Simple Harmonic Motion
Welcome to Oscillations & Waves. The motion in the topics so far went somewhere: along a straight line, along the curve of a projectile, or round and round a circle. This topic follows bodies that go back and forth about one place: a ball on a spring, a pendulum, a point on a guitar string, the air in front of a loudspeaker. A motion that repeats itself back and forth like this is an oscillation, also called a vibration, and a wave is an oscillation handed on from one point of a material to the next. This page sets up the simplest oscillation, a ball on a spring, and the quantities that describe it: the amplitude, the period and the frequency. The second page reads its graph and follows its energy, the third turns to pendulums and resonance, and the last two turn to waves.
The topic has had two or three questions in every paper so far, 9 of the 192. Two of them belong to this page and are the same question with the numbers swapped: the frequency of a spring oscillator from the time it takes for one full oscillation (December 2025, Q21; April 2026, Q21). A third, January 2026 Q18, asks which of four statements about an oscillator's amplitude, period, frequency and path is true. All three are worked through below. They need only f = 1/T and a clear picture of one oscillation, and the marks lost on them come from a few slips, which this page names as they come up: the period given as the frequency, the distance between the two ends taken as the amplitude, and a quarter of a period assumed to cover one amplitude from any starting point.
The spring oscillator
Fix one end of a light spring to a wall, attach a small ball to the other end, and rest the ball on a smooth horizontal surface. Left alone, the ball sits still at the point where the spring has its natural length, neither stretched nor compressed. Call this point O. Up and down, the ball's weight and the normal force balance; along the surface the only force is the spring's, and at O the spring exerts none. O is the equilibrium position: the place where the ball can stay at rest.
Now pull the ball to the right, to a point N, and let go. The stretched spring pulls it back toward O, and it speeds up. At O the spring's force is zero, but the ball is moving, and by the first law nothing is needed to keep it moving: it runs on past O. Beyond O the spring is compressed and pushes back toward O, so the ball slows down and stops for an instant at a point M on the left. On a smooth surface nothing takes energy away, and M turns out to be exactly as far from O as N is; the next page shows why. Then the spring pushes the ball back through O to N, and the whole trip repeats. A ball on a spring moving like this is a spring oscillator.
To say where the ball is, measure its position from O, taking right as positive. This is its displacement from the equilibrium position, x. At N, x is positive; at M it is negative; at O it is zero. It says how far the ball is from O and on which side, not how far the ball has travelled.
What pulls the ball back is the spring, and Hooke's law from Weight, Springs & Friction gives the size of its force. O is where the spring has its natural length, so when the ball is at displacement x the spring's change in length is |x|, and its force has size k|x|, where k is the spring constant. The direction depends on the side. To the right of O, x is positive, the spring is stretched and it pulls the ball to the left. To the left of O, x is negative, the spring is compressed and it pushes the ball to the right. Either way the force points back toward O, against the displacement. With right taken as positive, one equation says all of this:
F = −kx.
Check the signs. At x = +2 cm, F comes out negative: a force to the left, toward O. At x = −2 cm, F comes out positive: a force to the right, toward O again. The minus sign carries the physics: the force always points opposite to the displacement. A force that always points back toward the equilibrium position is called a restoring force. By the second law the ball's acceleration is a = F/m = −kx/m, so it too points toward O and grows in step with the distance from O.
Three things follow straight from F = −kx. At O, x = 0, so the restoring force and the acceleration are zero. At the two ends, where the ball is farthest from O, they are largest. And the force depends only on where the ball is, not on which way it is moving: at a point 2 cm to the right of O the ball feels the same pull to the left whether it is on its way out or on its way back.
That last point explains the shape of the motion. While the ball moves away from O, the force points against its motion, so it slows down, and it stops for an instant at the end. While it moves back toward O, the force points along its motion, so it speeds up. The ball is therefore fastest as it passes O and at rest for an instant at each end. The next page makes this exact with the graph of the motion and its energy; this page needs only the picture.
A spring oscillator has a spring constant of 100 N/m and a 0.5 kg ball. Find the restoring force on the ball and its acceleration when it is 4 cm to the left of O.
Step 1, the displacement in metres, with right as positive: x = −4 cm = −0.04 m.
Step 2, the force: F = −kx = −100 × (−0.04) = +4 N, so 4 N to the right, toward O. The spring is compressed by 4 cm and pushes the ball back.
Step 3, the acceleration: a = F/m = 4 ÷ 0.5 = 8 m/s², to the right.
The question did not say which way the ball was moving, and it did not need to. The force and the acceleration are the same on the way out and on the way back.
Amplitude and one full oscillation
The ball swings out to N on one side and to M on the other, the same distance from O each time. That largest distance from O is the amplitude, A: A = ON = OM. The two ends are therefore 2A apart, MN = 2A, so the amplitude is half the distance between the ends. The displacement keeps changing, anywhere from −A to +A, while the amplitude stays fixed: it is a size, always positive, and it measures how strongly the ball oscillates.
A full oscillation is one complete cycle of the motion: the ball comes back to the same point, moving the same way. Starting at N, that is N → O → M → O → N. Starting at O and moving right, it is O → N → O → M → O, ending at O moving right again. Two slips hide in this definition. Going from one end to the other, N to M, is only half an oscillation; the ball still has to come back. And the ball passes O twice in every full oscillation, once in each direction, so one pass through O is not a full oscillation either.
Period and frequency
Two more quantities describe how fast the ball oscillates. The period, T, is the time for one full oscillation, in seconds. The frequency, f, is the number of full oscillations per second, in hertz (Hz): 1 Hz is one oscillation per second, so 1 Hz = 1 s⁻¹. You met both in the Circular Motion & Gravitation topic, as the time for one turn and the number of turns per second, and they are linked in the same way here.
If one oscillation takes T seconds, then one second holds 1/T of them. An oscillator with T = 0.25 s makes 4 oscillations every second, so f = 4 Hz; one with T = 4 s makes only a quarter of an oscillation each second, so f = 0.25 Hz. So f = 1/T, and equally T = 1/f. A long period goes with a low frequency, and a short period with a high one.
One oscillation is too quick to time well by hand, so the period is usually measured over many. If the ball makes n full oscillations in a time t, each takes T = t/n, and f = n/t. Timing 20 oscillations instead of one makes the error from starting and stopping the watch 20 times smaller for each oscillation.
What sets the period? Not the amplitude. Pull the ball out 4 cm instead of 2 cm. Every point of the new path is twice as far from O as the matching point of the old one, so by F = −kx the force there, and the acceleration, are twice as large too. Twice the acceleration all the way along covers twice the distance in the same time, so the period is unchanged. The period is set by the oscillator itself. A heavier ball gets less acceleration from the same force and moves more sluggishly, so its period is longer; a stiffer spring pulls harder at the same displacement, so its period is shorter. This fixed period, and the frequency that goes with it, are the oscillator's natural period and natural frequency, which the third page uses.
A spring oscillator undergoes a simple harmonic motion. It completes one full oscillation in 2 s. Its frequency is ( ).
A. 0.5 Hz B. 0.2 Hz C. 5 Hz D. 2 Hz
Step 1, the period: the time for one full oscillation is the period, so T = 2 s.
Step 2, the frequency: f = 1/T = 1 ÷ 2 = 0.5 Hz. The answer is A.
Check: 0.5 Hz means half an oscillation each second, so one full oscillation every 2 s, as the stem says.
D, 2 Hz, is the trap: it copies the period across and gives it the unit of frequency. A period is a time and a frequency is a rate, and only when T = 1 s do the two share a number. B, 0.2 Hz, and C, 5 Hz, do not follow from any common slip.
The path: 4A in every period
Many questions ask how far the ball travels in a given time. This is its path, or distance travelled: the total length covered, counted whichever way the ball is going. It is not the displacement from O, which only says where the ball is. The Describing Motion topic drew the same line between distance and displacement.
Follow one full oscillation from N. The ball runs four legs, N to O, O to M, M to O and O to N, and each leg joins O to an end, so each is A long. In one period the ball travels 4A. Its displacement at the finish is the same as at the start: after a full oscillation it is back where it began, yet it has travelled 4A to get there.
The four legs also take equal times. The force law is the same on both sides of O, only reversed, so the trip from O out to M is the mirror image of the trip from O out to N, and each trip back is the trip out run in reverse. Each leg therefore takes a quarter of a period, T/4.
Half a period covers 2A, wherever the timing starts. The same mirror symmetry means that half a period after the ball is at some point, it is at the mirror point on the other side of O, moving the opposite way. Say the ball starts a distance d from O, heading out toward N. It travels A − d to reach N, then A + d back past O to the mirror point: (A − d) + (A + d) = 2A. Start it heading toward O instead, and it travels d + A to reach M, then A − d back out of the turn: 2A again. The starting point cancels out.
A quarter of a period is different. From O or from an end it covers exactly A, one whole leg. From anywhere else it does not, because the ball's speed changes along its path. A quarter period that carries the ball through O, where it moves fastest, covers more than A; one that carries it round an end, where it slows to a stop, covers less.
A ball on a spring oscillates with an amplitude of 3 cm and a period of 0.4 s. How far does it travel in 1.4 s? If it started at an end, how far is it from its starting point at the finish?
Step 1, the time in periods: 1.4 ÷ 0.4 = 3.5, so 1.4 s is 3 full periods and a half.
Step 2, the path: 3 periods cover 3 × 4A = 12A, and the half period covers 2A from any start, so the path is 14A = 14 × 3 = 42 cm. The starting point did not matter, because nothing shorter than a half period was left over.
Step 3, where it finishes: after 3 full periods the ball is back at the end it started from, and the extra half period carries it to the other end. It finishes 2A = 6 cm from its starting point, although it has travelled 42 cm.
Statements about one oscillator
January 2026, Q18 put the whole page into one question. It gives the distance between the ends and a number of oscillations in a timed interval, and asks which of four statements is true. The work is the same every time: find A, T and f first, then test each statement against them.
The paper printed AB = 1 cm and 5 full oscillations in 60 s, which make no option true, so the numbers below are corrected to 8 cm and 30 oscillations.
As shown in the figure, a spring oscillator performs a simple harmonic motion between A and B, with O being the equilibrium position. The distance between A and B is measured to be 8cm, and the ball completes 30 full oscillations in 60 seconds. Thus,
A. If the timing starts when the ball passes point O, the distance traveled within 3 seconds is 24cm
B. The ball travels 12cm in one complete oscillation
C. The period of the oscillator is 2s, with an amplitude of 6cm
D. The frequency of the oscillator is 2Hz
The paper's figure is the oscillator drawn at the top of this page: a wall on the left, a ball on a horizontal spring, and the ends called A, on the left, and B, on the right.
Step 1, the three numbers: the amplitude is half the distance between the ends, 8 ÷ 2 = 4 cm. The period is T = t/n = 60 ÷ 30 = 2 s, and the frequency is f = 1/T = 0.5 Hz.
Step 2, the statements. A: 3 s is 1.5 periods, one full period and a half, so the path is 4A + 2A = 6A = 24 cm. True. Starting the timing at O does no harm, but it is not even needed, because nothing shorter than a half period is left over. B: one full oscillation covers 4A = 16 cm, not 12 cm. C: the period is right, but the amplitude is 4 cm, not 6 cm. D: the frequency is 0.5 Hz, not 2 Hz. Only A is true, so the answer is A.
B, 12 cm, is only three of the four legs, 3A. C is only half true: a statement with two parts is false if either part is false, so check both before you choose it. D is the period, 2 s, given as the frequency, the same slip as in the December and April Q21.
Your turn! · Practice questions
Things to remember
| Concept | What to know |
|---|---|
| Spring oscillator | A ball on a spring on a smooth surface. O, the equilibrium position, is where the spring has its natural length and the ball can stay at rest. The displacement x is measured from O. |
| Restoring force | F = −kx: proportional to the distance from O and always pointing back toward O. a = −kx/m. Both are zero at O and largest at the ends. |
| Simple harmonic motion | Motion under a restoring force F = −kx. The ball slows on the way out, stops for an instant at each end, and is fastest as it passes O. |
| Amplitude | A, the largest distance from O. The ends are 2A apart, so A is half the distance between the ends. |
| Full oscillation | Back to the same point, moving the same way: N → O → M → O → N. The ball passes O twice. One end to the other is only half. |
| Period and frequency | T in s, f in Hz, f = 1/T. n full oscillations in t: T = t/n. December 2025 Q21: T = 2 s gives 0.5 Hz; April 2026 Q21: T = 0.5 s gives 2 Hz. |
| What sets the period | The ball's mass and the spring's stiffness, not the amplitude. A heavier ball: longer T; a stiffer spring: shorter T. |
| Path | 4A per period; 2A per half period from any start; A per quarter period only from O or an end. Take out whole and half periods first. The path is not the displacement. |
| Statement questions | Find A, T and f first, then test each statement. January 2026 Q18, corrected: A = 4 cm, T = 2 s, f = 0.5 Hz, and 3 s = 1.5T gives 6A = 24 cm. |
| The wrong options | The period given as the frequency; the distance between the ends taken as A; one end to the other counted as a full oscillation; A per quarter period from any start; a two-part statement that is only half true. |