AP Physics C Mechanics Quiz: Simple And Physical Pendulums

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Simple And Physical PendulumsQuestion 1 of 20

A torsion pendulum consists of a solid disk with rotational inertia $I$ attached to a wire with torsion constant $\kappa$ . It oscillates with a period $T_0$ . If the disk is replaced with a hoop of the same mass and radius, what is the new period? The rotational inertia of a hoop is twice that of a solid disk of the same mass and radius.

T_0/\sqrt{2}

T_0/2

\sqrt{2}T_0

2T_0

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Simple And Physical Pendulums

Practice Simple And Physical Pendulums in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Simple And Physical Pendulums, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$T_0/\sqrt{2}$
$T_0/2$
$\sqrt{2}T_0$ (correct answer)
$2T_0$

Explanation: The period of a torsion pendulum is

T = 2\pi\sqrt{I/\kappa}

. Since the period is proportional to the square root of the rotational inertia (

T \propto \sqrt{I}

), and the new rotational inertia is

I' = 2I

, the new period will be

T' = \sqrt{2}T_0

Question 2

A simple pendulum and a physical pendulum, consisting of a uniform rod of length $L$ pivoted at one end, both have the same length $L$ . How does the period of the rod, $T_{rod}$ , compare to the period of the simple pendulum, $T_{simple}$ ?

$T_{rod} < T_{simple}$ (correct answer)
$T_{rod} > T_{simple}$
$T_{rod} = T_{simple}$
The comparison depends on the mass of the rod and the pendulum bob.

Explanation: The period of the simple pendulum is

T_{simple} = 2\pi\sqrt{L/g}

. The period of the rod is

T_{rod} = 2\pi\sqrt{2L/3g}

. Since

\sqrt{2/3} \approx 0.816

, we have

T_{rod} \approx 0.816 \, T_{simple}

, so the rod's period is shorter.

Question 3

A torsion pendulum is constructed by suspending a horizontal rod of mass $M$ and length $L$ by a wire attached to its center. The rod has a rotational inertia $I_{rod} = \frac{1}{12}ML^2$ . Two small spheres, each of mass $m$ , are then attached to the ends of the rod. What is the new rotational inertia of the system?

$\frac{1}{12}ML^2 + 2m(L/2)^2$ (correct answer)
$\frac{1}{12}ML^2 + m(L/2)^2$
$\frac{1}{12}(M+2m)L^2$
$\frac{1}{12}ML^2 + 2mL^2$

Explanation: The total rotational inertia of the system is the sum of the rotational inertias of its components. The rod's inertia is given. Each sphere of mass

m

is at a distance

r=L/2

from the axis of rotation. Treating the spheres as point masses, their rotational inertia is

I_{sphere} = mr^2 = m(L/2)^2

. Since there are two spheres, their total contribution is

2m(L/2)^2

. The total inertia is

I_{total} = I_{rod} + I_{spheres} = \frac{1}{12}ML^2 + 2m(L/2)^2

Question 4

A uniform solid disk of radius $R$ is pivoted at its center and used as a torsion pendulum. It has a period $T_C$ . The disk is then re-pivoted at its rim to oscillate as a physical pendulum in Earth's gravitational field. Its period is $T_P$ . How do these periods depend on the mass of the disk?

Both periods depend on the mass of the disk.
Both periods are independent of the mass of the disk. (correct answer)
The period $T_C$ depends on mass, while $T_P$ is independent of mass.
The period $T_C$ is independent of mass, while $T_P$ depends on mass.

Explanation: For the torsion pendulum,

T_C = 2\pi\sqrt{I/\kappa}

, where

\kappa

is the torsion constant of the wire. While

I \propto M

, the torsion constant

\kappa

for a given wire and angular displacement is also proportional to

M

(since the restoring torque per unit angle depends on the inertial properties of the system). Thus the mass cancels and

T_C

is independent of mass. For the physical pendulum,

T_P = 2\pi\sqrt{I'/Mgd}

. Since both

I'

and

M

are proportional to mass, mass cancels and

T_P

is also independent of mass.

Question 5

A uniform rod of mass $M$ and length $L$ is pivoted at one end and oscillates as a physical pendulum. The rotational inertia of the rod about its end is $\frac{1}{3}ML^2$ . What is the period of the rod for small-amplitude oscillations?

$2\pi\sqrt{\frac{L}{g}}$
$2\pi\sqrt{\frac{2L}{3g}}$ (correct answer)
$2\pi\sqrt{\frac{L}{3g}}$
$2\pi\sqrt{\frac{L}{2g}}$

Explanation: The period of a physical pendulum is given by

T = 2\pi\sqrt{\frac{I}{mgd}}

. For a uniform rod pivoted at one end, the rotational inertia is

I = \frac{1}{3}ML^2

and the distance from the pivot to the center of mass is

d = L/2

. Substituting these values gives

T = 2\pi\sqrt{\frac{\frac{1}{3}ML^2}{Mg(L/2)}} = 2\pi\sqrt{\frac{2L}{3g}}

Question 6

A physical pendulum consists of a rigid body oscillating about a fixed pivot point. For the pendulum's motion to be accurately modeled as simple harmonic motion, which of the following conditions is required?

The mass of the object must be concentrated at its center of mass for the model to apply.
The period of oscillation must be independent of the mass of the object.
The amplitude of the oscillation must be small enough that $\sin\theta \approx \theta$ . (correct answer)
The pivot point must be located at one end of the object for the motion to be harmonic.

Explanation: The restoring torque for a physical pendulum is

\tau = -mgd\sin\theta

. Simple harmonic motion requires the restoring torque to be proportional to the angular displacement,

\tau = -k\theta

. This condition is met when the oscillation amplitude is small, allowing for the approximation

\sin\theta \approx \theta

Question 7

A simple pendulum of length $L$ has a period $T$ on Earth. The pendulum is moved to a planet with a radius twice that of Earth and a mass eight times that of Earth. What is the new period of the pendulum on this planet?

$T/2$
$T/\sqrt{2}$ (correct answer)
$\sqrt{2}T$
$2T$

Explanation: The acceleration due to gravity on a planet's surface is

g = GM/R^2

. The new gravity is

g' = G(8M)/(2R)^2 = G(8M)/(4R^2) = 2(GM/R^2) = 2g

. The period of a simple pendulum is

T = 2\pi\sqrt{L/g}

. The new period is

T' = 2\pi\sqrt{L/g'} = 2\pi\sqrt{L/(2g)} = (1/\sqrt{2})T

Question 8

A solid sphere of mass $M$ and radius $R$ is pivoted to oscillate as a physical pendulum about an axis tangent to its surface. The rotational inertia of a solid sphere about its center of mass is $\frac{2}{5}MR^2$ . What is the rotational inertia of the sphere about the pivot axis?

$\frac{2}{5}MR^2$
$MR^2$
$\frac{3}{5}MR^2$
$\frac{7}{5}MR^2$ (correct answer)

Explanation: The parallel-axis theorem states

I = I_{cm} + Md^2

. For a sphere pivoted at its surface, the distance

d

from the center of mass to the pivot is the radius

R

. Therefore,

I = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2

Question 9

The differential equation of motion for a physical pendulum is $I\frac{d^2\theta}{dt^2} = -mgd\sin\theta$ . For small oscillations, this equation is approximated to represent simple harmonic motion. What is the angular frequency $\omega$ of this simple harmonic motion?

$\sqrt{\frac{mgd}{I}}$ (correct answer)
$\sqrt{\frac{I}{mgd}}$
$\frac{mgd}{I}$
$\frac{I}{mgd}$

Explanation: For small angles,

\sin\theta \approx \theta

, so the equation becomes

I\frac{d^2\theta}{dt^2} = -mgd\theta

. This can be rewritten as

\frac{d^2\theta}{dt^2} = -(\frac{mgd}{I})\theta

. The general form for SHM is

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

. By comparison,

\omega^2 = \frac{mgd}{I}

, so

\omega = \sqrt{\frac{mgd}{I}}

Question 10

A pendulum clock keeps perfect time on Earth. If the clock is moved to a location on the Moon where the acceleration due to gravity is approximately one-sixth that of Earth, what would an observer on the Moon note about the clock's timekeeping over a long period?

The clock runs slow, taking more time to complete each oscillation. (correct answer)
The clock runs fast, taking less time to complete each oscillation.
The clock keeps the same time, as the period is independent of gravity.
The clock stops, as a pendulum cannot oscillate with such low gravity.

Explanation: The period of a pendulum is inversely proportional to the square root of the acceleration due to gravity (

T \propto 1/\sqrt{g}

). With lower gravity on the Moon, the period of oscillation will increase. A longer period means each swing takes more time, so the clock will run slow.

Question 11

A hoop of radius $R$ and mass $M$ is pivoted at its rim and oscillates as a physical pendulum. The rotational inertia of a hoop about its center is $MR^2$ . What is the period of the hoop for small-amplitude oscillations?

$2\pi\sqrt{\frac{R}{g}}$
$2\pi\sqrt{\frac{R}{2g}}$
$2\pi\sqrt{\frac{3R}{2g}}$
$2\pi\sqrt{\frac{2R}{g}}$ (correct answer)

Explanation: Using the parallel-axis theorem, the rotational inertia about the rim is

I = I_{cm} + Md^2 = MR^2 + MR^2 = 2MR^2

, where

d=R

. The period is

T = 2\pi\sqrt{\frac{I}{Mgd}} = 2\pi\sqrt{\frac{2MR^2}{MgR}} = 2\pi\sqrt{\frac{2R}{g}}

Question 12

A physical pendulum has a period $T$ . If the entire apparatus is placed in an elevator that is accelerating upward with a constant acceleration of magnitude $a$ , what is the new period of oscillation, $T'$ ?

$T' = T \sqrt{\frac{g}{g+a}}$ (correct answer)
$T' = T \sqrt{\frac{g+a}{g}}$
$T' = T \frac{g}{g+a}$
$T' = T$

Explanation: In the accelerating frame of the elevator, the effective gravitational acceleration is

g_{eff} = g+a

. The period of a physical pendulum is

T = 2\pi\sqrt{I/(mgd)}

, so it is inversely proportional to the square root of

g

. The new period will be

T' = 2\pi\sqrt{I/(mg_{eff}d)} = 2\pi\sqrt{I/(m(g+a)d)} = T\sqrt{g/(g+a)}

Question 13

A simple pendulum oscillates with a period $T$ when its maximum angular displacement is $5^\circ$ . According to the idealized model of simple harmonic motion, what would be the period if the maximum angular displacement were increased to $10^\circ$ ?

$T/2$
$T$ (correct answer)
$\sqrt{2}T$
$2T$

Explanation: In the model for simple harmonic motion of a pendulum (which relies on the small-angle approximation), the period is independent of the amplitude of oscillation. Therefore, changing the maximum displacement from

5^\circ

10^\circ

(both of which are considered small angles) does not change the period.

Question 14

A uniform rectangular plate of mass $M$ , width $W$ , and height $H$ is pivoted at its top left corner to swing as a physical pendulum. The rotational inertia about its center of mass is $I_{cm} = \frac{1}{12}M(W^2+H^2)$ . What is the distance $d$ from the pivot to the center of mass?

$\frac{1}{2}\sqrt{W^2+H^2}$ (correct answer)
$\frac{1}{2}(W+H)$
$\sqrt{W^2+H^2}$
$H/2$

Explanation: The center of mass of a uniform rectangular plate is at its geometric center. The pivot is at a corner. The coordinates of the center can be taken as

(W/2, H/2)

relative to the corner pivot at

(0,0)

. The distance

d

is found using the Pythagorean theorem:

d = \sqrt{(W/2)^2 + (H/2)^2} = \frac{1}{2}\sqrt{W^2+H^2}

Question 15

A uniform meter stick is pivoted at one end. A small object of the same mass as the meter stick is attached to the other end. The system oscillates as a physical pendulum. Let L be the length of the stick. The rotational inertia of the stick about the pivot is $\frac{1}{3}ML^2$ . What is the rotational inertia of the combined system about the pivot?

$\frac{1}{3}ML^2$
$\frac{2}{3}ML^2$
$ML^2$
$\frac{4}{3}ML^2$ (correct answer)

Explanation: The total rotational inertia is the sum of the inertias of the stick and the small object. The stick's inertia is

I_{stick} = \frac{1}{3}ML^2

. The object can be treated as a point mass

M

at a distance

L

from the pivot, so its inertia is

I_{object} = ML^2

. The total inertia is

I_{total} = I_{stick} + I_{object} = \frac{1}{3}ML^2 + ML^2 = \frac{4}{3}ML^2

Question 16

A uniform rod of length $L=1.0$ m is pivoted at a distance $d=0.25$ m from its center and oscillates as a physical pendulum. The rotational inertia of a rod about its center is $\frac{1}{12}ML^2$ . Which of the following expressions correctly gives the period of small oscillations?

$2\pi\sqrt{\frac{(1/12)L^2}{g(0.25)}}$
$2\pi\sqrt{\frac{(1/12)L^2 + (0.25)^2}{g(0.25)}}$ (correct answer)
$2\pi\sqrt{\frac{(1/12)L^2 + (0.25)^2}{gL}}$
$2\pi\sqrt{\frac{L^2}{g(0.25)}}$

Explanation: The period is

T = 2\pi\sqrt{I/Mgd}

. The distance from the pivot to the center of mass is given as

d=0.25

m. By the parallel-axis theorem, the rotational inertia about the pivot is

I = I_{cm} + Md^2 = \frac{1}{12}ML^2 + M(0.25)^2

. Substituting this into the period formula and cancelling M gives

T = 2\pi\sqrt{\frac{(1/12)L^2 + (0.25)^2}{g(0.25)}}

Question 17

A physical pendulum has a certain period of oscillation. If the mass of the pendulum is doubled, but its shape, size, and pivot point remain exactly the same, what is the effect on its period?

The period is halved because it is inversely proportional to mass.
The period is doubled because it is directly proportional to mass.
The period remains the same because the effects of mass in inertia and torque cancel. (correct answer)
The period increases by a factor of $\sqrt{2}$ because it is proportional to the square root of mass.

Explanation: The period of a physical pendulum is

T = 2\pi\sqrt{I/(mgd)}

. The rotational inertia

I

is directly proportional to the mass

m

(e.g.,

I = \beta m R^2

for some constant

\beta

). Therefore, the mass

m

in the numerator (within

I

) cancels with the mass

m

in the denominator, making the period independent of mass.

Question 18

A simple pendulum oscillates with period $T_0$ when the amplitude is very small. When the maximum angular displacement is increased to $30°$ , the period becomes approximately $T$ . Which of the following best represents the relationship between $T$ and $T_0$ ?

$T = T_0$
$T = 1.04 T_0$ (correct answer)
$T = 0.96 T_0$
$T = 1.15 T_0$

Explanation: For large amplitude oscillations, the period of a simple pendulum increases beyond the small angle approximation value. The exact formula involves elliptic integrals, but the first-order correction gives

T \approx T_0(1 + \frac{1}{16}\theta_{max}^2)

where

\theta_{max}

is in radians. For

30° = \frac{\pi}{6}

radians ≈ 0.524 radians, we get

T \approx T_0(1 + \frac{1}{16} \times 0.524^2) ≈ T_0(1 + 0.017) ≈ 1.04 T_0

. Choice A assumes the small angle approximation still holds. Choice C incorrectly suggests the period decreases. Choice D overestimates the correction factor.

Question 19

A simple pendulum of length $L$ oscillates with small amplitude in a region where the gravitational acceleration is $g$ . If the pendulum bob is replaced with one having twice the mass and the length is increased by a factor of 4, what is the ratio of the new period to the original period?

$\frac{1}{2}$
$1$
$2$ (correct answer)
$4$

Explanation: The period of a simple pendulum is

T = 2\pi\sqrt{\frac{L}{g}}

, which is independent of the mass of the bob. When the length increases by a factor of 4, the new period becomes

T_{new} = 2\pi\sqrt{\frac{4L}{g}} = 2 \cdot 2\pi\sqrt{\frac{L}{g}} = 2T_{original}

. The mass change has no effect on the period. Choice A results from incorrectly thinking the period decreases with length. Choice B assumes both mass and length changes cancel out. Choice D incorrectly applies the length factor directly without taking the square root.

Question 20

A compound pendulum consists of a uniform disk of radius $R$ and mass $M$ that can pivot about a horizontal axis passing through a point on its rim. If the disk oscillates with small amplitude, what is the length of the equivalent simple pendulum that would have the same period?

$\frac{R}{2}$
$R$
$\frac{3R}{2}$ (correct answer)
$2R$

Explanation: For a physical pendulum, the equivalent simple pendulum length is

L_{eq} = \frac{I}{md}

, where

I

is the moment of inertia about the pivot,

m

is the mass, and

d

is the distance from pivot to center of mass. For a disk pivoting at its rim,

d = R

. The moment of inertia about the rim is

I = I_{center} + MR^2 = \frac{MR^2}{2} + MR^2 = \frac{3MR^2}{2}

. Therefore,

L_{eq} = \frac{3MR^2/2}{M \cdot R} = \frac{3R}{2}

. Choice A uses only the center-of-mass moment of inertia. Choice B assumes the equivalent length equals the distance to center of mass. Choice D incorrectly doubles the radius without proper calculation.