AP Physics 1: Algebra-Based 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

Which of the following equations correctly describes the relationship between the period (T) and frequency (f) of an object in simple harmonic motion?

A)T = fB)T = 1/fC)T = 2πfD)T = f²

All Questions (10)

Which of the following equations correctly describes the relationship between the period (T) and frequency (f) of an object in simple harmonic motion?

A) T = f

B) T = 1/f

C) T = 2πf

D) T = f²

Correct Answer: B

As stated in the provided content, the period of SHM is the reciprocal of the frequency, given by the equation T = 1/f.

An object 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 Tₛ = 2π√(m/k). Since the period is proportional to the square root of the mass, multiplying the mass by 4 will multiply the period by √4, which is 2. The new period is 2T.

A simple pendulum of length *L* oscillates with a period *T*. If the length is increased to *9L*, what will be the new period of the pendulum?

A) T/3

B) 3T

C) 9T

D) T/9

Correct Answer: B

The period of a simple pendulum is given by Tₚ = 2π√(L/g). The period is proportional to the square root of the length. If the length is multiplied by 9, the period will be multiplied by √9, which is 3. The new period is 3T.

For a simple pendulum undergoing small-angle oscillations, which of the following factors does its period NOT depend on?

A) The length of the pendulum

B) The mass of the pendulum bob

C) The acceleration due to gravity (g)

D) The location of the pendulum (e.g., Earth vs. Moon)

Correct Answer: B

The equation for the period of a simple pendulum is Tₚ = 2π√(L/g). This equation shows that the period depends on the length (L) and the acceleration due to gravity (g), but it is independent of the mass of the bob.

An object-ideal-spring oscillator has a period of 4 seconds. What is the frequency of its motion?

A) 0.25 Hz

B) 0.5 Hz

C) 2 Hz

D) 4 Hz

Correct Answer: A

Frequency is the reciprocal of the period: f = 1/T. Given T = 4 s, the frequency is f = 1/4 s = 0.25 Hz.

A mass-spring system and a simple pendulum are both moved from Earth to a planet where the acceleration due to gravity is four times greater (g' = 4g). How do their respective periods change?

A) The spring's period is halved; the pendulum's period is halved.

B) The spring's period is unchanged; the pendulum's period is halved.

C) The spring's period is unchanged; the pendulum's period is doubled.

D) The spring's period is doubled; the pendulum's period is doubled.

Correct Answer: B

The period of a mass-spring system, Tₛ = 2π√(m/k), is independent of g, so it remains unchanged. The period of a simple pendulum, Tₚ = 2π√(L/g), is inversely proportional to the square root of g. If g becomes 4g, the new period will be T'ₚ = 2π√(L/4g) = (1/√4) * 2π√(L/g) = Tₚ/2.

An object attached to an ideal spring oscillates with a frequency *f*. To double the frequency of the oscillation, the spring constant *k* must be changed to:

A) k/4

B) k/2

C) 2k

D) 4k

Correct Answer: D

The period of a spring is Tₛ = 2π√(m/k). Frequency is f = 1/T, so f = (1/2π)√(k/m). This means frequency is proportional to the square root of the spring constant (f ∝ √k). To double the frequency (2f), the spring constant must be multiplied by a factor of 2², which is 4. The new spring constant must be 4k.

The period of an object-ideal-spring oscillator is determined by which of the following physical properties?

A) The mass of the object and the acceleration due to gravity.

B) The amplitude of oscillation and the spring constant.

C) The mass of the object and the spring constant.

D) The length of the spring and the mass of the object.

Correct Answer: C

Based on the formula Tₛ = 2π√(m/k), the period of an object-ideal-spring oscillator depends only on the mass of the object (m) and the spring constant (k).

A mass-spring system has a period *T*. If the mass is doubled and the spring constant is halved, what is the new period of the system?

A) T/2

B) T

C) 2T

D) 4T

Correct Answer: C

The original period is T = 2π√(m/k). The new mass is m' = 2m and the new spring constant is k' = k/2. The new period T' = 2π√(m'/k') = 2π√((2m)/(k/2)) = 2π√(4m/k) = 2 * (2π√(m/k)) = 2T.

A simple pendulum is constructed with a string of length *L* and has a period *T*. To create a mass-spring system with the same period *T*, using a mass *m*, what must the spring constant *k* be? (Assume the experiment is conducted where the acceleration due to gravity is *g*.)

A) k = mg/L

B) k = Lg/m

C) k = m/L

D) k = L/m

Correct Answer: A

To have the same period, we set Tₚ = Tₛ. This gives 2π√(L/g) = 2π√(m/k). We can cancel 2π from both sides and square both sides to get L/g = m/k. Solving for the spring constant k, we rearrange the equation to k = mg/L.