AP Physics 1 · Lesson 2

Period and frequency of springs and pendulums

Know Know it

The period T is the time for one full cycle, out and back. The frequency f is the number of cycles each second, in hertz: f=1/Tf = 1/T.

For a mass on a spring, T=2πm/kT = 2\pi\sqrt{m/k}: more mass or a softer spring means a longer period. For a simple pendulum swinging through small angles, T=2πL/gT = 2\pi\sqrt{L/g}: a longer string or weaker gravity means a longer period.

Neither period depends on the amplitude (for the pendulum, as long as the angle stays small). A pendulum’s period does not depend on the mass of its bob, and a spring’s period does not depend on g, so a hanging spring has the same period as the same spring and mass lying flat.

What a change does to the period
ChangeMass on a springSimple pendulum
Mass × 4period × 2no change
Spring constant × 4period ÷ 2does not apply
Length × 4does not applyperiod × 2
g ÷ 4no changeperiod × 2
Amplitude × 2 (angle still small)no changeno change

Calculator On the exam

On the exam you may use a four-function, scientific or graphing calculator on both sections, and the testing app also has scientific and graphing calculators built in. You also get an equation sheet with reference tables, but it will not tell you which equation fits a situation or what each symbol stands for there.

Worked example Easy

A plant basket hanging from a spring is bumped and bobs up and down. A student counts 14 complete cycles in 21 s. What are the period and the frequency of the motion?

  1. T = 0.67 s and f = 1.5 Hz
  2. T = 1.5 s and f = 0.67 HzAnswer
  3. T = 1.5 s and f = 4.2 Hz
  4. T = 21 s and f = 0.048 Hz

How to solve it

  1. Period = time ÷ number of cycles = 21 s ÷ 14 = 1.5 s.
  2. Frequency = cycles ÷ time = 14 ÷ 21 s = 0.67 Hz, which is also 1/T.

Why each choice is right or wrong

  • A. Incorrect. This swaps the two: 0.67 is cycles per second (the frequency), and 1.5 is seconds per cycle (the period).
  • B. Correct. The period is 21 s / 14 = 1.5 s, and the frequency is its reciprocal, 14 / 21 s = 0.67 Hz.
  • C. Incorrect. 4.2 is 2π/T, the angular frequency in radians per second, not the number of cycles each second.
  • D. Incorrect. 21 s is the time for all 14 cycles; the period is the time for just one cycle.

Trap Frequency is not 2π divided by the period

The frequency is f=1/Tf = 1/T, in cycles per second. The quantity 2π/T2\pi/T is the angular frequency, in radians per second, and it is 6.28 times larger. Also remember the two move opposite ways: when the period doubles, the frequency halves.

Explain Explain it

Task verbs: predict, explain

Use ratios instead of starting over. Because T∝mT \propto \sqrt{m}, multiplying the mass by 9 multiplies the period by 3. Because T∝LT \propto \sqrt{L}, a pendulum four times as long takes twice as long per cycle. Write the ratio first, then take its square root.

The period does not depend on the amplitude because a larger amplitude also means a larger restoring force and a larger top speed. The object has farther to go, but it moves faster, and the two effects cancel.

Worked example Medium

A cart on a horizontal spring oscillates with a period of 0.84 s when it carries a total mass of 0.30 kg. The load is changed so that the total mass is 0.75 kg, on the same spring. What is the new period?

  1. 2.1 s
  2. 0.53 s
  3. 0.84 s
  4. 1.3 sAnswer

How to solve it

  1. T∝mT \propto \sqrt{m} for the same spring, so T2=T1m2/m1T_2 = T_1\sqrt{m_2/m_1}.
  2. T2=0.840.75/0.30=0.842.5=1.3T_2 = 0.84\sqrt{0.75/0.30} = 0.84\sqrt{2.5} = 1.3 s.

Why each choice is right or wrong

  • A. Incorrect. 2.1 s multiplies the period by the full mass ratio, 2.5, but the period goes as the square root of the mass.
  • B. Incorrect. 0.53 s divides by the square root of 2.5; more mass makes the cart slower to turn around, so the period gets longer.
  • C. Incorrect. 0.84 s treats the period as independent of mass, which is true of a pendulum but not of a mass on a spring.
  • D. Correct. The mass grows by 0.75/0.30 = 2.5, so the period grows by 2.5=1.58\sqrt{2.5} = 1.58: (0.84)(1.58) = 1.3 s.

Explain Your turn

A student sets a block oscillating on a horizontal spring with an amplitude of 3 cm, then restarts it with an amplitude of 6 cm. She predicts the period will double, because the block now travels twice as far each cycle. Say whether her prediction is right, name the quantities the period does depend on, and explain what happens at the larger amplitude.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Says the period stays the same, so her prediction is wrong
  • Says the period depends only on the block’s mass and the spring constant
  • Says the block moves faster at the larger amplitude

Model answer

Her prediction is wrong: the period stays the same. For a block on a spring the period depends only on the block’s mass and the spring constant, T = 2π√(m/k), and neither has changed. At 6 cm the spring pulls twice as hard at each matching point, so the block speeds up more and moves faster; its maximum speed doubles. It covers twice the distance at twice the speed, so each cycle takes the same time.

Apply Apply it

Practice this topic. AP Practice questions on exactly what this lesson covers, with an explanation for every answer.