Frequency and Period of SHM: Study Guide (Physics 1 7.2)

This guide is for the algebra-based Physics 1 course. You need the SHM condition, F_net = −kx, and the result a = −(k/m)x from Topic 7.1. Both period equations here are on the course equation table; you use them, and you do not need to derive the 2π.

Period and frequency

The period, T, is the time for one complete cycle: from one position, out to both ends, and back to the same position moving the same way. It is measured in seconds.

The frequency, f, is the number of complete cycles per second. Its unit is the hertz (Hz): 1 Hz = 1 cycle per second = 1 s⁻¹.

T = 1/f and f = 1/T

A short period means a high frequency. A pendulum with T = 0.50 s has f = 2.0 Hz.

Measuring a period well

Do not time a single cycle. Your reaction time when starting and stopping a stopwatch is roughly 0.2 s. On one cycle of 0.84 s that is an uncertainty of about 24%. Time ten cycles instead: 10 cycles in 8.4 s gives T = 0.84 s, and the same 0.2 s is now spread over ten cycles, about 0.02 s each, or 2%.

Two more habits help:

• Start and stop timing as the object passes through equilibrium. It moves fastest there, so the moment it passes is easy to judge.
• Count “zero” as you start the watch, not “one”. Otherwise you time one cycle fewer than you count.

The object–spring oscillator

For an object of mass m on an ideal spring with spring constant k:

T_s = 2π√(m/k)

Read the equation as physics, not just algebra:

• More mass, longer period. More inertia means a smaller acceleration for the same spring force, so each cycle takes longer.
• Stiffer spring, shorter period. A larger k gives a larger restoring force at each displacement, so the object is turned around more quickly.
• No g. A vertical spring has the same period as a horizontal one (Topic 7.1 showed that gravity only shifts the equilibrium). So a spring oscillator measures mass, not weight. That is how mass can be measured in orbit, where objects are weightless.

Link this to Topic 7.1: a = −(k/m)x. A large k/m means a large acceleration for each metre of displacement, and a short period. T depends on m/k, the inverse of that ratio.

The simple pendulum

For a simple pendulum of length ℓ (pivot to centre of the bob), swinging through a small angle:

T_p = 2π√(ℓ/g)

• Longer string, longer period.
• Stronger gravity, shorter period. A pendulum would swing more slowly where g is smaller.
• No mass. In Topic 7.1 the restoring force along the arc was about (mg/ℓ)s. Divide by m to get the acceleration: a ≈ −(g/ℓ)s. The mass cancels. Compare with a = −(k/m)x: for a pendulum, g/ℓ plays the part of k/m. That is why ℓ/g replaces m/k in the period.
• Small angles only. The equation comes from the SHM model, which needs sin θ ≈ θ. At large angles the real period is a little longer.

Neither equation contains the amplitude. For SHM, a bigger swing does not take longer: the object travels further but its acceleration is bigger in the same proportion. Topic 7.3 returns to this.

Factor-of-change reasoning

Many questions ask how a period changes, not what it is. Write the equation as a proportion and change one thing at a time.

Change	Because	New period
Mass on spring × 4	T ∝ √m	× √4 = × 2
Spring constant × 4	T ∝ 1/√k	× 1/√4 = × ½
Mass × 2 and k × 2	m/k unchanged	unchanged
Pendulum length × 9	T ∝ √ℓ	× 3
g ÷ 4	T ∝ 1/√g	× 2
Bob mass × 3	m not in T_p	unchanged
Amplitude × 2 (still SHM)	A not in either equation	unchanged

The square root is the trap. To double a period you must quadruple the mass (or the length).

Worked example 1: k from a stopwatch

Question. A 0.300 kg block hangs from a spring and oscillates up and down. A student times 20 full cycles in 15.0 s. (a) Find the period and frequency. (b) Find the spring constant. (c) What mass would give a period of exactly 1.00 s on the same spring?

1. (a) T = 15.0 s ÷ 20 = 0.750 s. f = 1/T = 1 ÷ 0.750 s = 1.33 Hz.
2. (b) Square both sides of T = 2π√(m/k): T² = 4π²m/k, so k = 4π²m / T².
3. k = (39.48 × 0.300) ÷ 0.750² = 11.84 ÷ 0.5625 ≈ 21 N/m.
4. (c) Rearrange for mass: m = kT² / 4π² = (21.06 × 1.00²) ÷ 39.48 ≈ 0.53 kg.

Check by factor of change. The period must rise from 0.750 s to 1.00 s, a factor of 1.333. Mass goes with T², so it must rise by 1.333² = 1.78: 0.300 kg × 1.78 = 0.53 kg. Same answer, no k needed.

Interpretation. You found a spring constant with only a stopwatch and a known mass, no ruler. The block hangs vertically, but g does not appear anywhere in the calculation.

Worked example 2: g from a pendulum graph

Question. A student measures the period of a pendulum for five lengths, timing 10 swings each time and dividing by 10.

ℓ (m)	0.20	0.40	0.60	0.80	1.00
T (s)	0.90	1.27	1.56	1.79	2.01

(a) What should be plotted to get a straight line? (b) Plot it and use the slope to find g.

1. (a) Square T_p = 2π√(ℓ/g): T² = (4π²/g) ℓ. This has the form y = (slope) × x. Plot T² on the vertical axis against ℓ on the horizontal axis. The line should pass through the origin with slope 4π²/g. (A graph of T against ℓ is a curve, and its slope is not constant.)
2. (b) Square each period: T² = 0.81, 1.61, 2.43, 3.20 and 4.04 s².
3. Plot the points and draw one best-fit straight line (Figure 1). Pick two points on the line, far apart: (0.10 m, 0.41 s²) and (0.90 m, 3.63 s²).
4. Slope = (3.63 − 0.41) s² ÷ (0.90 − 0.10) m = 3.22 ÷ 0.80 ≈ 4.03 s²/m.
5. g = 4π² ÷ slope = 39.48 ÷ 4.03 ≈ 9.8 m/s².

Interpretation. The line passes through the origin, as the equation predicts, so the data support the model. Using the slope of five points is better than one calculation from one length: a single bad reading barely moves a best-fit line.

The same method works for springs. T_s² = (4π²/k) m, so plot T² against m: the slope is 4π²/k.

Worked example 3: correcting a slow pendulum clock

Question. A pendulum clock is built so that each full swing should take 2.00 s. It actually has T = 2.01 s. (a) How many minutes does it lose per day (86 400 s)? (b) Should the pendulum be lengthened or shortened, and by roughly how much?

1. (a) In one day the pendulum completes 86 400 ÷ 2.01 ≈ 42 985 swings. The clock counts each as 2.00 s, so it shows 85 970 s. It loses about 430 s, or 7.2 minutes, per day.
2. (b) The period is too long, so the pendulum must be shortened (T ∝ √ℓ).
3. Length goes with T²: ℓ_new / ℓ_old = (2.00 ÷ 2.01)² = 0.990.
4. The present length is ℓ = gT² / 4π² = 9.8 × 2.01² ÷ 39.48 ≈ 1.003 m. The new length is 0.990 × 1.003 ≈ 0.993 m, so shorten it by about 1.0 cm.

Check. A 0.5% change in period needs about a 1% change in length, because ℓ ∝ T². The answer agrees.

Designing an experiment

Suppose the question is “Does the period of a spring–mass system depend on the mass, and how?” A good plan:

1. Hang a known mass from the spring. Pull it a small distance below equilibrium and release it.
2. Time 10 (or 20) full cycles from a pass through equilibrium. Repeat and average.
3. Change only the mass. Keep the spring and the starting displacement the same.
4. Plot T² against m. A straight line through the origin supports T ∝ √m; the slope gives 4π²/k.

Name the variables: independent m, dependent T, controlled the spring (k), and the amplitude, kept small.

Common misconceptions

• “A heavier bob makes a pendulum swing slower.” Mass is not in T_p (FAQ above).
• “Gravity changes the period of a spring.” g is not in T_s; it only moves the equilibrium.
• “Double the mass, double the period.” T ∝ √m, so the period rises by √2 ≈ 1.41.
• “A bigger swing takes longer.” Not for SHM: amplitude is in neither equation.
• “Period and frequency are the same thing.” They are reciprocals: T = 1/f.
• “ℓ is the string length only.” Measure to the centre of the bob.
• “T against ℓ should be a straight line.” It curves. T² against ℓ is the straight line.

Where this leads

Topic 7.3, Representing and Analyzing SHM, uses the period to write position as a function of time, x = A cos(2πft), and reads the motion from graphs. Topic 7.4 adds energy. Try the practice questions now, then use the revision notes and the checklist to consolidate. You can also return to the course roadmap.