AP Physics C: Mechanics Practice Quiz: Frequency and Period of SHM

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

Test your understanding with short quizzes. This quiz has 10 questions to check your progress.

Question 1 of 10

An object is exhibiting simple harmonic motion (SHM). If its period of oscillation is T, what is its frequency, f?

A)1/TB)2π/TC)T/2πD)T

All Questions (10)

An object is exhibiting simple harmonic motion (SHM). If its period of oscillation is T, what is its frequency, f?

A) 1/T

B) 2π/T

C) T/2π

D) T

Correct Answer: A

The period (T) is the time for one full oscillation, while frequency (f) is the number of oscillations per unit time. They are reciprocals of each other, as shown by the relationship T = 1/f.

A block of mass *m* is attached to an ideal spring with spring constant *k* and oscillates with a period *T*. If the mass is quadrupled to *4m*, what is the new period of oscillation?

A) T/2

B) T

C) 2T

D) 4T

Correct Answer: C

The period of a mass-spring system is given by T_s = 2π√(m/k). The period is directly proportional to the square root of the mass. If the mass is quadrupled (multiplied by 4), the period will be multiplied by √4, which is 2. Therefore, the new period is 2T.

A simple pendulum of length *L* has a period *T* on Earth. If the pendulum is taken to the Moon, where the acceleration due to gravity is approximately one-sixth that of Earth, what will its new period be?

A) T/6

B) T/√6

C) √6 T

D) 6T

Correct Answer: C

The period of a simple pendulum is given by T_p = 2π√(ℓ/g). The period is inversely proportional to the square root of the acceleration due to gravity (g). If g becomes g/6, the new period will be T' = 2π√(ℓ/(g/6)) = (√6) * 2π√(ℓ/g) = √6 T.

Which of the following factors affects the period of a simple pendulum oscillating at a small angle?

A) The mass of the pendulum bob

B) The length of the pendulum

C) The amplitude of the swing

D) The material of the pendulum bob

Correct Answer: B

According to the equation T_p = 2π√(ℓ/g), the period of a simple pendulum depends on its length (ℓ) and the local acceleration due to gravity (g). It is independent of the mass and, for small angles, the amplitude.

An object in SHM has an angular frequency of ω. Which expression correctly represents the period, T, of the motion?

A) ω / 2π

B) 2π / ω

C) ω / f

D) 1 / (2πω)

Correct Answer: B

The content provides the direct relationship between period (T) and angular frequency (ω) as T = 2π/ω. Angular frequency represents the rate of change of angular displacement in radians per second, and one full cycle is 2π radians.

A mass attached to a spring oscillates with period T_s. A simple pendulum oscillates with period T_p. Both systems are moved from Earth to a space station in zero gravity. What happens to their periods?

A) T_s becomes zero and T_p becomes infinite.

B) T_s remains the same and T_p becomes infinite.

C) Both T_s and T_p become infinite.

D) T_s remains the same and T_p becomes zero.

Correct Answer: B

The period of a mass-spring system, T_s = 2π√(m/k), does not depend on gravity (g), so it remains unchanged. The period of a pendulum, T_p = 2π√(ℓ/g), does depend on gravity. As g approaches zero, the value of T_p approaches infinity, meaning it would not oscillate.

To double the period of an object-ideal-spring oscillator, one must change the mass to what factor of its original value?

A) 1/2

B) √2

C) 2

D) 4

Correct Answer: D

The period of a mass-spring system is T_s = 2π√(m/k). To double the period (T -> 2T), the term inside the square root, m, must be quadrupled (m -> 4m) because √4 = 2. The new period would be 2π√(4m/k) = 2 * (2π√(m/k)) = 2T.

Two simple pendulums, A and B, are at the same location. Pendulum A has a bob of mass *m* and length *L*. Pendulum B has a bob of mass *2m* and length *L*. How does the period of pendulum A (T_A) compare to the period of pendulum B (T_B)?

A) T_A = 2T_B

B) T_A = √2 T_B

C) T_A = T_B

D) T_A = T_B / 2

Correct Answer: C

The period of a simple pendulum is given by the equation T_p = 2π√(ℓ/g). This equation shows that the period is dependent on the length (ℓ) and the acceleration due to gravity (g), but it is independent of the mass of the bob. Since both pendulums have the same length and are at the same location (same g), their periods are equal.

A mass *m* oscillates on an ideal spring with a spring constant *k* and a period *T*. If the spring is replaced with one that has a spring constant of *k/4*, what is the new period of oscillation?

A) T/4

B) T/2

C) 2T

D) 4T

Correct Answer: C

The period of a mass-spring system is T_s = 2π√(m/k). The period is inversely proportional to the square root of the spring constant, k. If k is changed to k/4, the new period T' will be T' = 2π√(m/(k/4)) = 2π√(4m/k) = 2 * (2π√(m/k)) = 2T.

An object-spring system has a period *T* and an angular frequency *ω*. If the mass is decreased to one-ninth of its original value (m/9), what are the new period and angular frequency?

A) T/3 and 3ω

B) T/9 and 9ω

C) 3T and ω/3

D) 9T and ω/9

Correct Answer: A

The period of a mass-spring system is T_s = 2π√(m/k). If mass becomes m/9, the new period T' = T/√9 = T/3. The angular frequency is related to the period by ω = 2π/T. Since the new period is T/3, the new angular frequency ω' = 2π/(T/3) = 3 * (2π/T) = 3ω.