AP Physics C: Mechanics · Topic 7.2

Topic 7.2: Frequency and Period of SHM

Unit 7: Oscillations10-15% of the multiple-choice section

In AP Physics C the period of simple harmonic motion is two pi divided by the angular frequency, which also equals one over the frequency. You get the angular frequency by reading the constant that multiplies the position in the equation of motion, so one method gives every period in the unit.

AP Physics: Unit 7 (topics 7.2 Frequency and Period of SHM). Topic 7.2 of the current AP Physics C: Mechanics course and exam description, inside Unit 7, which is weighted 10 to 15% of the multiple-choice section at about 12 to 17 class periods. One learning objective, 7.2.A (describe the frequency and period of an object exhibiting SHM), and three essential-knowledge statements: 7.2.A.1 relating the period of SHM to the angular frequency by T = 2*pi/omega = 1/f; 7.2.A.1.i giving the object and ideal-spring period as 2*pi*sqrt(m/k); and 7.2.A.1.ii giving the period of a simple pendulum displaced by a small angle as 2*pi*sqrt(l/g). Suggested skills 1.C, 2.A, 2.D, 3.B. Topic 7.2 prints no boundary statement; Unit 7 prints exactly one, under Topic 7.3. Calculus differentiator against the identically titled AP Physics 1 Topic 7.2: the AP Physics 1 sheet prints only T = 1/f and never links the period to an angular frequency, while the C: Mechanics sheet prints T = 2*pi/omega = 1/f, so omega is the primary quantity here and is read off the coefficient of the second-order differential equation. That coefficient method is the only route to a torsion pendulum period, which the CED never prints, and to oscillators such as a cylinder rolling on a spring, whose period is 2*pi*sqrt(3m/2k) rather than the printed spring result. Equation-sheet facts verified against both Tables of Information rendered as images: the C sheet prints T = 1/f, T = 2*pi/omega = 1/f, T_s, T_p and T_phys, and defines omega as angular frequency or angular speed and phi as phase angle with no A in its variable list; the AP Physics 1 sheet prints T = 1/f, T_s and T_p only, defines omega as angular speed, has no phase angle, and lists A as amplitude or area. Sample multiple-choice question 5 aligns to 7.2.A and 7.2.A.1 with skill 2.C, keyed B: replacing 1 kg spheres with 2 kg spheres leaves the pendulum's period unchanged and increases the spring's. Sample instructional activities 2 and 3 sit on this topic, a ranking task over four to six spring cases and a clamped steel ruler whose spring constant is found from measured periods.

What Topic 7.2 requires

Topic 7.2 of AP Physics C: Mechanics carries one learning objective and three essential-knowledge statements. That is the whole of the required content.

Learning objective 7.2.A: describe the frequency and period of an object exhibiting SHM.

Statement	What it says
7.2.A.1	The period of SHM is related to the angular frequency, $\omega$ , of the object's motion by $T = \dfrac{2\pi}{\omega} = \dfrac{1}{f}$
7.2.A.1.i	The period of an object and ideal-spring oscillator is given by $T_s = 2\pi\sqrt{\dfrac{m}{k}}$
7.2.A.1.ii	The period of a simple pendulum displaced by a small angle is given by $T_p = 2\pi\sqrt{\dfrac{\ell}{g}}$

The suggested skills are 1.C (create qualitative sketches of graphs that represent features of a model or the behavior of the physical system), 2.A (derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway), 2.D (predict new values or factors of change of physical quantities using functional dependence between variables) and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim).

Skill 2.A appears here and did not appear in Topic 7.1. That is the signal that this topic is where the derivations start, and it matches what the equations look like: none of the three statements above is presented as something to memorise, and the framework labels none of them a Derived equation either. All three are printed on the equation sheet verbatim.

Topic 7.2 prints no boundary statement. Unit 7 prints exactly one across all five topics and it sits under Topic 7.3.

One structural note that only makes sense in this course. Statement 7.2.A.1.ii states the simple-pendulum period here and does not justify it. The justification arrives two topics later: 7.5.A.3 makes the simple pendulum a special case of a physical pendulum, and 7.5.A.2 derives the physical-pendulum period from Newton's second law in rotational form. So Topic 7.2 gives you the result and Topic 7.5 tells you where it came from. AP Physics 1 never closes that loop, because it has no Topic 7.5.

Angular frequency is the primary quantity in this course, not the period

This is the difference that shows up on the equation sheet, and it changes what a Topic 7.2 answer looks like.

Statement 7.2.A.1 is written as a relation from the angular frequency: the period of SHM is related to the angular frequency of the object's motion by $T = 2\pi/\omega = 1/f$ . Read the order of that sentence literally. You obtain $\omega$ first, from the physics, and $T$ is a conversion.

Where does $\omega$ come from? From the coefficient in the differential equation of statement 7.3.A.2:

\frac{d^2x}{dt^2} = -\omega^2 x

Write Newton's second law for the restoring force, rearrange until the second derivative is alone, and whatever multiplies the position is $-\omega^2$ . Then $T = 2\pi/\omega$ turns it into a period and $f = 1/T$ into a frequency. That is the whole procedure, and it is what makes skill 2.A a suggested skill for this topic.

Run it once, on the spring, to see the printed formula fall out rather than be recalled. The sheet prints $\vec{F}_s = -k\Delta\vec{x}$ , so

m\frac{d^2x}{dt^2} = -kx \quad \Longrightarrow \quad \frac{d^2x}{dt^2} = -\frac{k}{m}x \quad \Longrightarrow \quad \omega^2 = \frac{k}{m}

and therefore

T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{k/m}} = 2\pi\sqrt{\frac{m}{k}}

which is exactly 7.2.A.1.i. The formula is printed, so you may quote it. What the derivation buys you is every oscillator the exam invents that is not a block on a spring, because for those there is no printed formula and the coefficient is all you have.

Beware one collision of notation. In this unit $\omega$ is an angular frequency in radians per second, and it describes an oscillation that is not going round anything. In Units 5 and 6 the same symbol was an angular speed. The C: Mechanics sheet's variable list is explicit about carrying both meanings: it reads " $\omega$ = angular frequency or angular speed". Do not read a physical rotation into an oscillating block.

One method, four periods

The reason to work from the coefficient rather than from a list is that the list would need four entries and the method needs one. Every row below is the same three steps: write the second law, isolate the second derivative, read the coefficient.

Oscillator	Equation of motion	$\omega^2$	Period
Object on an ideal spring	$\dfrac{d^2x}{dt^2} = -\dfrac{k}{m}x$	$\dfrac{k}{m}$	$T_s = 2\pi\sqrt{\dfrac{m}{k}}$ , printed
Simple pendulum, small angle	$\dfrac{d^2\theta}{dt^2} = -\dfrac{g}{\ell}\theta$	$\dfrac{g}{\ell}$	$T_p = 2\pi\sqrt{\dfrac{\ell}{g}}$ , printed
Physical pendulum, small angle	$\dfrac{d^2\theta}{dt^2} = -\dfrac{mgd}{I}\theta$	$\dfrac{mgd}{I}$	$T_{\text{phys}} = 2\pi\sqrt{\dfrac{I}{mgd}}$ , printed
Torsion pendulum	$\dfrac{d^2\theta}{dt^2} = -\dfrac{k}{I}\theta$	$\dfrac{k}{I}$	$2\pi\sqrt{\dfrac{I}{k}}$ , not printed

The last row is the one that shows why the method matters. Statement 7.5.A.4 gives the torsion pendulum's Derived equation as $I\alpha = -k\Delta\theta$ and gives no period at all. There is nothing to quote, and reading $\omega^2 = k/I$ off the coefficient is the only route to a number.

The three angular rows use the angular twin of the differential equation, $d^2\theta/dt^2 = -\omega^2\theta$ , which statement 7.5.A.2.iii prints. Note that $\theta$ in those rows is the oscillating variable and $\omega$ is the angular frequency of the oscillation. They are not the same kind of angular quantity, and $\omega$ here is emphatically not $d\theta/dt$ .

Worked example 1 adds a fifth row the CED does not print anywhere: a cylinder that rolls on a spring rather than sliding on it. Its period is not $2\pi\sqrt{m/k}$ , and no formula on the sheet will tell you what it is.

What moves each period, and what does not

Read the three printed formulas for what each one contains, since a large fraction of Topic 7.2 questions are answered by noticing which symbol is missing.

$T_s = 2\pi\sqrt{m/k}$ contains mass and force constant. Quadruple the mass and the period doubles. Quadruple the force constant and the period halves. It contains no $g$ , so a spring oscillator has the same period vertical or horizontal, on Earth or on the Moon, and Topic 7.1's vertical-spring derivation shows exactly why: gravity relocates the equilibrium and cancels out of the equation of motion.

$T_p = 2\pi\sqrt{\ell/g}$ contains length and gravitational field strength. It contains no mass at all, so the bob's mass does not matter. Quadruple the length and the period doubles. Take the pendulum somewhere with weaker gravity and the period lengthens.

Neither contains the amplitude. Statement 7.3.A.5 says plainly that changing the amplitude of a system exhibiting SHM will not change its period.

That last claim has one conditional attached, and stating it is the difference between a correct answer and a lucky one. For a pendulum, amplitude independence holds only inside the small-angle approximation. Statement 7.2.A.1.ii already restricts itself, saying the period "of a simple pendulum displaced by a small angle". The restoring torque actually goes as $\sin\theta$ , and $\sin\theta$ is always smaller than $\theta$ , so the true restoring effect is weaker than the model and the true period is longer than the formula gives, by more at larger amplitudes. Statement 7.5.A.2.ii is where the CED makes the approximation explicit. A spring is not conditional in the same way: within the ideal-spring assumption the force is exactly proportional at any extension, and the exam conventions box states that springs and strings are assumed to be ideal unless otherwise stated.

And the mass contrast is the CED's own sample multiple-choice question. Question 5, aligned to objective 7.2.A and essential knowledge 7.2.A.1 with skill 2.C, sets a 1 kg sphere on a light string beside a 1 kg sphere on a spring, at equal equilibrium lengths, oscillating with the same period. Replace both spheres with 2 kg spheres at unchanged amplitudes. The keyed answer is B: the pendulum's period remains the same while the spring's period increases. Both halves come straight from which symbol appears in which formula.

Functional dependence, which is what skill 2.D actually asks

Skill 2.D is "predict new values or factors of change of physical quantities using functional dependence between variables", and it carries 10 to 15% of the multiple-choice section. On this topic it takes a recognisable shape: change one input by a factor, and report the factor by which each output changes. Doing it by computing two numbers and dividing wastes time and invites arithmetic errors. Do it by exponents.

Every Unit 7 quantity for a spring oscillator is a power of $m$ , $k$ and $A$ :

Quantity	Expression	Depends on
Angular frequency	$\omega = \sqrt{k/m}$	$k^{1/2}m^{-1/2}$ , not amplitude
Period	$T = 2\pi\sqrt{m/k}$	$m^{1/2}k^{-1/2}$ , not amplitude
Frequency	$f = 1/T$	$m^{-1/2}k^{1/2}$ , not amplitude
Maximum speed	$v_{\max} = A\omega$	$A k^{1/2} m^{-1/2}$
Maximum acceleration	$a_{\max} = A\omega^2$	$A k m^{-1}$
Maximum force	$F_{\max} = kA$	$kA$ , not mass
Total energy	$E = \frac{1}{2}kA^2$	$kA^2$ , not mass

The two derived maxima are statement 7.3.A.3.ii and are not printed on the equation sheet. The total energy is statement 7.4.A.4.ii and is not printed either. Only the top three rows come off the sheet.

The unit's second sample instructional activity is a ranking task built on this table. It asks teachers to give students four to six cases of a mass on a spring, differing in mass, force constant and amplitude, and have students rank them by period, frequency, maximum speed, maximum acceleration, maximum force and total energy. Worked example 2 runs a four-case version of exactly that.

The trap the table exposes: amplitude is absent from the first three rows and present in the last four. So a question that changes only the amplitude is asking whether you can hold the period fixed while everything about the energy and the speeds moves. And a question that changes only the mass is asking whether you can hold the maximum force and the total energy fixed while the period moves.

The row the AP Physics 1 sheet does not have

Checked line by line against the Table of Information in the AP Physics C: Mechanics course and exam description and the corresponding appendix of the AP Physics 1 course and exam description, both rendered as images.

Line	C: Mechanics sheet	AP Physics 1 sheet
$T = \dfrac{1}{f}$	yes	yes
$T = \dfrac{2\pi}{\omega} = \dfrac{1}{f}$	yes	no
$T_s = 2\pi\sqrt{m/k}$	yes	yes
$T_p = 2\pi\sqrt{\ell/g}$	yes	yes
$T_{\text{phys}} = 2\pi\sqrt{I/(mgd)}$	yes	no
$x = x_{\max}\cos(\omega t + \phi)$	yes	no
$x = A\cos(2\pi ft)$ and $x = A\sin(2\pi ft)$	no	yes, both
Variable list entry for $\omega$	angular frequency or angular speed	angular speed
Variable list entry for $\phi$	phase angle	absent
Variable list entry for $A$	absent	amplitude or area

The second row is the one to notice on this topic. The AP Physics 1 sheet never links the period to an angular frequency at all. It prints $T = 1/f$ and nothing else, so in that course $\omega$ has no route into an oscillation and the two $2\pi ft$ position functions carry the frequency directly. On the C: Mechanics sheet the $2\pi/\omega$ form is printed, the $2\pi ft$ forms are not, and every equation in the unit is phrased in $\omega$ .

That is not a cosmetic difference. It is why $\omega$ is the quantity you solve for on this exam and $T$ is the last line of the answer, and it is why a student who learned SHM in the algebra-based course arrives here looking for a line that is not there.

Two more absences worth carrying: $v_{\max} = A\omega$ and $a_{\max} = A\omega^2$ are not printed on either sheet, and neither is $a = -\omega^2 x$ .

Measuring a period, and the two Topic 7.2 lab activities

Two of the unit's five sample instructional activities sit on this topic, and both are timing tasks.

Activity 3 is a desktop experiment: obtain a steel ruler or yardstick, clamp it to a table, attach various masses to the end with the hole in it, have students measure the period of oscillation for each mass, then use the data to determine the spring constant of the steel ruler. That is 7.2.A.1.i run backwards.

The route the exam wants is a linearization, not a single measurement. Square the printed formula:

T^2 = \frac{4\pi^2}{k}m

So a graph of $T^2$ against $m$ is a straight line through the origin with slope $4\pi^2/k$ , and $k = 4\pi^2/\text{slope}$ . Plotting $T$ against $m$ instead gives a curve, and reading a force constant off a curve is not a thing you can do. Linearization is named in the CED's own list of error-reducing procedures for Unit 7.

Two cautions that apply to any period measurement, and both are worth a sentence on an experimental-design answer:

Time many oscillations and divide. Reaction-time uncertainty is a fixed number of milliseconds per timing, so dividing it across 20 or 25 cycles divides its effect on $T$ by the same factor. Worked example 3 puts numbers on how much that buys.
A non-zero intercept means something is oscillating that you did not put there. For the steel ruler, the ruler's own mass moves too. The slope is still the honest route to $k$ , which is one reason a slope beats a single point.

The CED's fifth activity, on Topic 7.5, names the three error-reducing components it wants to see: timing multiple periods, linearizing the data, and very precisely finding the centre of mass of the bob. All three transfer to this topic. Because the accepted value of $g$ is itself worth care when you compare a pendulum result to it, the is g 9.8 or 10 guide is worth reading first.

How Topic 7.2 is tested

Unit 7 is weighted 10 to 15% of the multiple-choice section of the AP Physics C: Mechanics exam, at about 12 to 17 class periods. The exam is 3 hours: 42 multiple-choice questions in 85 minutes for half the score, then 4 free-response questions in 95 minutes for the other half, always in the order Mathematical Routines, Translation Between Representations, Experimental Design and Analysis, and Qualitative/Quantitative Translation. A four-function, scientific, or graphing calculator is allowed on both sections.

On the multiple-choice section, skill 2.A is weighted 25 to 30%, the heaviest of the six, and 2.D is 10 to 15%, and both are suggested skills for this topic. Skill 3.B is 15 to 25%. Science Practice 1 is not assessed on that section at all, so this topic's 1.C points at the free-response section, where Practice 1 carries 20 to 35%.

The framework's exam-weighting page adds that required course content can be assessed with any skill, so treat the suggested skills as a teaching signal rather than a promise.

One of the CED's fifteen sample multiple-choice questions aligns to this topic: question 5, on the mass dependence of the two periods, keyed B with skill 2.C and essential knowledge 7.2.A.1. It is worked through in the section above. None of the four sample free-response questions aligns to Unit 7 at all, which is a reminder that with four questions and seven units a Unit 7 free-response question is not guaranteed in a given year. The reliable figure is the multiple-choice weighting.

Because skill 2.A is the heaviest on the exam and is suggested here, expect the derivation to be asked symbolically. An answer of the form "the period is $2\pi$ times the square root of the total inertia over the total restoring coefficient, so here it is $2\pi\sqrt{3m/2k}$ " is what a Mathematical Routines question wants. A number without the symbolic pathway is a partial answer.

For the step-by-step routines rather than the framework, the simple harmonic motion guide owns the procedures, and the period compared with frequency page is the two-minute version of the first line of 7.2.A.1.

If you are in AP Physics 1, this is not your page

AP Physics 1 has a Topic 7.2 with the same title, and the two printed period formulas are word for word the same. The honest split is about what surrounds them.

The [AP Physics 1 Topic 7.2 page](/ap-physics-1/unit-7-oscillations/7-2-frequency-and-period-of-shm) is for students in the algebra-based course, whose equation sheet gives the period only as one over the frequency and who will only ever meet the two formulas as formulas. This page is for students in AP Physics C: Mechanics, whose sheet also prints the period as two pi over the angular frequency, who are expected to derive both results from a differential equation, and who will be handed oscillators for which no formula is printed. If you are in Physics 1, the physical pendulum and the torsion pendulum are not on your exam and the coefficient method is not something you need.

AP Physics C: Mechanics is a calculus-based, college-level course, equivalent to a first course in an introductory college sequence in calculus-based physics. Its stated prerequisite is that students have taken or are concurrently taking calculus. AP Physics 1 weights its own Unit 7 at 5 to 8% of its multiple-choice section against 10 to 15% here, and gives it four topics rather than five.

Where to go next: Topic 7.1 is where the differential equation comes from, Topic 7.3 solves it and carries the unit's only boundary statement, Topic 7.4 is where amplitude finally matters, and Topic 7.5 derives the pendulum period this topic simply states. The rolling oscillator in worked example 1 leans on Topic 6.5. The Unit 7 hub lists all five topics.

A cylinder that rolls on a spring, where no printed formula applies

A uniform solid cylinder of mass $m = 1.6$ kg and radius $R = 0.10$ m rolls without slipping on a horizontal surface. A spring of force constant $k = 24\ \mathrm{N/m}$ is attached to its axle and to a wall. Its rotational inertia about its centre is $\frac{1}{2}mR^2$ . Find (a) the equation of motion, (b) the angular frequency and period, and (c) how the period compares with the same cylinder sliding on a frictionless surface with the same spring.

Declare the convention: $x$ is the displacement of the cylinder's centre from the spring's equilibrium position, positive away from the wall, and rolling without slipping ties the angular speed to the centre's speed by $v_{\text{cm}} = R\omega_{\text{roll}}$ .
(a) Take the energy route, which is the shortest one available in this course. The sheet prints $K = \frac{1}{2}mv^2$ , $K_{\text{rot}} = \frac{1}{2}I\omega^2$ and $U_s = \frac{1}{2}k(\Delta x)^2$ , so the total mechanical energy is $E = \frac{1}{2}mv^2 + \frac{1}{2}I\omega_{\text{roll}}^2 + \frac{1}{2}kx^2$ .
Substitute $I = \frac{1}{2}mR^2$ and $\omega_{\text{roll}} = v/R$ : the rotational term becomes $\frac{1}{2}\left(\frac{1}{2}mR^2\right)\dfrac{v^2}{R^2} = \frac{1}{4}mv^2$ , so $E = \frac{3}{4}mv^2 + \frac{1}{2}kx^2$ .
Rolling without slipping does no work at the contact point, so $E$ is constant. Differentiate with respect to time: $\dfrac{dE}{dt} = \frac{3}{2}mv\dfrac{dv}{dt} + kx\dfrac{dx}{dt} = 0$ .
With $dx/dt = v$ and $dv/dt = a$ , divide through by $v$ : $\frac{3}{2}ma + kx = 0$ , so $\dfrac{d^2x}{dt^2} = -\dfrac{2k}{3m}x$ .
That has the shape of 7.3.A.2, so the motion is simple harmonic and $\omega^2 = \dfrac{2k}{3m}$ . Notice that the printed $T_s = 2\pi\sqrt{m/k}$ does not apply: the spring is the same, but a third of the kinetic energy is now going into rotation.
The force route gives the same answer. Translation gives $-kx + f = ma$ with $f$ the friction force, and taking torques about the centre of mass, where the spring exerts none, gives $f = -\frac{1}{2}ma$ . Substituting, $-kx = \frac{3}{2}ma$ , identical.
(b) $\omega^2 = \dfrac{2(24)}{3(1.6)} = \dfrac{48}{4.8} = 10.0\ \mathrm{rad^2/s^2}$ , so $\omega = \sqrt{10.0} = 3.1623\ \mathrm{rad/s}$ .
The sheet prints $T = 2\pi/\omega$ , so $T = \dfrac{2\pi}{3.1623} = 1.9869 \approx 1.99\ \mathrm{s}$ . For the record, $I = \frac{1}{2}(1.6)(0.10)^2 = 8.0 \times 10^{-3}\ \mathrm{kg \cdot m^2}$ , though it never had to be evaluated separately.
(c) Sliding, the printed formula applies: $\omega_{\text{slide}} = \sqrt{k/m} = \sqrt{24/1.6} = \sqrt{15} = 3.8730\ \mathrm{rad/s}$ , so $T_{\text{slide}} = 2\pi/3.8730 = 1.6223 \approx 1.62\ \mathrm{s}$ .
The ratio is $1.9869/1.6223 = 1.2247$ , which is $\sqrt{3/2}$ exactly, since $\omega^2$ fell from $k/m$ to $\frac{2}{3}(k/m)$ . Rolling slows the oscillator by a fixed factor set only by $I/(mR^2)$ , and the radius, the mass and the force constant all cancel out of that factor.

(a) $d^2x/dt^2 = -\dfrac{2k}{3m}x$ , derived by differentiating the total energy. (b) $\omega = 3.16\ \mathrm{rad/s}$ and $T = 1.99$ s. (c) Sliding gives 1.62 s, so rolling lengthens the period by a factor of exactly $\sqrt{3/2} = 1.22$ , and no printed formula covers the rolling case.

Ranking four spring oscillators by six quantities at once

Take a baseline oscillator: a block of mass $m = 0.50$ kg on an ideal spring of force constant $k = 200\ \mathrm{N/m}$ , amplitude $A = 0.040$ m. Compare it with three variants: Case 2 has four times the mass, Case 3 has four times the force constant, and Case 4 has twice the amplitude. Everything else is unchanged in each case. Rank all four by period, frequency, maximum speed, maximum acceleration, maximum force and total energy.

Do it by exponents, not by four separate arithmetic runs. The relevant dependences are $\omega = \sqrt{k/m}$ , $T = 2\pi/\omega$ , $v_{\max} = A\omega$ , $a_{\max} = A\omega^2$ , $F_{\max} = kA$ and $E = \frac{1}{2}kA^2$ . Only the first two are printed; the rest are statements 7.3.A.3.ii and 7.4.A.4.ii, and 7.1.A.2 for the force.
Baseline: $\omega_0 = \sqrt{200/0.50} = \sqrt{400} = 20\ \mathrm{rad/s}$ , so $T_0 = 2\pi/20 = 0.3142$ s and $f_0 = 3.183$ Hz. Then $v_{\max} = (0.040)(20) = 0.80\ \mathrm{m/s}$ , $a_{\max} = (0.040)(400) = 16\ \mathrm{m/s^2}$ , $F_{\max} = (200)(0.040) = 8.0$ N and $E = \frac{1}{2}(200)(0.040)^2 = 0.16$ J.
Case 2, mass $\times 4$ : $\omega$ halves to $10\ \mathrm{rad/s}$ , so $T = 0.6283$ s and $f = 1.592$ Hz. $v_{\max} = 0.40\ \mathrm{m/s}$ and $a_{\max} = (0.040)(100) = 4.0\ \mathrm{m/s^2}$ . $F_{\max}$ and $E$ contain no mass, so they stay at 8.0 N and 0.16 J.
Case 3, force constant $\times 4$ : $\omega$ doubles to $40\ \mathrm{rad/s}$ , so $T = 0.1571$ s and $f = 6.366$ Hz. $v_{\max} = 1.60\ \mathrm{m/s}$ , $a_{\max} = (0.040)(1600) = 64\ \mathrm{m/s^2}$ , $F_{\max} = (800)(0.040) = 32$ N and $E = \frac{1}{2}(800)(0.0016) = 0.64$ J.
Case 4, amplitude $\times 2$ : $\omega$ is unchanged at $20\ \mathrm{rad/s}$ , so $T = 0.3142$ s and $f = 3.183$ Hz, identical to the baseline. $v_{\max} = (0.080)(20) = 1.60\ \mathrm{m/s}$ , $a_{\max} = (0.080)(400) = 32\ \mathrm{m/s^2}$ , $F_{\max} = (200)(0.080) = 16$ N and $E = \frac{1}{2}(200)(0.0064) = 0.64$ J.

Quantity	Case 1 (base)	Case 2 ( $4m$ )	Case 3 ( $4k$ )	Case 4 ( $2A$ )
$T$ (s)	0.314	0.628	0.157	0.314
$f$ (Hz)	3.18	1.59	6.37	3.18
$v_{\max}$ (m/s)	0.80	0.40	1.60	1.60
$a_{\max}$ (m/s $^2$ )	16	4.0	64	32
$F_{\max}$ (N)	8.0	8.0	32	16
$E$ (J)	0.16	0.16	0.64	0.64

Period, longest to shortest: Case 2, then Cases 1 and 4 tied, then Case 3. Frequency is the exact reverse, because $f = 1/T$ .
Maximum speed: Cases 3 and 4 tied at the top, then Case 1, then Case 2. That tie is worth staring at, because the two cases reach the same top speed by different routes: Case 3 covers the same distance faster, Case 4 covers twice the distance in the same time.
Maximum acceleration: Case 3, then Case 4, then Case 1, then Case 2. The tie from the previous ranking has broken, because $a_{\max}$ carries $\omega^2$ and $v_{\max}$ carries only $\omega$ .
Maximum force: Case 3, then Case 4, then Cases 1 and 2 tied. Total energy: Cases 3 and 4 tied at 0.64 J, then Cases 1 and 2 tied at 0.16 J.
Cross-check Case 2's force and energy against the physics rather than the algebra. Adding mass does not change how hard the spring is pulled at a given extension, so $F_{\max} = kA$ cannot move, and the energy stored at maximum displacement is entirely the spring's, so $E$ cannot move either. What extra mass does is make the block slower to respond, which is the period lengthening.

Period: Case 2 (0.628 s) > Cases 1 and 4 (0.314 s) > Case 3 (0.157 s); frequency reverses that order. Maximum speed: Cases 3 and 4 tied at 1.60 m/s > Case 1 (0.80) > Case 2 (0.40). Maximum acceleration: 64, 32, 16, 4.0 m/s $^2$ for Cases 3, 4, 1, 2. Maximum force: 32, 16, 8.0, 8.0 N for Cases 3, 4, 1, 2. Total energy: Cases 3 and 4 tied at 0.64 J, Cases 1 and 2 tied at 0.16 J. Adding mass moves the period and leaves the force and energy alone.

Measuring the length of a long string with a stopwatch

A cable hangs from a high ceiling with a small heavy bob on the end, and its length cannot be reached with a tape measure. Displaced through a small angle, it completes 25 full oscillations in 148.0 s. Take $g = 9.8\ \mathrm{m/s^2}$ . Find (a) the period, (b) the length, (c) how much better timing 25 cycles is than timing one, and (d) three things that would bias the answer.

This is the CED's own essential question for Unit 7, which asks how you could measure the length of a long string with a stopwatch. The model is 7.2.A.1.ii, the simple pendulum at small angular displacement.
(a) $T = \dfrac{148.0\ \mathrm{s}}{25} = 5.920\ \mathrm{s}$ . The angular frequency, which is the quantity this course works in, is $\omega = 2\pi/T = 2\pi/5.920 = 1.0614\ \mathrm{rad/s}$ .
(b) Rearrange the printed $T_p = 2\pi\sqrt{\ell/g}$ symbolically before substituting anything, which is what skill 2.A asks for: square both sides to get $T^2 = \dfrac{4\pi^2\ell}{g}$ , so $\ell = \dfrac{gT^2}{4\pi^2}$ .
$T^2 = (5.920)^2 = 35.046\ \mathrm{s^2}$ , so $\ell = \dfrac{(9.8)(35.046)}{39.478} = \dfrac{343.45}{39.478} = 8.6998 \approx 8.70\ \mathrm{m}$ .
Check by going forward: $2\pi\sqrt{8.6998/9.8} = 2\pi\sqrt{0.88774} = 2\pi(0.94220) = 5.9200\ \mathrm{s}$ , matching the measured period to every digit carried.
(c) Since $\ell \propto T^2$ , a fractional error in $T$ doubles into $\ell$ . Suppose each stopwatch reading carries about 0.2 s of reaction-time uncertainty. Timing a single period, that is $0.2/5.920 = 3.4\%$ in $T$ and about $6.8\%$ in $\ell$ , which is $\pm 0.6$ m on an 8.7 m cable.
Timing 25 periods, the same 0.2 s is spread over 148.0 s: $0.2/148.0 = 0.135\%$ in $T$ and about $0.27\%$ in $\ell$ , which is $\pm 0.02$ m. The uncertainty fell by a factor of 25 for no extra equipment, which is exactly why the CED's error-reducing list opens with timing multiple periods.
(d) Three biases, each in a known direction. First, amplitude: the true restoring torque goes as $\sin\theta$ , which is smaller than $\theta$ , so a large swing gives a period longer than the model and an answer for $\ell$ that is too big. Keep the angle small and say so.
Second, where you measure the length to: statement 7.5.A.3 defines $\ell$ as the distance from the pivot to the point mass, so the length in the formula runs to the bob's centre of mass, not to the top of the bob. Finding that centre of mass precisely is the third item on the CED's error-reducing list.
Third, the model itself: a bob with appreciable size is a physical pendulum, not a simple one, and its period is set by $I/(mgd)$ rather than by $\ell$ alone. That correction is Topic 7.5's business, and it is small only when the bob is small compared with the cable.
Sanity check the size of the answer against something familiar. A 1.00 m pendulum has $T = 2\pi\sqrt{1.00/9.8} = 2.007$ s, and periods scale as $\sqrt{\ell}$ , so a 5.92 s period needs $(5.920/2.007)^2 = 8.70$ times that length. Consistent.

(a) $T = 5.920$ s, so $\omega = 1.06\ \mathrm{rad/s}$ . (b) $\ell = gT^2/(4\pi^2) = 8.70$ m, confirmed by substituting back. (c) Timing 25 cycles instead of one cuts the length uncertainty from roughly 7% to roughly 0.3%, because reaction time is a fixed number of seconds per timing. (d) A large amplitude biases the length high, measuring to the wrong point on the bob biases it either way, and a large bob makes the simple-pendulum model itself wrong.

Frequently asked questions

What is the formula for the period of simple harmonic motion in AP Physics C?

Essential knowledge 7.2.A.1 gives the period as two pi divided by the angular frequency, which also equals one over the frequency, and that line is printed on the AP Physics C: Mechanics equation sheet. Two specific cases are printed with it: the period of an object on an ideal spring is two pi times the square root of the mass over the force constant, and the period of a simple pendulum at small angular displacement is two pi times the square root of the length over the gravitational field strength. The sheet also prints the physical pendulum period, which is two pi times the square root of the rotational inertia over the product of mass, gravitational field strength and the distance from the axis to the centre of mass.

How do you find the period of an oscillator that has no formula on the equation sheet?

Write Newton's second law for the system, in translational or rotational form as appropriate, rearrange until the second derivative of the oscillating variable is alone on one side, and compare with the framework's derived equation, which sets that second derivative equal to minus the angular frequency squared times the variable. Whatever multiplies the variable is the angular frequency squared. Take its square root and use the printed relation that the period is two pi over the angular frequency. This is the only route for a torsion pendulum, for a rolling object attached to a spring, and for any oscillator invented in the question stem, and it reproduces every printed formula as a special case.

Does the mass of the bob change a pendulum's period?

No. The printed period of a simple pendulum contains only the length and the gravitational field strength, so the bob's mass cancels out entirely. The physical reason is that gravity supplies both the restoring effect and the inertia that resists it, and both scale with mass. The same cancellation holds for a physical pendulum, where the rotational inertia and the product of mass and centre-of-mass distance are both proportional to mass. A block on a spring is the opposite case: its printed period contains the mass explicitly, so quadrupling the mass doubles the period. The AP Physics C: Mechanics course description uses exactly that contrast as sample multiple-choice question 5.

Does amplitude affect the period of simple harmonic motion?

No. Essential knowledge 7.3.A.5 says plainly that changing the amplitude of a system exhibiting simple harmonic motion will not change its period, and neither printed period formula contains an amplitude. There is one conditional attached, and it belongs to pendulums rather than springs. A pendulum's restoring torque is proportional to the sine of the angle, not to the angle, so it is only simple harmonic under the small-angle approximation of statement 7.5.A.2.ii. Outside that range the period does grow with amplitude, and it grows upward, because the sine of an angle is always smaller than the angle so the real restoring effect is weaker than the model.

What is angular frequency in simple harmonic motion, and how is it different from frequency?

Angular frequency is measured in radians per second and frequency is measured in cycles per second, so the two differ by the two pi radians in one cycle: the angular frequency is two pi times the frequency. On the AP Physics C: Mechanics equation sheet the two are linked through the printed line that sets the period equal to two pi over the angular frequency and equal to one over the frequency. Angular frequency is the quantity that appears in the differential equation, in the cosine solution, and in the derived expressions for maximum speed and maximum acceleration, which is why it is the one to solve for first. Nothing is physically rotating in an oscillating block, despite the name.

How do you find a spring constant from measured periods?

Square the printed spring-period formula so that the period squared equals four pi squared times the mass divided by the force constant. That makes a graph of period squared against mass a straight line through the origin whose slope is four pi squared over the force constant, so the force constant is four pi squared divided by the slope. Use a slope over several masses rather than one measurement, and time many oscillations for each mass rather than one, since reaction-time uncertainty is a fixed number of seconds per timing. A second, independent route needs no timing at all: hang a known mass from a vertical spring and the static stretch gives the force constant as the weight divided by the stretch.

Why does a spring oscillator have the same period on the Moon as on Earth?

Because the printed period of a spring oscillator contains only the mass and the force constant, and no gravitational field strength. For a vertical spring, gravity does one job: it moves the equilibrium position to wherever the spring is stretched by the weight. Measure displacement from that new equilibrium and the gravitational force cancels exactly against the extra spring force, leaving the same equation of motion as a horizontal spring. So a block hangs lower on Earth than on the Moon and oscillates about its own hanging position at the same rate. A pendulum is the opposite case: its printed period goes as one over the square root of the gravitational field strength, so it runs slower where gravity is weaker.