Spring vs Pendulum Period: What Is the Difference?

Mass moves one period and not the other

Both oscillators are treated as simple harmonic motion in this course, both periods are $2\pi$ times a square root, and the two square roots hold opposite answers to the same question. The spring period depends on the oscillating mass. The pendulum period does not depend on the bob's mass.

The AP Physics 1 CED states both at essential knowledge 7.2.A.1, under learning objective 7.2.A, describe the frequency and period of an object exhibiting SHM.

• EK 7.2.A.1.i gives the period of an ideal-spring oscillator as $T_s = 2\pi \sqrt{\dfrac{m}{k}}$.
• EK 7.2.A.1.ii gives the period of a simple pendulum displaced by a small angle as $T_p = 2\pi \sqrt{\dfrac{\ell}{g}}$.

Read the two symbol lists rather than the shapes. The first contains $m$ and the second does not. Nothing else about the two formulas is worth remembering as hard as that.

The consequence is testable in a way most physics facts are not. Hang a steel bob and then a cork bob of the same size on the same string, and the swing takes the same time. Put a $0.45$ kg block and then a $1.80$ kg block on the same spring, and the second takes twice as long. Worked example one runs both experiments side by side and prints the four numbers.

This page is about that one difference. If you want the routines for solving period problems, the simple harmonic motion guide has them, and the SHM practice set has problems to work.

Spring period vs pendulum period, side by side

Four rows are worth unpacking.

The mass row and the length row are mirror images. Quadruple the mass on a spring and the period doubles. Quadruple the length of a pendulum and the period doubles. Each formula has exactly one quantity you can adjust to tune the period, and they are different quantities.

The $g$ row cuts both ways. $g$ appears in the pendulum formula and nowhere in the spring formula, so a spring oscillator keeps its period on the Moon while an identical pendulum slows down. That is not a subtlety about hanging the spring vertically: a hanging spring's equilibrium position shifts, but $k$ and $m$ are unchanged, so $T_s$ is unchanged.

The amplitude row hides different reasons. For the spring, amplitude independence follows from $\vec{F}_s = -k \Delta \vec{x}$ being linear in the displacement, which is EK 7.1.A.2's condition for SHM: the magnitude of the restoring force is proportional to the displacement from equilibrium. For the pendulum, amplitude independence is a property of the small-angle model, not of the real pendulum. A pendulum released from a large angle really does take longer per swing.

The free-fall row is the cleanest test of the distinction. It is also one of the AP Physics 1 Unit 7 essential questions: how can an astronaut be weighed in space? Worked example two answers it, and the answer is a spring.

The case that separates them: change the mass, time the swing

Set up two oscillators that start with the same $0.45$ kg object.

• A block of mass $0.45$ kg on a horizontal spring of stiffness $k = 20$ N/m.
• A bob of mass $0.45$ kg on a string of length $\ell = 0.45$ m, released from a small angle.

Now swap in a $1.80$ kg object, four times heavier, and change nothing else.

Same mass change, same factor of four, and one oscillator responds while the other does not move at all. A stopwatch pointed at these two setups is an instrument that can tell mass from length.

The arithmetic is in worked example one. What matters here is that the two columns are not close: the spring period changed by $0.94$ s, which is most of a second on a swing that started under a second, while the pendulum period changed by nothing that a stopwatch could resolve. This is not a small correction that gets ignored for convenience. The mass is absent from $T_p = 2\pi\sqrt{\ell/g}$, so within the model it has no effect whatsoever.

The reverse experiment is just as clean, and it is the one that catches people who have memorised only half the rule. Keep the masses fixed and change the geometry instead: shorten the string and the pendulum speeds up, while stretching or compressing the spring further before release does nothing to the spring's period at all. The CED's own Topic 7.2 sample activity does exactly this, asking students to clamp a swinging pendulum's string between their fingers and pull it through so the length shortens while the bob oscillates, then explain why the period decreases and the amplitude angle increases.

Why the mass cancels for the pendulum and not for the spring

The cancellation is not a coincidence of algebra. It happens because gravity plays two roles in the pendulum and only one of them survives.

Start with what SHM requires. EK 7.1.A.2 says SHM results when the magnitude of the restoring force exerted on an object is proportional to that object's displacement from its equilibrium position, and the CED gives the derived equation $ma_x = -k\Delta x$. So the period is set by the ratio of the restoring stiffness to the inertia.

For the spring, those two are unrelated quantities. The stiffness $k$ is a property of the spring. The inertia $m$ is a property of the block. Nothing forces them to move together, so the ratio $k/m$ changes whenever either one changes, and $T_s = 2\pi\sqrt{m/k}$ carries both.

For the pendulum, gravity supplies both. The restoring force on the bob is the component of the gravitational force along the arc, and the gravitational force on the bob is proportional to the bob's mass, EK 2.6.A.3 writing it as $\text{Weight} = F_g = mg$. So a heavier bob is pulled back harder, in exact proportion to how much harder it is to accelerate. Writing the restoring force for a small displacement $x$ along the arc of a string of length $\ell$:

The $m$ on the left is inertial mass and the $m$ on the right is gravitational mass. They cancel because those two masses are the same number, which is not a definition but an experimental result: EK 2.6.D.3 states that inertial mass and gravitational mass have been experimentally verified to be equivalent. The pendulum's mass independence is that equivalence made audible with a stopwatch.

What is left after the cancellation, $g/\ell$, contains no reference to the bob. That is the square of the angular frequency, so $T_p = 2\pi\sqrt{\ell/g}$.

The same accounting explains the free-fall row in the table above. In a freely falling frame there is no restoring force on the bob at all, so there is nothing to divide by the inertia and no oscillation happens. The spring is untouched by any of this, because a stretched spring pulls whether or not anything is falling.

The small-angle condition, and why neither formula is exact

The pendulum formula carries a restriction that the spring formula does not, and the CED states it in the equation's own sentence rather than in a footnote.

EK 7.2.A.1.ii, in full: the period of a simple pendulum displaced by a small angle is given by the equation $T_p = 2\pi\sqrt{\ell/g}$. The words "displaced by a small angle" are part of the statement of the equation. Drop them and you have quoted something the CED did not say.

EK 7.1.A.2.iii carries the same restriction in the other direction, explaining why a pendulum belongs in this unit at all: the motion of a pendulum with a small angular displacement can be modeled as simple harmonic motion because the restoring torque is proportional to the angular displacement. Can be modeled as is the operative phrase. A pendulum is not a simple harmonic oscillator; it is a system that a simple harmonic oscillator models well over a limited range.

AP Physics C: Mechanics spells out the step that gets taken. At EK 7.5.A.2.ii, under Topic 7.5, Simple and Physical Pendulums, the CED says that for small amplitudes of motion the small-angle approximation can be applied to the restoring torque, with the derived equations $\sin\theta \approx \theta$ and $\tau = -mgd\theta = I\alpha$. The approximation is the whole bridge from the true restoring torque $\tau = -mgd\sin\theta$, at EK 7.5.A.2.i, to a torque that is proportional to the angle and therefore produces SHM. See /glossary/small-angle-approximation for the approximation on its own.

Now the part that is easy to get backwards. The spring formula is not exact either, and saying so is not pedantry. It rests on assumptions the exam makes for you rather than on an approximation you apply:

• The spring obeys $\vec{F}_s = -k\Delta \vec{x}$ over the whole range of the motion. Real springs stop doing that when they are overstretched or fully compressed.
• The spring's own mass is negligible compared with the oscillating mass. A spring with mass oscillates more slowly than $2\pi\sqrt{m/k}$ predicts.
• Nothing dissipates energy.

The AP Physics 1 exam grants the first two by convention. The Table of Information's conventions box states that springs and strings are assumed to be ideal unless otherwise stated, and that air resistance is assumed to be negligible unless otherwise stated. So the honest way to state the difference is not exact against approximate. It is that the pendulum's restriction is on the motion, which you control, and the spring's restrictions are on the apparatus, which the exam idealises for you. A pendulum released from 40 degrees breaks its formula while sitting on a perfect apparatus; a spring oscillator breaks its formula only if the spring itself is not ideal.

What each oscillator can measure that the other cannot

Because the two periods depend on different things, each one is a measuring instrument for what the other ignores. The CED's Topic 7.2 sample activities are built on exactly this split.

The pendulum measures $g$. Solve $T_p = 2\pi\sqrt{\ell/g}$ for the field strength and you get $g = 4\pi^2 \ell / T_p^2$, which needs only a length and a time. One CED sample activity for Topic 7.2 asks students to use a pendulum to determine the acceleration due to gravity, refining from a single-trial calculation to an average and then to a graph of linearized data. The bob's mass never enters, which is precisely why the method works with whatever is on the end of the string.

The spring measures mass. Solve $T_s = 2\pi\sqrt{m/k}$ for the mass and you get $m = k T_s^2 / 4\pi^2$. Another Topic 7.2 sample activity asks students to determine the spring constant of a spring twice, first with known masses and a meterstick, then with known masses and a stopwatch only. The second method is the one that keeps working when there is no weight to hang: it uses the period, not the stretch.

That second point is the answer to the unit's own essential question about weighing an astronaut in space. A balance compares weights and needs a gravitational field. A spring oscillator compares inertias and does not. Worked example two takes a spring-mounted chair through the measurement.

One more Topic 7.2 sample activity puts the two side by side and is the cleanest demonstration on this page: have students choose a song, find its tempo in beats per minute, then build a pendulum that swings on each beat, and separately find the mass that makes a given spring oscillate on each beat. Same target period, and the two knobs you turn are different quantities. Worked example three tunes both to the same song and then shows what happens when you turn the wrong knob.

Where the confusion costs a mark

Each of these is a specific scoring error rather than a general caution.

• Saying a heavier bob swings faster, or slower. It does neither. This wrong answer comes in both directions, depending on which half of the reasoning the student stopped at.
• Saying a heavier block on a spring keeps the same period. The symmetric error, made by students who learned the pendulum result first and generalised it. $m$ is in $T_s$.
• Putting $g$ into a spring problem. A vertically hanging spring shifts its equilibrium position by $mg/k$, and that shift changes nothing about the period. Writing $T = 2\pi\sqrt{m/(k+mg)}$ or anything like it invents a term.
• Using the pendulum formula outside the small angle. EK 7.2.A.1.ii applies to a simple pendulum displaced by a small angle. A question that specifies a large release angle is telling you the formula does not apply, and it usually wants a conservation-of-energy answer instead.
• Confusing the string length with the arc or the drop height. $\ell$ is the distance from the pivot to the center of mass of the bob, not the length of the swing.
• Using $\ell$ where the mass belongs, or the reverse. Under exam pressure the two square roots blur into one. The units settle it: $\sqrt{\text{kg} / (\text{N/m})}$ and $\sqrt{\text{m} / (\text{N/kg})}$ both reduce to seconds, but only one of them accepts a mass.
• Doubling the period when you double the mass. The dependence is on $\sqrt{m}$, so doubling the mass multiplies $T_s$ by $\sqrt{2} \approx 1.41$. You need four times the mass to double the period.
• Changing the amplitude and expecting the period to move. It does not, for either oscillator, within the models this course uses. Amplitude changes the maximum speed and the energy, not the timing.
• Treating a physical pendulum as a simple one. A swinging rod or a hoop is not a point mass on a string, and $T_p = 2\pi\sqrt{\ell/g}$ gives the wrong answer for it. That case is simple vs physical pendulum, and only AP Physics C: Mechanics requires it.
• Quoting the pendulum formula as exact. It is a small-angle result, and a free-response answer that says so is answering the question the rubric asked.

What the CED requires, and which booklets print the two formulas

Both oscillators sit in Unit 7, Oscillations, in the two mechanics courses, and the two courses weight that unit very differently.

AP Physics 1 Unit 7, Oscillations: 5 to 8 percent of the multiple-choice section, about 5 to 10 class periods. Topic 7.1, Defining Simple Harmonic Motion, carries LO 7.1.A with EK 7.1.A.1 and 7.1.A.2 and the three sub-statements 7.1.A.2.i through 7.1.A.2.iii. Topic 7.2, Frequency and Period of SHM, carries LO 7.2.A with EK 7.2.A.1 and its two sub-statements, which are the two formulas on this page. Suggested skills are 1.C, 2.A, 2.B and 3.B for Topic 7.1, and 1.B, 2.A, 2.D, 3.A and 3.C for Topic 7.2.

AP Physics C: Mechanics Unit 7, Oscillations: 10 to 15 percent, about 12 to 17 class periods. Its Topic 7.2 uses the same EK numbering and the same two formulas, with the period statement written as $T = \dfrac{2\pi}{\omega} = \dfrac{1}{f}$ at EK 7.2.A.1. Suggested skills for that topic are 1.C, 2.A, 2.D and 3.B. C: Mechanics then adds Topic 7.5, Simple and Physical Pendulums, which AP Physics 1 does not have.

On the equation sheets, checked against the appendix pages of all four Course and Exam Descriptions:

The last row surprises people. The AP Physics C: Electricity and Magnetism booklet carries a full MECHANICS table alongside its electricity and magnetism table, and the three period formulas are in it. The two Physics C booklets print the same mechanics page.

The symbol key on the sheets defines $\ell$ as length, $k$ as spring constant, $m$ as mass and $T$ as period, so the letters in the two formulas are not left to guess. The constants box prints $g = 9.8\ \text{m/s}^2$ and, on a separate line, $g = 9.8\ \text{N/kg}$, both on all four booklets. Note that EK 2.6.B.2 says that near the surface of Earth the strength of the gravitational field is $g \approx 10$ N/kg, so the CED body and the printed table give different roundings; is g 9.8 or 10 works through which to use and when.

For the CED framing of each topic, see Topic 7.1 and Topic 7.2, or the C: Mechanics versions at Topic 7.2 and Topic 7.5. The full sheet is at the AP Physics 1 formula page.