AP Physics C: Mechanics Flashcards: Frequency and Period of SHM
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Review key ideas with interactive flashcards. This set includes 10 cards to help you master important concepts.
State the equation for the period of a simple pendulum displaced by a small angle.
The period of a simple pendulum is given by the equation T_p = 2π√(ℓ/g), where ℓ is the length and g is the acceleration due to gravity.
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State the equation for the period of a simple pendulum displaced by a small angle.
The period of a simple pendulum is given by the equation T_p = 2π√(ℓ/g), where ℓ is the length and g is the acceleration due to gravity.
State the equation for the period of an object-ideal-spring oscillator.
The period of an object-ideal-spring oscillator is given by the equation T_s = 2π√(m/k), where m is the mass and k is the spring constant.
For a simple pendulum in SHM, how does increasing the length (ℓ) affect the period (T_p)?
Increasing the length will increase the period, as the period is directly proportional to the square root of the length (T_p ∝ √ℓ).
What is the Frequency (f) in the context of Simple Harmonic Motion (SHM)?
Frequency is the number of complete cycles or oscillations an object completes per unit of time. It is related to period by the equation f = 1/T.
What is the Period (T) in the context of Simple Harmonic Motion (SHM)?
The period is the time it takes for an object to complete one full cycle of its motion. It is related to frequency by the equation T = 1/f.
What is the equation that relates the period (T) of SHM to the angular frequency (ω)?
The period of SHM is related to the angular frequency by the equation T = 2π/ω.
For a mass-spring system in SHM, how does increasing the mass (m) affect the period (T_s)?
Increasing the mass will increase the period, as the period is directly proportional to the square root of the mass (T_s ∝ √m).
What condition must be met for the simple pendulum period equation, T_p = 2π√(ℓ/g), to be accurate?
The equation for the period of a simple pendulum is only accurate when it is displaced by a small angle.
For an object-ideal-spring oscillator, what happens to the period if the spring is replaced with a stiffer one (larger k)?
The period will decrease because the period is inversely proportional to the square root of the spring constant (T_s ∝ 1/√k).
If a simple pendulum is moved to a location with a stronger gravitational field (larger g), how will its period be affected?
The period will decrease because the period is inversely proportional to the square root of the acceleration due to gravity (T_p ∝ 1/√g).