Topic 7.2: Frequency and Period of SHM

What Topic 7.2 requires

One learning objective, one essential knowledge statement, and two equations under it. LO 7.2.A asks you to describe the frequency and period of an object exhibiting SHM.

EK 7.2.A.1 says that the period of SHM is related to the frequency $f$ of the object's motion by the equation

Two sub-statements supply the specific systems. EK 7.2.A.1.i gives the period of an oscillator made of an object and an ideal spring as

and EK 7.2.A.1.ii gives the period of a simple pendulum displaced by a small angle as

That is the whole of the required content. Topic 7.2 has no boundary statement, and neither does any other topic in Unit 7.

The five suggested skills say more about the topic than the equation count does: 1.B (create quantitative graphs with appropriate scales and units, including plotting data), 2.A (derive a symbolic expression), 2.D (predict new values or factors of change using functional dependence between variables), 3.A (create experimental procedures appropriate for a given scientific question), and 3.C (justify or support a claim using evidence). Not one of the five is about substituting into a printed equation, which is why so much of this topic is tested without any numbers at all.

Everything here assumes the system already satisfies the definition in Topic 7.1. The period equations describe systems exhibiting SHM, and applying them to a system that does not qualify is the most expensive mistake available in this unit.

Period and frequency are one fact stated two ways

The period $T$ is the time for one complete cycle, in seconds. One complete cycle means all the way out, back past the start to the other side, and back again to where it began moving the way it started. A pendulum that swings left to right has completed half a cycle, not a whole one, which is a quiet source of factor-of-two errors in lab work.

The frequency $f$ is the number of complete cycles per second, in hertz. One hertz is one cycle per second, so the unit is really an inverse second.

EK 7.2.A.1 ties them with $T = 1/f$, and the equation sheet prints it in exactly that direction. There is no separate $f = 1/T$ line, because the same relation solved the other way is the same relation. If a problem hands you 2.5 Hz, the period is $1/2.5 = 0.40$ s.

A vocabulary warning. Most college courses write these equations using angular frequency, $\omega = 2\pi f$, and quote $T = 2\pi/\omega$ and $a = -\omega^2 x$. AP Physics 1 does not. Every Unit 7 statement in the CED and every oscillation line on the equation sheet is written in terms of $f$, and $\omega$ appears on the AP Physics 1 sheet only as the angular velocity of a rotating body in the Unit 5 equations. Angular frequency is a legitimate tool if you already know it, but you cannot cite it as given.

The spring period: what changes it and what does not

Read $T_s = 2\pi\sqrt{m/k}$ as a list of what is allowed to matter. Exactly two quantities appear: the oscillating mass and the spring constant. Anything not printed in the equation does not change the period.

Mass is the dependence people get wrong by carrying it over from the pendulum. Doubling the mass on a spring does change the period, by a factor of $\sqrt{2}$, roughly 1.41. Quadrupling it doubles the period. That is the opposite of the pendulum result in the next section, and problems are built out of exactly that contrast.

The absence of $g$ has consequences worth stating plainly. A spring oscillator keeps the same period on the Moon, in orbit, and in a lift accelerating upward, because none of those change $m$ or $k$. A hanging spring oscillator and a horizontal one with the same mass and spring share a period too. Gravity moves the equilibrium position down by $mg/k$, and Topic 2.8 works through why, but once displacement is measured from that new position the restoring force is the same $k$ times the same displacement, so the period cannot know the difference.

One assumption hides in the phrase object and ideal spring: the mass in $T_s$ is the attached object's, and the ideal-spring model gives the spring itself negligible mass.

The pendulum period: what changes it and what does not

Now read $T_p = 2\pi\sqrt{\ell/g}$ the same way. Two quantities appear: the length of the pendulum and the gravitational field strength at that location. The mass of the bob is absent, and so is the amplitude.

The mass result is the one that reliably surprises people, so have the reason ready rather than the fact alone. A heavier bob is pulled harder by gravity, so the restoring torque grows in proportion to $m$. But the heavier bob also resists angular acceleration more, and its rotational inertia grows in proportion to $m$ as well. The two effects sit on opposite sides of the same relationship and cancel exactly, which is the same cancellation that makes all objects fall at the same rate. Swap a lead bob for a cork bob on the same string and a stopwatch cannot tell the difference.

The $g$ dependence is what makes a pendulum a measuring instrument rather than a curiosity, and the second worked example below turns it into a measurement of $g$.

Use $g = 9.8 \ \mathrm{m/s^2}$ from the Table of Information when you need a number. Be aware that the CED also states, in a Topic 1.3 boundary statement, that the exam will use $g \approx 10 \ \mathrm{m/s^2}$ wherever a numerical value for $g$ is required, and that the same statement says students will not be penalized for correctly using $9.81 \ \mathrm{m/s^2}$ or $9.8 \ \mathrm{m/s^2}$. Both figures are genuinely in the source. Pick one, say which you used, and stay with it for the whole problem.

Why amplitude drops out, and the one place that claim is conditional

Amplitude appears in neither period equation. That is easy to see and easy to state carelessly, so it is worth separating what is definitely true from what is true only under a condition.

The CED's own version of the claim is EK 7.3.A.2, over in Topic 7.3: changing the amplitude of a system exhibiting SHM will not change the period of that system. Read the qualifier. It is a statement about systems that are exhibiting SHM, which by Topic 7.1 means systems whose restoring force is proportional to displacement.

The physical reason is a cancellation. Double the amplitude and the object has twice as far to travel each cycle. But at every corresponding point the displacement is also twice as large, so by the definition of SHM the restoring force and the acceleration are twice as large too. The object covers twice the distance at twice the speed, and the time comes out identical. The proportionality in the definition is exactly what makes the two effects cancel, which is why amplitude independence is a consequence of SHM rather than an extra fact about it.

For a mass on an ideal spring the claim is unconditional inside the model, because the spring force stays proportional at any extension the ideal-spring assumption covers. Pull it twice as far and the period is unchanged.

For a pendulum the claim inherits the small-angle condition. $T_p$ is itself a small-angle result: the restoring torque goes as $\sin\theta$, and treating it as proportional to $\theta$ is what produced the equation. Release a pendulum from a large angle and the system is no longer exhibiting SHM, so EK 7.3.A.2 does not apply to it and neither does $T_p$. The exact motion takes longer per cycle than $T_p$ predicts, by roughly 3 percent at an amplitude of 40 degrees and by well under 1 percent below 20 degrees.

So the honest summary is one sentence with two halves: amplitude does not affect the period of simple harmonic motion, and a pendulum is simple harmonic motion only while its angle stays small. Both halves belong in a justification, and answers giving only the first are the ones that lose the point when a question sets the amplitude at 45 degrees on purpose.

Deriving instead of substituting

Skill 2.A, derive a symbolic expression from known quantities, is listed for this topic in both the topic sidebar and the unit at-a-glance table, and it is worth 15 to 20 percent of the multiple-choice section. In practice it means rearranging the two period equations before any number goes in.

Both rearrangements start the same way. Square both sides to clear the root, then isolate.

From $T_s = 2\pi\sqrt{m/k}$, squaring gives $T_s^2 = 4\pi^2 m/k$, and therefore

From $T_p = 2\pi\sqrt{\ell/g}$, squaring gives $T_p^2 = 4\pi^2 \ell/g$, and therefore

Four expressions, one algebraic move. The constant $4\pi^2 = 39.478$ shows up in every one of them, which is worth recognizing on sight in a multiple-choice answer.

The squared period is not just algebra. It is why the graph work in this topic uses $T^2$ rather than $T$ on an axis, and why every one of these expressions is linear in the quantity being measured. That connection is the content of the next section.

For the other half of skill 2.D, predicting a factor of change without ever computing a period, the simple harmonic motion guide runs the ranking drill with a table of worked cases.

The Topic 7.2 lab: linearize, then take a slope

Skills 1.B and 3.A both sit on this topic, and the CED backs them up heavily: of the five sample instructional activities it lists for Unit 7, four are attached to Topic 7.2 and only one to another topic. They are determining a spring constant from known masses using a meterstick only and then using a stopwatch only; using a pendulum to determine the acceleration due to gravity, refining from a single trial to an average to a graph of linearized data; predicting and then explaining what happens to the period and the amplitude angle when a swinging pendulum's string is pulled shorter mid-swing; and finding a song's tempo in beats per minute, then building a pendulum that swings on each beat and working out what mass on a given spring would do the same.

The method behind the measuring ones is the same, and it is the reason $T^2$ keeps appearing. A graph of $T$ against $m$, or of $T$ against $\ell$, is a curve, and you cannot read a physical constant off the slope of a curve. Square the period and both relationships turn into straight lines through the origin:

Three procedural points earn marks in the design half of a laboratory question and cost them when missing.

Time many cycles, not one. A stopwatch carries a reaction-time uncertainty of a couple of tenths of a second whether you time one cycle or twenty. Time 20 and divide by 20, and that uncertainty is divided by 20 too.

Vary one thing. The CED describes a scientifically sound procedure for the laboratory free-response question as varying a single parameter and measuring how that change affects a single characteristic. Changing the mass and the spring at once produces data no slope can interpret.

Say what the intercept should be. Both linearized plots should pass through the origin, because zero mass or zero length means zero period. A best fit line with a clear nonzero intercept is telling you something is wrong with the procedure rather than with the physics, and saying so is a 3.C justification.

Count cycles carefully too. Starting the stopwatch at the bob's lowest point and stopping it the next time the bob passes that point counts a half cycle, because the bob is moving the other way.

Where these equations sit on the equation sheet

Checked line by line against the AP Physics 1 equation sheet, the oscillations content is five lines, all of them in the group headed Mechanics and Fluids (rotational, oscillations, and fluids):

• $T = \dfrac{1}{f}$
• $T_s = 2\pi\sqrt{\dfrac{m}{k}}$
• $T_p = 2\pi\sqrt{\dfrac{\ell}{g}}$
• $x = A\cos(2\pi f t)$
• $x = A\sin(2\pi f t)$

Everything Topic 7.2 requires is on that list, so nothing here has to be memorized. Both position functions are printed as well, which is worth knowing before you spend exam minutes reconstructing them.

Two further lines that oscillation problems reach for are printed in the translational group instead, one block up the page: $\vec{F}_s = -k\Delta\vec{x}$ and $U_s = \frac{1}{2}k(\Delta x)^2$. Hunting for them among the oscillation equations is a common way to waste half a minute.

The omissions matter just as much. There is no $\omega = 2\pi f$ on the sheet. There is no $E_{\text{total}} = \frac{1}{2}kA^2$, even though the CED prints that as a relevant equation under EK 7.4.A.4.ii in Topic 7.4. And there is no rearranged $k = 4\pi^2 m / T^2$ or $g = 4\pi^2 \ell / T^2$: those you produce yourself, which is the point of skill 2.A being attached to this topic.

The symbol the sheet uses for pendulum length is $\ell$, a script letter, the same one used for length elsewhere in the mechanics equations. It is not the number one.

How Topic 7.2 is tested, and what costs points

A four-function, scientific, or graphing calculator is allowed on both sections of the exam, and both period equations are printed, so a question that only wanted a substitution would be free marks. Questions are built around the dependences instead.

The skills back that up. Skill 2.D, predicting new values or factors of change using functional dependence between variables, is 10 to 15 percent of the multiple-choice section, and it is the skill behind every question of the form what happens to the period if the mass is tripled. Skill 1.B is not assessed in the multiple-choice section at all, since Science Practice 1 is free-response only, where it appears in the Experimental Design and Analysis question. That question is 10 points with a suggested time of 25 to 30 minutes, and its listed skills are 1.B, 2.B, 2.D and 3.A.

Six failures account for most of the lost marks on period questions.

• Using the pendulum result on the spring. Mass genuinely does change the spring period, as $\sqrt{m}$, and genuinely does not change the pendulum period. Problems pair the two deliberately.
• Forgetting the square root. Quadrupling the mass on a spring doubles the period, it does not quadruple it.
• Counting a half cycle as a whole one. Left to right is half a period. This doubles or halves a lab answer while every subsequent step looks fine.
• **Plotting $T$ instead of $T^2$.** You cannot read $k$ or $g$ off the slope of a curve.
• Stating amplitude independence with no condition attached. True for SHM, and a pendulum is only SHM at small angles.
• **Using $g$ in a spring problem.** It is not in $T_s$. A vertical spring oscillator has the same period as a horizontal one with the same mass and spring.