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AP Physics C: Mechanics Practice Quiz: Simple and Physical Pendulums

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 9 questions to check your progress.

Question 1 of 9

A simple pendulum is described as a special case of a physical pendulum. This is because the simple pendulum model assumes that the hanging object's...

All Questions (9)

A simple pendulum is described as a special case of a physical pendulum. This is because the simple pendulum model assumes that the hanging object's...

A) mass is concentrated at a single point.

B) motion has a very large amplitude.

C) mass is negligible compared to the string.

D) period is independent of gravity.

Correct Answer: A

The provided content explicitly states that for a simple pendulum, 'the hanging object can be modeled as a point mass at a distance, ℓ, from the pivot point.' This is the key simplifying assumption that makes it a special case of a more general physical pendulum.

A student creates a simple pendulum of length ℓ and measures its period T. If the student wants to create a new simple pendulum with a period of 2T, what should the length of the new pendulum be?

A) √2 ℓ

B) 2ℓ

C) 4ℓ

D) ½ ℓ

Correct Answer: C

The period of a simple pendulum is given by T_p = 2π√(ℓ/g). This shows that the period T is proportional to the square root of the length ℓ (T ∝ √ℓ). To double the period (T → 2T), the term inside the square root must be quadrupled. Therefore, the length must be increased to 4ℓ, because √(4ℓ) = 2√ℓ.

The period equations for both simple and physical pendulums are derived from the application of which fundamental principle for small amplitudes of motion?

A) Conservation of Energy

B) Newton’s Second Law in rotational form

C) Conservation of Linear Momentum

D) The Ideal Gas Law

Correct Answer: B

The provided text directly states: 'For small amplitudes of motion, the period of a physical pendulum is derived from the application of Newton’s second law in rotational form.' Since the simple pendulum is a special case, its equation is also derived from this principle.

A uniform meter stick is allowed to swing back and forth about a pivot at its top end. Which model and corresponding equation should be used to most accurately determine its period of oscillation for small amplitudes?

A) Simple pendulum, using T_p = 2π√(ℓ/g), because the stick has a defined length.

B) Physical pendulum, using T_phys = 2π√(I/mgd), because the mass is distributed along the length of the stick.

C) Simple pendulum, using T_p = 2π√(ℓ/g), because the mass of the stick is uniform.

D) Physical pendulum, using T_phys = 2π√(I/mgd), because the amplitude of motion must be small.

Correct Answer: B

A meter stick is a rigid body with its mass distributed over its entire volume, not concentrated at a single point. Therefore, it must be treated as a physical pendulum. The simple pendulum model is an idealization that would not be accurate for this scenario.

According to the equation T_p = 2π√(ℓ/g), if the mass of the bob of a simple pendulum is doubled while its length ℓ is kept constant, the new period will be...

A) half the original period.

B) the same as the original period.

C) double the original period.

D) four times the original period.

Correct Answer: B

The equation for the period of a simple pendulum, T_p = 2π√(ℓ/g), does not include the mass 'm'. This indicates that for an idealized simple pendulum, the period is independent of the mass of the bob.

The equation for a simple pendulum can be derived from the physical pendulum equation T_phys = 2π√(I/mgd). This is achieved by modeling the simple pendulum's bob as a point mass, for which the moment of inertia is I = mℓ² and the distance to the center of mass is d = ℓ. Which of the following shows the correct substitution of these values into the physical pendulum equation?

A) T = 2π√((mℓ²)/(mgℓ))

B) T = 2π√(I/(mgℓ))

C) T = 2π√((mℓ²)/(mgd))

D) T = 2π√(ℓ²/(gℓ))

Correct Answer: A

To derive the simple pendulum equation, both I and d in the physical pendulum equation must be replaced with their point-mass equivalents. Substituting I = mℓ² and d = ℓ into T_phys = 2π√(I/mgd) yields T = 2π√((mℓ²)/(mgℓ)), which simplifies to the simple pendulum equation T_p = 2π√(ℓ/g).

A physical pendulum has a period T on Earth. If this pendulum is taken to a planet where the acceleration due to gravity is 4g, what will its new period be?

A) 4T

B) 2T

C) T/2

D) T/4

Correct Answer: C

The period of a physical pendulum is given by T_phys = 2π√(I/mgd). This shows that the period T is inversely proportional to the square root of g (T ∝ 1/√g). If g becomes 4g, the new period T' will be proportional to 1/√(4g), which is (1/2) * (1/√g). Therefore, the new period will be T/2.

According to the provided equations, the period of a physical pendulum depends on its moment of inertia, I, while the period of a simple pendulum does not explicitly show this dependence. This implies that the period of a physical pendulum, unlike a simple pendulum, is dependent on the...

A) local acceleration due to gravity.

B) total length from the pivot.

C) object's total mass.

D) distribution of the object's mass about the pivot.

Correct Answer: D

Moment of inertia (I) is a measure of an object's resistance to rotational acceleration and depends on how its mass is distributed relative to the axis of rotation. The simple pendulum model assumes all mass is at a single point, so its 'distribution' is fixed by the length ℓ. The physical pendulum model accounts for the actual shape and mass distribution through the variable I.

Consider a physical pendulum made from a flat, uniform, rigid material of a specific shape, pivoted at a certain point. If this pendulum is replaced by another one made of a denser material but with the exact same shape and dimensions, how will the period of oscillation change?

A) The period will increase because the mass m increases.

B) The period will decrease because the moment of inertia I increases.

C) The period will remain unchanged.

D) The change in period cannot be determined without knowing the specific shape.

Correct Answer: C

The period is T_phys = 2π√(I/mgd). For a rigid body of a given shape, the moment of inertia I is directly proportional to the mass (e.g., I = kmr², where k is a constant based on shape). Therefore, if the mass changes from m to m', the moment of inertia will change from I to I' such that the ratio I/m remains constant. Since g and d (the distance to the center of mass, determined by the object's shape) also remain constant, the entire term under the square root does not change, and the period remains the same.