AP Physics 1: Algebra-Based 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.
What two physical properties determine the period of an object–ideal-spring oscillator?
The period of a mass-spring system is determined by the mass of the object (m) and the spring constant (k).
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What two physical properties determine the period of an object–ideal-spring oscillator?
The period of a mass-spring system is determined by the mass of the object (m) and the spring constant (k).
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).
Comparing the period equations for a spring and a pendulum, what variable affects a spring's period but not a pendulum's?
The mass (m) of the oscillating object affects the period of a mass-spring system but does not affect the period of a simple pendulum.
An oscillator has a frequency of 5 Hz. What is its period?
The period is the inverse of the frequency (T = 1/f), so the period is 1/5 Hz = 0.2 seconds.
What is the mathematical relationship between the period (T) and frequency (f) of an object in Simple Harmonic Motion (SHM)?
The period of SHM is the inverse of the frequency of the object's motion, represented by the equation T = 1/f.
What two factors determine the period of a simple pendulum?
The period of a simple pendulum is determined by its length (L) and the local acceleration due to gravity (g).
How would the period of a simple pendulum change if it were moved to a planet where the acceleration due to gravity (g) is four times that of Earth's?
The period would be halved, as the period of a pendulum (T_p = 2π √(L/g)) is inversely proportional to the square root of g.
According to the equation T_s = 2π √(m/k), how would the period of a mass-spring system change if the mass (m) were quadrupled?
If the mass is quadrupled, the period will double because the period is proportional to the square root of the mass.
To halve the period of a simple pendulum, how must its length (L) be changed?
To halve the period, the length (L) must be reduced to one-fourth of its original value, since T is proportional to the square root of L.
State the equation for the period of an object attached to an ideal spring.
The period of an object–ideal-spring oscillator is given by the equation: T_s = 2π √(m/k).