Defining Simple Harmonic Motion (SHM): Study Guide (Physics 1 7.1)

This guide is for the algebra-based Physics 1 course. You need Hooke’s law (Topic 2.8), Newton’s second law (Topic 2.5) and torque (Topic 5.3). Unit 7 uses no new laws of physics. It applies forces, torques and energy to motion that repeats.

Periodic motion and what makes SHM special

Many things move back and forth: a swing, a guitar string, a car on its suspension. Any motion that repeats itself in equal time intervals is periodic motion. The time for one full repeat is the period (Topic 7.2 deals with it in detail).

Simple harmonic motion (SHM) is one special kind of periodic motion, defined by the force that causes it. First, two definitions.

Equilibrium position and restoring force

The equilibrium position is the place where the net force on the object (or system) is zero. If you put the object there at rest, it stays at rest.

• For a block on a smooth horizontal surface attached to a spring, equilibrium is where the spring is relaxed.
• For a block hanging from a vertical spring, equilibrium is where the spring’s upward pull equals the weight. That is lower than the end of the relaxed spring (Worked example 2).
• For a pendulum, equilibrium is with the string hanging straight down.

Measure the displacement, x, from the equilibrium position, and state your axis, for example “+x to the right, x = 0 at equilibrium”.

A restoring force is a force that points opposite to the displacement, back toward equilibrium. Pull the block to +x and the restoring force points toward −x. Push it to −x and the restoring force points toward +x. It does not matter which way the object is moving at that instant.

A restoring force alone does not make motion simple harmonic. What matters is how its size changes with displacement.

The condition for SHM

An object moves in SHM when the size of the restoring force is proportional to the size of its displacement from equilibrium. With signs, along one axis:

F_net = −kx

Here k is a positive constant in N/m: the spring constant for a block on an ideal spring, or whatever constant links force and displacement in another system. The minus sign says the force is a restoring force. The direct proportion says that twice as far from equilibrium means twice the restoring force.

Now apply Newton’s second law, a = F_net / m:

a = −(k/m)x

This is the key relationship for this topic. Read it in words:

• The acceleration always points toward equilibrium (opposite sign to x).
• Its size is proportional to the distance from equilibrium.
• At equilibrium (x = 0) the acceleration is zero. At the ends of the motion, where the displacement is largest, the acceleration is largest.
• A larger k/m means a larger acceleration for the same displacement. Topic 7.2 shows this makes the oscillation faster.

You may also see this written as a = −ω²x, with ω² = k/m. It is the same relationship; Topic 7.2 links ω to the frequency.

A graph of a against x (or F_net against x) for SHM is a straight line through the origin with a negative slope. That is the easiest test for SHM from data.

SHM compared with other periodic motion

Each system below is periodic. Check the force before calling it SHM.

System	Restoring force	SHM?
Block on an ideal horizontal spring, no friction	F = −kx	Yes, for any displacement the spring can take
Block on a vertical spring	−k × (displacement from the new equilibrium)	Yes (Worked example 2)
Pendulum, small angle	about proportional to angle	Approximately (see below)
Block sliding in a frictionless V-shaped trough	constant size, mg sin θ along each side	No
Ball bouncing on a hard floor	constant weight mg in the air; a large brief push at the floor	No
Puck sliding between two walls	zero except at the walls	No

The pendulum: a restoring torque

A simple pendulum is a small, heavy bob on a light string of length ℓ, swinging from a fixed pivot. Its displacement is an angle, θ, measured from the vertical.

Take torques about the pivot. The tension in the string acts along the string, through the pivot, so its torque is zero. The weight mg acts at the bob. Its lever arm (the perpendicular distance from the pivot to the weight’s line of action) is ℓ sin θ. So the size of the torque is

τ = mgℓ sin θ

This torque always turns the bob back toward the vertical, so it is a restoring torque. With signs: τ = −mgℓ sin θ.

For SHM we need the restoring torque to be proportional to the angle, θ. It is proportional to sin θ instead. But for small angles, measured in radians, sin θ ≈ θ. Then

τ ≈ −mgℓθ

which is proportional to θ. So a pendulum with a small angular displacement can be modelled as SHM. Worked example 3 shows how small “small” is.

Worked example 1: is it SHM? Testing force data

Question. Take +x to the right, with x = 0 at equilibrium. A 0.40 kg glider sits on a level air track between two springs. A force sensor records the net horizontal force on it at five positions:

x (cm)	−6.0	−3.0	0	+3.0	+6.0
F_net (N)	+1.49	+0.76	0.00	−0.74	−1.51

(a) Do the data support the claim that the glider moves in SHM? (b) Find the glider’s acceleration when it is at x = +4.0 cm.

1. Direction. At every negative x the force is positive, and at every positive x it is negative. The force always points toward x = 0, so it is a restoring force.
2. Proportion. Convert to metres and find −F_net / x for each non-zero point: 24.8, 25.3, 24.7 and 25.2 N/m. They are all close to one constant, k ≈ 25 N/m. A straight line through (−0.060 m, +1.50 N) and (+0.060 m, −1.50 N) passes through the origin with slope −3.00 N ÷ 0.120 m = −25 N/m.
3. (a) The force is opposite to the displacement and proportional to it, F_net ≈ −(25 N/m)x. The data support the SHM claim, at least over ±6 cm.
4. (b) a = −(k/m)x = −(25 ÷ 0.40) × 0.040 = −62.5 × 0.040 = −2.5 m/s² (2.5 m/s² to the left, toward equilibrium).

Contrast. A 0.050 kg block slides without friction in a V-shaped trough whose sides slope at 10°. On either side, the net force along the slope has size mg sin 10° = 0.050 × 9.8 × 0.174 ≈ 0.085 N, whatever the distance from the bottom. The force is a restoring force, but its size does not grow with displacement (Figure 1b). The block’s motion is periodic, but not SHM.

Worked example 2: a vertical spring

Question. A 0.25 kg mass hangs from an ideal spring with k = 49 N/m. Take +y downward. (a) Find how far the spring is stretched at equilibrium. (b) The mass is pulled 3.0 cm below equilibrium. Find the net force and acceleration. (c) Show that the net force is −ky, where y is the displacement from equilibrium.

1. (a) At equilibrium the net force is zero, so the spring force balances the weight: kd = mg. d = mg ÷ k = (0.25 × 9.8) ÷ 49 = 2.45 N ÷ 49 N/m = 0.050 m (5.0 cm).
2. (b) The total stretch is now 5.0 + 3.0 = 8.0 cm. Spring force = 49 × 0.080 = 3.92 N upward. Weight = 2.45 N downward. Net force = 3.92 − 2.45 = 1.47 N upward, so F_net = −1.47 N with +y downward. a = 1.47 ÷ 0.25 ≈ 5.9 m/s² upward, toward equilibrium.
3. Check 3.0 cm above equilibrium. Stretch = 2.0 cm, spring force = 0.98 N up, weight 2.45 N down, net = 1.47 N downward. Same size, opposite direction: a restoring force again.
4. (c) At displacement y from equilibrium, the stretch is d + y. With +y downward: F_net = mg − k(d + y) = (mg − kd) − ky. Because kd = mg, the bracket is zero, so F_net = −ky.

Interpretation. Gravity only shifts the equilibrium down by mg/k. Measured from there, the net force has the same form as for a horizontal spring, so a vertical spring–mass system also moves in SHM. Measuring y from the relaxed spring instead gives −ky + mg, which is not proportional to y.

Worked example 3: how small is a small angle?

Question. A pendulum has length ℓ = 0.80 m and a 0.20 kg bob. Compare the true restoring torque, mgℓ sin θ, with the SHM model, mgℓθ, at 5°, 15°, 30° and 45°.

1. mgℓ = 0.20 × 9.8 × 0.80 = 1.568 N·m.
2. Convert each angle to radians (multiply by π/180) and work out both torques:

θ	θ (rad)	sin θ	true τ (N·m)	model τ (N·m)	model too large by
5°	0.0873	0.0872	0.137	0.137	0.1%
15°	0.2618	0.2588	0.406	0.411	1.2%
30°	0.5236	0.5000	0.784	0.821	4.7%
45°	0.7854	0.7071	1.109	1.232	11%

3. Conclusion. Up to about 15° the restoring torque is proportional to θ to within about 1%, so SHM is a good model. At 45° it is not: the real torque grows more slowly than θ, so a graph of torque against θ bends away from the straight line at large angles. (Figure 1c bends the other way, more steeply, but the lesson is the same: only the part near zero is close to a straight line.)

Check. The angle must be in radians for sin θ ≈ θ. In degrees, “θ = 5” and “sin θ = 0.087” are nowhere near each other.

The same result works with forces. For small angles the bob moves along an arc of length s = ℓθ, and the restoring force along the arc is mg sin θ ≈ mgθ = (mg/ℓ)s. That is “force = −constant × displacement”, with constant mg/ℓ = 2.45 N/m for this pendulum.

Common misconceptions

• “Any back-and-forth motion is SHM.” It is periodic, but SHM also needs F_net = −kx (the trough, bouncing-ball and wall examples).
• “The restoring force points the way the object is moving.” It points toward equilibrium, whichever way the object moves.
• “Zero force at equilibrium means the object stops there.” Zero force means zero acceleration, not zero velocity. The object passes through equilibrium moving (Topic 7.3).
• “The acceleration is largest at equilibrium.” It is zero there, and largest at the ends, where |x| is largest.
• “For a vertical spring, measure x from the relaxed length.” Measure it from the equilibrium position, where the net force is zero (Worked example 2).
• “Any pendulum is SHM.” Only for small angles, in radians, where sin θ ≈ θ (Worked example 3).
• “The minus sign in F = −kx means the force is negative in size.” The minus sign only gives direction: opposite to the displacement.

Where this leads

Once you know a motion is SHM, you can predict how long each cycle takes. The next topic, Frequency and Period of SHM, gives the period of a spring–mass system and of a small-angle pendulum. Topics 7.3 and 7.4 add graphs of position, velocity and acceleration against time, and the energy of an oscillator. Try the practice questions now, then use the revision notes and the checklist to consolidate. You can also return to the course roadmap.