AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/7/7-2)
Unit 7 · Topic 7.2
7.2 Frequency and Period of SHM
The period T is the time for one complete cycle and the frequency f is the number of cycles per second, so T = 1/f. A block on a spring has T = 2π√(m/k), and a small-angle pendulum has T = 2π√(L/g). Neither period depends on the amplitude, and the pendulum's doesn't depend on the bob's mass.
Key terms
- period
- frequency
- hertz
- spring constant
- pendulum length
Period and frequency
The period T is the time for one full cycle: out, back, out the other way and back to the start. It's measured in seconds. The frequency f is how many cycles happen each second, measured in hertz (1 Hz = 1 cycle per second).
They are reciprocals: T = 1/f and f = 1/T. A pendulum that takes 2.0 s per swing cycle has a frequency of 0.50 Hz.
A full cycle is not one trip from side to side. Going from one turning point to the other is only half a period.
Mass on a spring
For a block on an ideal spring, T = 2π√(m/k). A heavier block responds more sluggishly, so the period gets longer. A stiffer spring (bigger k) pulls back harder, so the period gets shorter.
The period does not depend on the amplitude: pull the block twice as far and it moves twice as fast, covering the extra distance in the same time. It also doesn't depend on g, so the same spring and block have the same period on the Moon, and the same period hanging vertically as lying flat.
Simple pendulum
For a pendulum with small swings, T = 2π√(L/g), where L is measured from the pivot to the center of the bob. A longer pendulum has a longer period, and weaker gravity also lengthens it.
The bob's mass does not appear. A heavier bob feels a bigger gravitational force, but it also has more inertia, and the two effects cancel exactly. Amplitude doesn't matter either, as long as the angle stays small.
Predicting changes
Because of the square roots, quadrupling a quantity only doubles (or halves) the period. Work these by ratio rather than plugging in numbers.
| Change | Spring T = 2π√(m/k) | Pendulum T = 2π√(L/g) |
|---|---|---|
| Mass × 4 | T × 2 | no change |
| k × 4 | T × ½ | (not used) |
| Length × 4 | (not used) | T × 2 |
| Amplitude × 2 | no change | no change (small angles) |
| Taken to the Moon (g ÷ 6) | no change | T × √6 ≈ 2.4 |
Measuring a period in the lab
Time several full cycles, say 10, and divide by the number of cycles. That spreads your reaction-time error over many cycles. Start and stop the timer as the object passes through equilibrium, where it moves fastest and the moment is easiest to judge.
To test or use the period equation, make the graph a straight line. Squaring the pendulum equation gives T² = (4π²/g)L, so a graph of T² against L is a line through the origin with slope 4π²/g. For a spring, T² against m has slope 4π²/k. Find the slope from a best-fit line, then solve for g or k.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Period of a spring–block oscillator
A 0.25 kg block oscillates on a horizontal spring with k = 40 N/m. Find its period and frequency.
Show the solutionHide the solution
- Step 1: T = 2π√(m/k) = 2π√(0.25/40) = 2π√(0.00625) = 2π(0.0791) ≈ 0.50 s.
- Step 2: f = 1/T = 1/0.497 s ≈ 2.0 Hz.
- Step 3: Notice that the amplitude wasn't given, and you didn't need it.
Answer: T ≈ 0.50 s, f ≈ 2.0 Hz
- Example 2Calculator allowed
Changing mass and amplitude (classic trap)
A block on a spring has a period of 0.80 s. The block is replaced with one four times as massive, and it's pulled out twice as far before release. What is the new period? If the same change of mass were made to a pendulum bob with a 0.80 s period, what would the new period be?
Show the solutionHide the solution
- Step 1: Spring: T ∝ √m, so four times the mass gives √4 = 2 times the period.
- Step 2: Amplitude doesn't appear in T = 2π√(m/k), so doubling it changes nothing. The trap is to double T again for the bigger amplitude.
- Step 3: New spring period: 2 × 0.80 s = 1.6 s.
- Step 4: Pendulum: T = 2π√(L/g) has no mass in it, so a heavier bob leaves the period at 0.80 s.
Answer: Spring: 1.6 s. Pendulum: still 0.80 s.
- Example 3Calculator allowed
Finding g from a pendulum graph
A student times small swings of pendulums of different lengths and records L = 0.20, 0.40, 0.60, 0.80 and 1.00 m with periods T = 0.90, 1.27, 1.55, 1.80 and 2.01 s. Explain what to graph to get a straight line, and use it to find g.
Show the solutionHide the solution
- Step 1: Square the period equation: T² = (4π²/g)L. So graph T² on the vertical axis against L on the horizontal axis. The theory predicts a straight line through the origin with slope 4π²/g.
- Step 2: Squared periods: 0.81, 1.61, 2.40, 3.24 and 4.04 s². These rise by about 0.81 s² for every 0.20 m, and a best-fit line has slope about 4.04 s²/m with an intercept very close to zero.
- Step 3: Solve slope = 4π²/g for g: g = 4π²/slope = 39.48/4.04 ≈ 9.8 m/s².
- Step 4: Using the slope of a best-fit line, rather than a single data point, averages out the random error in each timing.
Answer: Graph T² against L; slope ≈ 4.04 s²/m, so g ≈ 9.8 m/s²
Common mistakes
- Thinking a bigger amplitude means a longer period. For SHM the period is the same at any amplitude.
- Putting mass into the pendulum period. T = 2π√(L/g) doesn't depend on the bob's mass.
- Forgetting the square root: doubling the mass on a spring multiplies T by √2 ≈ 1.41, not by 2.
- Counting one swing from one side to the other as a full period. That's only half a cycle.
On the exam
- Factor-of-change questions are very common: one quantity is doubled or quadrupled and you predict the new period or frequency. Reason from the square root and say which quantities don't matter.
- In experimental design questions, describe timing many cycles, varying one quantity while holding the others fixed, and graphing T² against L or m so the slope gives g or k.
Connected topics
Videos
Check yourself
4 questions on 7.2 Frequency and Period of SHM. Pick an answer to see if you got it, and why.
A student counts 20 complete oscillations of a mass on a spring in 8.0 s. What is the frequency of the oscillation?
A 0.20 kg block on a horizontal ideal spring with spring constant 80 N/m oscillates on a frictionless surface. What is its period?
What length should a simple pendulum have so that its period is 2.0 s for small swings? Use g = 9.8 m/s².
The mass hanging on an ideal spring is made four times as large. How does the period of oscillation change?
0 of 4 answered