7.2 Frequency and period

How long does one swing take? For a spring it depends on mass and stiffness. For a pendulum it depends on length and gravity. Surprisingly, it does not depend on how far you pull it.

Little picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. A longer rope swings slower

2. A heavier rider keeps the same time

Two swings have the same rope length. A heavier rider sits on one. Its period is:

3. The grandfather clock

Read the story as text

An old grandfather clock keeps time with a swinging pendulum. Each swing moves the hands forward by the same amount. Over a week the swings get a little smaller as friction takes energy away, but the clock still keeps good time. If the clock runs slow, the owner does not add weight to the pendulum. They turn a small nut that moves the bob up a few millimetres.

The question: what sets the time for one swing, and why do mass and swing size not matter for a pendulum?

A pendulum takes 2.0 s for one swing there and back. What is its frequency?

4. Check yourself

Think of your answer first, then tap to see it.

a) A swing's rope is made 4 times longer. What happens to the time for one swing there and back?

Show answer

It becomes 2 times longer: T = 2π√(L/g), and √4 = 2.

b) A grown-up takes the place of a small kid on the same swing, with the same small push. Is the swing faster, slower or the same?

Show answer

The same. The mass is not in T = 2π√(L/g). More weight pulls harder, but more mass is also harder to speed up.

c) One swing there and back takes 0.50 s. What is the frequency?

Show answer

f = 1/T = 1/0.50 s = 2.0 Hz: two full swings every second.

Already know this?

Answer these three. All right on the first try? You can skip ahead.

One swing there and back takes T = 0.25 s. What is the frequency?

The mass on a spring is made 4 times bigger. The period:

Which of these changes the period of a simple pendulum?

The idea: period and frequency

In pictures

  • The period T is the time for one whole trip: there and back to the start.
  • The frequency f is how many whole trips happen each second.
  • A longer rope swings slower. A heavier rider keeps the same time. A smaller swing keeps the same time too.
  • For a spring, a heavier block is slower and a stiffer spring is faster.

Below is the same idea in words, and then with numbers.

The period T is the time for one full cycle: out, back, and to the start again. It is measured in seconds. The frequency f is how many cycles happen each second, in hertz (Hz = 1/s). They are flips of each other:

T = 1/f     f = 1/T

Mass on a spring

Ts = 2π √(m/k)

A bigger mass is harder to speed up, so it is slower: bigger T. A stiffer spring (bigger k) pushes harder, so it is faster: smaller T. Because of the square root, you need 4 times the mass to double the period.

Simple pendulum (small angles)

Tp = 2π √(L/g)

L is the length from the pivot to the centre of the bob. A longer pendulum is slower. Stronger gravity makes it faster. This works for small angles (up to about 15°).

What does NOT change the period

Changes TDoes not change T
Springmass m, spring constant kamplitude A, g (a hanging spring has the same T)
Pendulumlength L, gravity gamplitude (small angles), mass of the bob

Why amplitude does not matter: pull a spring twice as far and the force is twice as big (F = −kx). The block has twice as far to go, but it also goes twice as fast. The two effects cancel, so the time is the same. For a pendulum, a heavier bob feels a bigger pull of gravity but is also harder to speed up; those cancel too.

With numbers

Show all steps as text
  1. Spring: m = 0.50 kg, k = 20 N/m. T = 2π√(0.50/20) = 2π√0.025 = 2π(0.158) = 0.99 s. Put mass in kg and k in N/m; then √(kg/(N/m)) = √(s²) = s.
  2. Frequency: f = 1/T = 1/0.99 s ≈ 1.0 Hz. About one full cycle every second.
  3. Same spring, mass 2.0 kg (4 times bigger): T = 2π√(2.0/20) = 2π(0.316) = 1.99 s. 4 × mass → √4 = 2 × period.
  4. Pendulum: L = 0.80 m on Earth. T = 2π√(0.80/9.8) = 2π(0.286) = 1.80 s. Only L and g go in. The mass of the bob is not in the formula.
  5. The same pendulum on the Moon (g = 1.62 m/s²): T = 2π√(0.80/1.62) = 2π(0.703) = 4.42 s. Weaker gravity → weaker restoring force → slower swings.

Play: spring and pendulum side by side

Each one has its own controls. First predict how many cycles happen in the 8 second run, then press Play. Try to make the two have the same period. Then change only the amplitude, or only the bob's mass, and check that the period stays the same.

positionvelocityacceleration

Mass on a spring

T = …

Simple pendulum

T = …

For the pendulum, the graphs show the position along the arc, s = Lθ (positive to the right), and its velocity and acceleration.

Worked examples

From frequency to period basic

A guitar-string-like toy vibrates at 2.5 Hz. What is its period, and how long do 20 cycles take?

Show solution
Show all steps as text
  1. T = 1/f = 1/(2.5 Hz) = 0.40 s.Period and frequency are reciprocals.
  2. 20 cycles take 20 × 0.40 s = 8.0 s.Each cycle takes one period.

A stiff spring basic

A 0.20 kg block is on a spring with k = 180 N/m. Find the period.

Show solution
Show all steps as text
  1. T = 2π√(m/k) = 2π√(0.20/180) = 2π√(0.00111) = 2π(0.0333) = 0.21 s.Stiff spring, small mass → fast oscillation.

A seconds pendulum medium

How long must a pendulum be for its period to be 2.0 s on Earth?

Show solution
Show all steps as text
  1. Square both sides of T = 2π√(L/g): T² = 4π² L/g.Get rid of the square root first.
  2. L = gT²/(4π²) = (9.8)(2.0)²/(39.5) = 39.2/39.5 = 0.99 m.About one metre: an easy fact to remember and to check answers with.

Comparing two springs medium

System A is mass m on a spring of constant k. System B is mass 2m on a spring of constant k/2. How does TB compare with TA?

Show solution
Show all steps as text
  1. TB = 2π√(2m/(k/2)) = 2π√(4m/k) = 2 × 2π√(m/k).Dividing by k/2 is the same as multiplying by 2/k.
  2. So TB = 2TA.Both changes make B slower: more mass and a softer spring.

Finding k from timing data AP

A student times 10 oscillations for different masses on one spring and divides by 10 to get T.

m (kg)0.200.400.600.80
T (s)0.630.891.091.26

(a) Why time 10 oscillations? (b) What should be graphed to get a straight line? (c) Find k.

Show solution
Show all steps as text
  1. (a) Starting and stopping a stopwatch has a reaction-time error of about 0.2 s. Spread over 10 cycles, the error in each period is only about 0.02 s.Timing many cycles makes the percent uncertainty smaller.
  2. (b) Square: T² = (4π²/k) m. Graph T² (vertical) against m (horizontal); it is a straight line through the origin with slope 4π²/k.T against m is a curve (square root), which is hard to read a constant from.
  3. T² values: 0.40, 0.79, 1.19, 1.59 s². Slope = (1.59 − 0.40)/(0.80 − 0.20) = 1.19/0.60 ≈ 1.98 s²/kg.Use the best fit line, with points far apart.
  4. (c) k = 4π²/slope = 39.5/1.98 ≈ 20 N/m.Rearrange slope = 4π²/k.

Practice

  1. A block on a spring oscillates with amplitude A and period T. It is set going again with amplitude 2A. The new period is

    Show answer

    (B). T = 2π√(m/k) has no A in it. Twice as far, but twice the force and twice the top speed, so the time is the same.

  2. The mass on a spring is made 4 times larger. The period

    Show answer

    (B). T ∝ √m, and √4 = 2.

  3. A pendulum is taken to the Moon, where g is about 1/6 of Earth's. Its period becomes about

    Show answer

    (C). T ∝ 1/√g. g becomes g/6, so T becomes √6 ≈ 2.45 times as long.

  4. Which change makes a simple pendulum's period longer?

    Show answer

    (B). T = 2π√(L/g). Mass is not in the formula, and at small angles the amplitude does not matter.

  5. A 0.50 kg block is on a spring with k = 50 N/m. What is the frequency?

    Show answer

    (B). T = 2π√(0.50/50) = 2π(0.10) = 0.628 s. f = 1/T = 1.59 Hz. (A) is the period with the wrong unit; (D) is √(k/m), which is not the frequency in hertz.

  6. The same block and spring are used horizontally on a smooth table and hanging vertically. Compared with the horizontal period, the vertical period is

    Show answer

    (C). Gravity only moves the equilibrium point down by mg/k. Around that point the net force is still −kx, so T = 2π√(m/k) for both.

  7. Short answer. A pendulum clock runs slow: it shows 59 minutes when an hour has passed. Should the bob be moved up or down? Explain using the period formula.

    Show answer

    The clock counts swings. Running slow means each swing takes too long, so T is too big. T = 2π√(L/g), so reduce L: move the bob up (closer to the pivot). Adding mass would not help because mass is not in the formula.

    Scoring idea: 1 point for "period too long", 1 point for "shorten L / move bob up" with the formula as the reason.

  8. Short answer. A 0.40 kg block on a spring makes 15 full oscillations in 12.0 s. Find T, f and k.

    Show answer
    1. T = 12.0 s / 15 = 0.80 s.Total time divided by number of cycles.
    2. f = 1/T = 1.25 Hz.Or 15 cycles / 12.0 s.
    3. k = 4π²m/T² = (39.5)(0.40)/(0.64) = 24.7 N/m (about 25 N/m).Square T = 2π√(m/k) and solve for k.

AP question types for this topic

Here are four short free response questions, one of each AP Physics 1 type, all about this topic. Try each one on paper first, then open the worked answer.

1. Mathematical Routines

Lina hangs a block of mass m on a spring of unknown spring constant k, sets it bouncing, and times N full oscillations as taking a total time t.

  1. Derive an expression for k in terms of m, N, t and physical constants.
  2. For m = 0.25 kg she times 10 oscillations in 6.3 s. Calculate the period, the frequency and k.
  3. What mass should she hang on the same spring to make the period exactly 1.0 s?
Show worked answer and scoring

(a)

  1. One period is the total time over the number of cycles: T = t/N.
  2. T = 2π√(m/k). Square both sides: T² = 4π²m/k, so k = 4π²m/T² = 4π²mN²/t².

(b)

  1. T = 6.3/10 = 0.63 s; f = 1/T = 1/0.63 = 1.59 Hz.
  2. k = 4π²(0.25)/(0.63)² = 9.87/0.3969 = 24.9 N/m.

(c)

  1. Rearrange T = 2π√(m/k): m = kT²/(4π²) = (24.9)(1.0)²/(4π²) = 0.63 kg.
  2. Check: the period went from 0.63 s to 1.0 s, a factor 1.59; mass goes up by the square, 2.52 × 0.25 = 0.63 kg. ✓T goes with √m, so m goes with T².

AP-style scoring

  • 1 point for T = t/N, and 1 point for correctly solving T = 2π√(m/k) for k.
  • 1 point for T = 0.63 s, f ≈ 1.6 Hz and k ≈ 25 N/m with units.
  • 1 point for m ≈ 0.63 kg.

2. Translation Between Representations

Dmitri hangs different masses m from the same spring and measures the period T of each. He wants to sketch T against m before he takes any data.

Four possible graphs of period T (up) against mass m (across):

mT
(A) straight line rising from the origin
mT
(B) curve rising from the origin, getting flatter
mT
(C) horizontal line
mT
(D) curve rising from the origin, getting steeper
  1. Which graph best shows T against m?
  2. Justify your choice from the equation for the period, and say in words why heavier blocks are slower.
  3. What should Dmitri plot to get a straight line? If k = 16 N/m, what is the slope of that line?
Show worked answer and scoring

(a)

  1. Graph B: rising from the origin and getting flatter.

(b)

  1. T = 2π√(m/k), so T ∝ √m. A square-root curve starts at zero, rises, and flattens: 4 times the mass only doubles the period.
  2. In words: the same spring force must speed up more mass, so each trip takes longer. Not A (that would be T ∝ m), not C (mass does change T for a spring), not D (that would be T ∝ m²).

(c)

  1. Square the equation: T² = (4π²/k)·m. Plot T² against m: a straight line through the origin.
  2. Slope = 4π²/k = 39.5/16 = 2.47 s²/kg.

AP-style scoring

  • 1 point for graph B.
  • 1 point for T ∝ √m from T = 2π√(m/k), and 1 point for a physical reason (more inertia, same restoring force).
  • 1 point for T² against m with slope 4π²/k ≈ 2.5 s²/kg.

3. Experimental Design and Analysis

Amara wants to measure g in her classroom using a simple pendulum. She has string, a small metal bob, a clamp stand, a metre stick, a protractor and a stopwatch.

Her data:

Length L (m)Time for 10 swings (s)T (s)T² (s²)
0.209.00.900.810
0.4012.71.271.613
0.6015.51.552.403
0.8018.01.803.240
1.0020.02.004.000
  1. Describe her procedure, including how she keeps the motion close to simple harmonic and how she reduces timing error.
  2. What should she graph to get a straight line? How does she get g from it?
  3. Use the data to find g.
Show worked answer and scoring

(a)

  1. Measure L from the pivot to the centre of the bob with the metre stick. Pull the bob aside to a small angle (under about 15°, checked with the protractor) and release.
  2. Time 10 full swings and divide by 10, then repeat for at least five lengths.Timing many swings makes the reaction-time error a small fraction of the total.

(b)

  1. T = 2π√(L/g) gives T² = (4π²/g)·L. Graph T² against L: a straight line through the origin.
  2. Slope = 4π²/g, so g = 4π²/slope.

(c)

  1. Best-fit line through the five (L, T²) points: slope ≈ 4.00 s²/m (intercept ≈ +0.011 s², close to 0).
  2. g = 4π²/slope = 39.48/4.00 ≈ 9.9 m/s², close to the accepted 9.8 m/s².

AP-style scoring

  • 1 point for measuring L to the bob's centre and using small angles, and 1 point for timing several swings.
  • 1 point for graphing T² against L and stating slope = 4π²/g.
  • 1 point for g ≈ 9.8 to 9.9 m/s² from the slope of a best-fit line (not from one row).

4. Qualitative/Quantitative Translation

Yusuf swaps a 0.10 kg bob for a 0.40 kg bob on a pendulum, and also swaps the same two masses on a spring. He claims: "Changing the mass changes the spring's period but not the pendulum's, because for a pendulum the pull back home grows with the mass, and for a spring it does not."

  1. Explain in words why Yusuf's reasoning is correct.
  2. For a pendulum displaced a small arc distance s, the restoring force is about mg·s/L. Use this to derive the pendulum's period and show the mass cancels.
  3. Calculate the periods for L = 0.50 m and for a spring with k = 10 N/m, with each mass, and say how the numbers match your words.
Show worked answer and scoring

(a)

  1. Period depends on how strong the pull home is compared with the inertia. For a pendulum the pull home is part of gravity, which is proportional to m, so 4 times the mass gives 4 times the pull and 4 times the inertia: they cancel. A spring's pull depends only on k and the stretch, so more mass just means slower.

(b)

  1. F = −(mg/L)s has the form F = −keffs with keff = mg/L.
  2. T = 2π√(m/keff) = 2π√(m/(mg/L)) = 2π√(L/g): m cancels.

(c)

  1. Pendulum: T = 2π√(0.50/9.8) = 1.42 s for both masses.
  2. Spring: 0.10 kg gives 2π√(0.10/10) = 0.63 s; 0.40 kg gives 2π√(0.40/10) = 1.26 s, exactly 2 times longer.
  3. Match: the pendulum's keff grows with m (the “pull grows with mass” idea), so T is unchanged; the spring's k is fixed, so 4 × m gives √4 = 2 × T.

AP-style scoring

  • 1 point for explaining the cancellation of pull and inertia for the pendulum and the fixed k for the spring.
  • 1 point for keff = mg/L, and 1 point for T = 2π√(L/g) with m cancelled.
  • 1 point for 1.42 s (both), 0.63 s and 1.26 s, linked back to the words.

Common mistakes

The mistake: "Pull it further and it takes longer to come back."

Why it is wrong: further away means a bigger restoring force and a faster trip. In SHM the period does not depend on the amplitude.

How to spot it: look for A in the formula. It is not there.

The mistake: "A heavier pendulum bob swings more slowly."

Why it is wrong: more mass means more weight pulling it back, but also more inertia. They cancel. Mass matters for a spring, not for a pendulum.

How to spot it: spring → m and k. Pendulum → L and g.

The mistake: forgetting the square root: "double the mass, double the period".

Why it is wrong: T ∝ √m. Double the mass → T × √2 ≈ 1.41. You need 4× the mass to double T.

How to spot it: do ratio problems with the formula: Tnew/Told = √(mnew/mold).

The mistake: mixing up T and f, or counting half a swing as a period.

Why it is wrong: a period is a full cycle: out and back to the same place moving the same way. From +A to −A is only half a period.

How to spot it: check units (T in s, f in Hz) and that T × f = 1.

The mistake: measuring the pendulum length to the top of the bob, or using big angles.

Why it is wrong: L goes to the centre of the bob, and T = 2π√(L/g) is only accurate for small angles (under about 15°).

How to spot it: in a lab question, say "measure from pivot to centre of mass, keep the angle small".