Frequency and Period of SHM: Study Guide (Physics C: Mechanics 7.2)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 7.2 called Frequency and Period of SHM. This guide is the calculus-based one. It derives the period formulas from Newton’s second law written as a differential equation. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

Period, frequency and angular frequency

One cycle is a full there-and-back: the object returns to the same position, moving in the same direction.

• Period T: the time for one cycle, in seconds.
• Frequency f: the number of cycles per second, in hertz (1 Hz = 1 s⁻¹). f = 1/T.
• Angular frequency ω: 2π radians per cycle, in rad/s. ω = 2πf = 2π/T.

For example, an oscillation at f = 4.0 Hz has T = 0.25 s and ω = 2π × 4.0 = 25 rad/s.

From the equation of motion to the period

In Topic 7.1, Newton’s second law for SHM came out as d²x/dt² = −(k/m)x. Write the positive constant as ω²:

d²x/dt² = −ω²x

Topic 7.3 shows that the solution is a cosine (or sine) of ωt. A cosine repeats each time its argument grows by 2π, so the motion repeats after ωT = 2π:

T = 2π/ω = 1/f

This gives a general method that works for any system:

1. Choose a variable measured from equilibrium (a position, an arc length or an angle).
2. Apply Newton’s second law (or τ = Iα) and rearrange to d²(variable)/dt² = −(constant) × (variable).
3. The constant is ω². Then T = 2π/ω.

If the equation will not take this form, the motion is not SHM and these formulas do not apply.

The object–spring oscillator

For a mass m on an ideal spring of constant k, the constant is k/m:

ω = √(k/m), T = 2π√(m/k), f = (1/2π)√(k/m)

Read the dependence carefully:

• A larger mass has more inertia: it accelerates less for the same force, so T is longer.
• A stiffer spring gives a bigger force for the same displacement, so T is shorter.
• The amplitude does not appear. A bigger pull gives a bigger force, and the two effects cancel.
• Gravity does not appear. A vertical spring has the same period as a horizontal one, because gravity only shifts the equilibrium (Topic 7.1).

When several springs act together, use the effective spring constant from Topic 2.8: k_eff = k₁ + k₂ side by side (parallel), and 1/k_eff = 1/k₁ + 1/k₂ end to end (series).

A useful result for a vertical spring: at equilibrium, kd = mg, where d is the static stretch. So m/k = d/g and T = 2π√(d/g). You can predict the period from how far the spring sags.

The simple pendulum

A simple pendulum is a point bob of mass m on a light, inextensible string of length ℓ. Let θ be the angle from the vertical, positive counterclockwise (Figure 1).

The tension acts along the string, through the pivot, so it has no torque. The weight has lever arm ℓ sin θ, so τ = −mgℓ sin θ. With I = mℓ² and τ = I d²θ/dt²:

d²θ/dt² = −(g/ℓ) sin θ

For small angles in radians, sin θ ≈ θ, and the equation takes the SHM form with ω² = g/ℓ:

ω = √(g/ℓ), T = 2π√(ℓ/g)

• The mass cancels. Weight (which drives the motion) and inertia (which resists it) are both proportional to m.
• A longer string gives a longer period; a stronger gravitational field gives a shorter one.
• The result needs a small angle. At larger angles, the true period is a little longer than 2π√(ℓ/g).

A rigid body swinging about a pivot (a physical pendulum) follows the same method with a different I and lever arm. That is Topic 7.5.

Factors of change

Because T depends on square roots, a factor n in the quantity under the root changes T by √n.

Change	Spring T = 2π√(m/k)	Pendulum T = 2π√(ℓ/g)
mass × 4	T × 2	no change
k × 4	T × ½	not relevant
length × 4	not relevant	T × 2
g × 4	no change	T × ½
amplitude × 2 (still small)	no change	no change

Frequency changes by the inverse factor, since f = 1/T.

Worked example 1: from the equation of motion to T and f

Question. Take +y upward, origin at equilibrium. A 2.0 kg block hangs from two identical ideal springs, each with k = 400 N/m, joined end to end. (a) Write the equation of motion and find ω, T and f. (b) Find T if the same springs are used side by side instead.

1. (a) In series: 1/k_eff = 1/400 + 1/400, so k_eff = 200 N/m.
2. Newton’s second law about equilibrium: d²y/dt² = −(200 ÷ 2.0)y = −(100 s⁻²)y.
3. So ω² = 100 s⁻², ω = 10 rad/s. T = 2π/10 = 0.63 s. f = 1/T = 1.6 Hz.
4. (b) Side by side: k_eff = 800 N/m, ω = √(800 ÷ 2.0) = 20 rad/s, T = 0.31 s.

Check. The parallel spring is four times stiffer than the series pair, so T should halve: 0.63 s → 0.31 s. It does. Also, the static stretch in (a) is d = mg/k = 19.6 ÷ 200 = 0.098 m, and 2π√(0.098 ÷ 9.8) = 0.63 s, as the sag rule predicts.

Worked example 2: k from a graph of T² against m

Question. A student hangs different masses m on one spring and finds the period from 20 cycles each time:

m (kg)	0.100	0.200	0.300	0.400	0.500
T (s)	0.435	0.589	0.711	0.814	0.906

(a) What should be plotted to give a straight line? (b) Find k from the graph. (c) Explain the intercept.

1. (a) Square T = 2π√(m/k): T² = (4π²/k)m. A graph of T² against m should be a straight line with slope 4π²/k.
2. (b) T² values: 0.189, 0.347, 0.506, 0.663, 0.821 s². The best-fit line (Figure 2) has slope 1.58 s²/kg and intercept 0.031 s². So k = 4π² ÷ 1.58 = 25.0 N/m.
3. (c) The theory predicts a line through the origin. A positive intercept means some mass moves that is not in m. Dividing intercept by slope gives about 0.020 kg of extra effective mass, for example the hanger or the spring itself. (Background, beyond the course: for a uniform spring, theory gives an effective mass of about one third of the spring’s mass.)

Why the graph is better than one reading. Using only m = 0.300 kg in k = 4π²m/T² gives 23.4 N/m, because it ignores the extra mass. The slope is not affected by it.

Worked example 3: a pendulum that keeps time with a spring

Question. You want a simple pendulum with the same period as the oscillator in Worked example 1(a). (a) Find its length. (b) Compare it with the static stretch of the springs. (c) The 2.0 kg block is replaced by a 4.0 kg block, and the pendulum bob by one twice as heavy. What happens to each period?

1. (a) Equal periods need equal ω: √(g/ℓ) = 10 rad/s, so ℓ = 9.8 ÷ 100 = 0.098 m.
2. (b) The static stretch was also d = 0.098 m. That is no coincidence: T = 2π√(d/g) for the spring and T = 2π√(ℓ/g) for the pendulum, so they match when ℓ = d.
3. (c) Spring: T = 2π√(4.0 ÷ 200) = 0.89 s, a factor √2 longer. Pendulum: mass cancels, so T stays 0.63 s. The two no longer keep time.

Condition. The pendulum is short, so keep the swing small: an arc of 1.0 cm from the lowest point is about 0.10 rad (about 6°), where sin θ ≈ θ is good.

Measuring a period well

• Time many cycles (10 or 20) and divide. Your reaction-time error is shared across all of them.
• Start timing as the object passes equilibrium, where it moves fastest and the instant is sharpest.
• Keep pendulum angles small, and check that the period does not change when you halve the amplitude.
• Linearise before fitting: T² against m, T² against ℓ, or ln T against ln m to test the power.

Common misconceptions

• Confusing f and ω. ω = 2πf. A value in rad/s is not a frequency in Hz.
• “Heavier pendulum bobs swing more slowly.” Mass cancels for a simple pendulum.
• “A vertical spring has a different period.” Gravity shifts equilibrium only.
• “Bigger swings take longer.” Not in SHM; for a pendulum, only once the angle is no longer small.
• Forgetting the square root. Doubling m multiplies T by √2, not 2.
• Using one data point when the graph has an intercept. Use the slope.
• Degrees in sin θ ≈ θ. The approximation needs radians.

Where this leads

Topic 7.3 writes the motion as x = A cos(ωt + φ) and finds velocity and acceleration from it, using the ω you found here. Topic 7.4 adds energy, and Topic 7.5 applies the same method to physical and torsion pendulums. Next, try the practice questions, then use the revision notes and the checklist. When you are ready, move on to Topic 7.3, Representing and Analyzing SHM, or return to the course roadmap.