

matter so much?

The minus is what points the acceleration back toward home; with a plus sign (

\overset{x}{¨} = + ω^{2} x

) the solution would be growing exponentials that fly off to infinity — no oscillation at all.

Why does an SHM system need exactly two initial conditions?

Because the equation involves a second derivative, integrating it twice introduces two unknown constants; you pin them down with two facts — where you start (

x_{0}

) and how fast (

v_{0}

) — encoded as

A

and

ϕ

Why is the max speed at the centre even though the force there is zero?

The particle has spent the whole inward journey being accelerated by the restoring force; by the time the force drops to zero at the centre, all that pushing has accumulated into maximum speed — force zero, but velocity peak.

Why can we picture SHM as the shadow of something going round a circle?

A point on a circle rotating at steady rate

ω

has a horizontal projection

A cos (ω t + ϕ)

— literally our solution — so SHM is uniform circular motion seen edge-on. See Reference circle and SHM as projection.

Why does the restoring force being proportional to displacement (not just opposing it) guarantee a constant period?

Only strict proportionality makes acceleration

= - ω^{2} x

with a single constant

ω

; that single

ω

sets one fixed period independent of how far you pull, which is why SHM is isochronous.

Edge cases

What is the motion if you place the mass exactly at

x = 0

with zero velocity?

It stays put forever — with

x_{0} = 0

and

v_{0} = 0

the amplitude is

A = 0

, so there is no oscillation; equilibrium is a valid (trivial) solution.

What happens to the period as the spring constant

k \to 0

ω = k / m \to 0

, so

T = 2 π / ω \to \infty

— an infinitely floppy spring gives an infinitely long period, i.e. no return force, no oscillation.

If you start at maximum displacement with the largest possible push outward, what happens?

You cannot —

x = A

is by definition the turning point where speed is zero; any outward velocity there would carry you past

A

, contradicting that

A

is the maximum. The amplitude simply grows to accommodate whatever energy you inject.

At the turning points

x = \pm A

, what are the velocity and acceleration?

Velocity is zero (momentarily at rest) while acceleration is at its maximum magnitude

A ω^{2}

pointing back to centre — the instant of standstill is the instant of strongest pull.

What does the solution predict as

t \to \infty

for ideal (undamped) SHM?

It oscillates forever with unchanged amplitude

A

— no energy is lost. Real systems shrink over time because of friction; that requires an extra term, giving Damped oscillations.

If the restoring force came from stretch alone (a rubber band that goes slack when compressed), is the motion still SHM?

No — a band that only pulls when stretched gives a force that isn't a single linear

- k x

across all

x

, so

\overset{x}{¨} = - ω^{2} x

fails on the slack side and the motion is not pure SHM. Compare with Hooke's Law and the spring force, which assumes a two-sided linear spring.