Oscillations & simple harmonic motion

The short answer

Work out where an oscillator is fastest, where it accelerates hardest, and what does — and does not — change its period.

Written and checked by GAMSAT tutors — not AI-generated.

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Try the reasoning style

Section I · Humanities & Social SciencesIllustrative example

We treat forgetting as a failure — a lapse to be patched with reminders and records. Yet a mind that kept everything could not think; it would drown in the undifferentiated noise of every moment it had ever lived. To forget is not so much to lose information as to decide, mostly without our noticing, what was never worth keeping.

The author's argument relies most directly on which unstated assumption?

Pick an option to see how the tutor reasons to the answer — not just whether you were right.

How to reason to the answer

Not quite — the answer is B.

Work backwards from the conclusion: a mind that ‘kept everything’ supposedly ‘could not think.’ That only follows if thinking means leaving most of experience out — so B is the premise the argument quietly rests on. A raises reliability, which the passage never weighs; C contradicts ‘mostly without our noticing’; D smuggles in a claim about intellect the passage never makes. The question rewards finding the hidden premise, not recalling a fact.

Section III hands you an oscillator you have never met — a vibrating bond, a buoy on a swell — and asks what happens when one thing about it changes. The question is almost always proportional.

The whole topic in one line

$a = - ω^{2} x$ , with $ω = 2 π / T$ : acceleration proportional to displacement, pointing the other way. A stimulus announces it: the restoring force is proportional to displacement, $F = - k x$ . So the acceleration is not constant and SUVAT never applies, and setting $a = - (k / m) x$ beside that line gives $T = 2 π m / k$ .

displacement
velocity
acceleration

Displacement, velocity and acceleration through one cycle (period 4.0 s)

One cycle, released from rest at maximum displacement. Velocity leads displacement by a quarter cycle; acceleration is the displacement trace flipped, because a = −ω²x. The marked speed peak sits at a trough: speed is size, not sign.

Reading a displacement–time trace

Velocity is the gradient

Read steepness, don't differentiate. Steepest at the axis crossings, so the object is fastest there; flat at crest and trough, so the velocity is zero; sloping downwards, so the velocity is negative.

Acceleration is the trace flipped

Because $a = - ω^{2} x$ , flip the displacement trace about the axis and rescale.

Then scale, rather than re-derive

Peak speed is $ω A$ , peak acceleration $ω^{2} A$ . Doubling the amplitude doubles both; halving the period doubles $ω$ , so peak speed doubles and peak acceleration quadruples.

The trap: fastest and hardest-accelerating are never the same place

At the two ends the object is momentarily stationary — exactly where the acceleration is greatest; at the centre it is fastest, and there the acceleration is zero. Acceleration always points back towards equilibrium.

One change at a time, with every other quantity held fixed. Mass on a spring:

T = 2 π m / k

. Simple pendulum:

T = 2 π L / g

Change made	Mass on a spring	Simple pendulum
Amplitude doubled	unchanged	unchanged
Oscillating mass × 4	doubled	unchanged
Spring constant × 4	halved	not applicable
String length × 4	not applicable	doubled
Taken to the Moon (g ÷ 6)	unchanged	longer, by a factor of about 2.4

Both formulas weigh inertia against restoring stiffness. Amplitude is in neither (for a pendulum, while the arc stays small): twice as far to travel, but twice as fast everywhere, so the same time. Amplitude changes the energy instead, which goes as A². Moon row: dividing g by six lengthens the period by √6 ≈ 2.4, not by 6.

Worked example

Isolator 1 carries a 2.0 kg instrument on a spring of stiffness $k$ , with a period of 0.60 s. Isolator 2 carries 8.0 kg on stiffness $2 k$ . Which is slower, by what factor — and at equal displacement, which takes the greater peak acceleration?

Check yourself

A simple pendulum swinging through a small arc takes 2.0 s to complete one full oscillation, over and back. For the next run the bob is replaced with one four times as heavy, the string is shortened to one quarter of its original length, and the bob is released from twice its previous (still small) angle. The new period is closest to:

Key takeaways

$F = - k x$ defines it, hence $a = - ω^{2} x$ . Acceleration is not constant, so SUVAT never applies.
Fastest at the centre where acceleration is zero; hardest-accelerating at the ends where speed is zero.
On a trace: velocity is the gradient, acceleration is the trace flipped. Peak speed $ω A$ , peak acceleration $ω^{2} A$ .
Period ignores amplitude; energy does not. Spring: mass matters. Pendulum: mass cancels.
Reason in ratios: the square root turns a four-fold change into a doubling.

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