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Frequency and Period of SHM

AP Course

AP Physics 1

Updated for the AP Physics 1 framework effective fall 2024, including mechanics, rotation, oscillations, and fluids.

Study every current unit through original concept lessons, formula derivations, experiments, quizzes, and practice sets.

AP Physics 1 · Unit 7 · Topic 7.2

Frequency and Period of SHM

Period tells how long one complete oscillation takes; frequency tells how many oscillations occur per unit time. For ideal springs and small-angle pendulums, the restoring interaction determines these timing quantities in predictable ways.

Titration response · Frequency and Period of SHMThe equivalence region produces the steepest change; its location depends on stoichiometry and acid-base strength.

Learning Goals

After this lesson, you should be able to:

  • identify one complete oscillation from motion or a graph;
  • calculate period and frequency from cycle counts and elapsed time;
  • relate \(T\), \(f\), and angular frequency \(\omega\);
  • derive the ideal mass–spring period from the SHM acceleration model;
  • derive the small-angle pendulum period;
  • predict factor changes using square-root proportionalities;
  • identify which variables do and do not affect each ideal period;
  • extract timing information from position–time data;
  • linearize spring and pendulum period data; and
  • design a timing experiment with reduced random uncertainty.

1. What Counts as One Cycle?

One complete oscillation returns the object to the same position while moving in the same direction. Starting at \(+A\), one cycle is

\[+A\rightarrow0\rightarrow-A\rightarrow0\rightarrow+A\]

Travel from \(+A\) to \(-A\) is only half a cycle. Returning to the same position with the opposite velocity is also not a full cycle.

2. Period and Frequency

If \(N\) cycles occur during elapsed time \(\Delta t\),

\[\boxed{T=\frac{\Delta t}{N}},\qquad \boxed{f=\frac{N}{\Delta t}}\]

Therefore,

\[\boxed{T=\frac1f},\qquad \boxed{f=\frac1T}\]

Period is measured in seconds per cycle. Frequency is measured in hertz:

\[1\,\mathrm{Hz}=1\,\mathrm{cycle/s}=1\,\mathrm{s^{-1}}\]

3. Detailed Visual Model

Frequency and Period of SHMTopic-matched fine pixel-art model · use the lesson's equations and labeled diagrams for exact scientific meaning.

4. Angular Frequency

One cycle corresponds to \(2\pi\) radians of phase. Angular frequency is

\[\boxed{\omega=2\pi f=\frac{2\pi}{T}}\]

Its units are radians per second. Frequency \(f\) counts cycles per second, while \(\omega\) counts radians of phase per second. They describe the same repetition rate using different units.

5. Worked Example: Count Many Cycles

An oscillator completes \(45\) cycles in \(30.0\,\mathrm s\).

\[f=\frac{45}{30.0}=1.50\,\mathrm{Hz}\]
\[T=\frac{30.0}{45}=0.667\,\mathrm s\]
\[\omega=2\pi f=2\pi(1.50)=9.42\,\mathrm{rad/s}\]

Answer: \(f=1.50\,\mathrm{Hz}\), \(T=0.667\,\mathrm s\), and \(\omega=9.42\,\mathrm{rad/s}\).

6. Deriving the Mass–Spring Period

For a mass \(m\) attached to an ideal spring of force constant \(k\),

\[F_{\mathrm{net}}=-kx\]

Newton's second law gives

\[a=-\frac{k}{m}x\]

The general SHM form is \(a=-\omega^2x\), so

\[\omega^2=\frac{k}{m},\qquad \omega=\sqrt{\frac{k}{m}}\]

Using \(T=2\pi/\omega\),

\[\boxed{T_s=2\pi\sqrt{\frac{m}{k}}}\]

This model assumes a negligibly massive ideal spring, a linear force range, and negligible damping.

7. What Controls a Spring Oscillator?

ChangePeriod responseFrequency response
Increase \(m\)\(T\propto\sqrt m\): increases\(f\propto1/\sqrt m\): decreases
Increase \(k\)\(T\propto1/\sqrt k\): decreases\(f\propto\sqrt k\): increases
Change amplitude \(A\)No change in ideal linear SHMNo change in ideal linear SHM
Turn horizontal setup verticalNo change about shifted equilibriumNo change
Change local \(g\)No change for ideal spring modelNo change

Amplitude independence holds only while the restoring force remains linear and damping is negligible.

8. Worked Example: Ideal Spring Oscillator

A \(0.80\,\mathrm{kg}\) block oscillates on a spring with \(k=50\,\mathrm{N/m}\).

\[T_s=2\pi\sqrt{\frac{0.80}{50}}=0.795\,\mathrm s\]
\[f=\frac1T=1.26\,\mathrm{Hz}\]

If the amplitude doubles but the spring remains ideal, the period is still \(0.795\,\mathrm s\). The energy changes, but the timing does not.

9. Factor-Change Reasoning for Springs

For two ideal spring oscillators,

\[\boxed{\frac{T_2}{T_1}=\sqrt{\frac{m_2}{m_1}\frac{k_1}{k_2}}}\]

Examples:

  • If mass becomes \(4m\), then \(T\) becomes \(2T\).
  • If stiffness becomes \(9k\), then \(T\) becomes \(T/3\).
  • If mass doubles and stiffness doubles, the period is unchanged.
  • To triple the period at fixed \(k\), mass must become \(9m\).

Apply the square root after forming the ratio; do not assume direct proportionality.

10. Effective Spring Constants

When a system of springs can be replaced by one linear restoring relation \(F=-k_{\mathrm{eff}}x\), use \(k_{\mathrm{eff}}\) in the period equation.

\[T=2\pi\sqrt{\frac{m}{k_{\mathrm{eff}}}}\]

For two ideal springs attached in parallel to the same mass and stretched by the same displacement, their restoring forces add:

\[k_{\mathrm{eff}}=k_1+k_2\]

A larger effective stiffness shortens the period. Always derive \(k_{\mathrm{eff}}\) from the system's force–displacement relation rather than guessing.

11. Deriving the Small-Angle Pendulum Period

For a pendulum of length \(L\), the restoring torque about the pivot is

\[\tau=-mgL\sin\theta\]

For small \(|\theta|\) in radians, \(\sin\theta\approx\theta\). With \(I=mL^2\) for a point-mass bob,

\[I\alpha=-mgL\theta\]
\[\alpha=-\frac{g}{L}\theta\]

Compare with \(\alpha=-\omega^2\theta\):

\[\omega=\sqrt{\frac{g}{L}},\qquad \boxed{T_p=2\pi\sqrt{\frac{L}{g}}}\]

12. What Controls a Simple Pendulum?

ChangeSmall-angle period responseReason
Increase \(L\)Increases as \(\sqrt L\)Longer pendulum changes direction more slowly
Increase \(g\)Decreases as \(1/\sqrt g\)Stronger restoring interaction
Change bob massNo changeMass cancels from torque equation
Small change in amplitudeApproximately no changeSmall-angle linear model
Large angular amplitudeIdeal formula becomes inaccurate\(\sin\theta\not\approx\theta\)

13. Worked Example: Small-Angle Pendulum

A simple pendulum has length \(L=0.90\,\mathrm m\) where \(g=9.8\,\mathrm{m/s^2}\).

\[T_p=2\pi\sqrt{\frac{0.90}{9.8}}=1.90\,\mathrm s\]
\[f=\frac1{1.90}=0.525\,\mathrm{Hz}\]

Doubling the bob mass leaves these values unchanged. Increasing the length to \(3.60\,\mathrm m=4L\) doubles the period to \(3.80\,\mathrm s\).

14. Factor-Change Reasoning for Pendulums

For two small-angle simple pendulums,

\[\boxed{\frac{T_2}{T_1}=\sqrt{\frac{L_2}{L_1}\frac{g_1}{g_2}}}\]

Examples:

  • If length becomes \(9L\), period becomes \(3T\).
  • If gravitational field strength becomes \(g/4\), period becomes \(2T\).
  • If both \(L\) and \(g\) double, period is unchanged.
  • To halve the period at fixed \(g\), length must become \(L/4\).

15. Reading Period from a Graph

On an \(x\)-versus-\(t\) graph, measure the horizontal separation between repeated states:

  • successive maxima;
  • successive minima;
  • upward equilibrium crossings; or
  • downward equilibrium crossings.

The time from a maximum to the next minimum is \(T/2\). The time from a maximum to the next equilibrium crossing is \(T/4\). Two equilibrium crossings in opposite directions are separated by \(T/2\), not \(T\).

16. Linearizing Spring Data

Square the spring-period equation:

\[\boxed{T^2=\frac{4\pi^2}{k}m}\]

A graph of \(T^2\) vertically versus \(m\) horizontally should be linear with

\[\text{slope}=\frac{4\pi^2}{k},\qquad \boxed{k=\frac{4\pi^2}{\text{slope}}}\]

If \(T\) is plotted directly against \(m\), the graph is a square-root curve, not a straight line.

17. Linearizing Pendulum Data

Square the pendulum-period equation:

\[\boxed{T^2=\frac{4\pi^2}{g}L}\]

A graph of \(T^2\) vertically versus \(L\) horizontally should be linear with

\[\text{slope}=\frac{4\pi^2}{g},\qquad \boxed{g=\frac{4\pi^2}{\text{slope}}}\]

A significant nonzero intercept may indicate timing bias, an incorrect effective length, sensor delay, or a model assumption that is not satisfied.

18. Experimental Timing Strategy

Measuring one short period with a handheld stopwatch makes reaction time a large fraction of the result. Instead, time many cycles:

\[T=\frac{\Delta t_{\mathrm{many\ cycles}}}{N}\]
  1. Choose a clear repeated state, such as a leftmost turning point.
  2. Start and stop timing at the same state and direction.
  3. Time at least \(10\)–\(20\) cycles when practical.
  4. Repeat trials and calculate a mean period.
  5. Keep amplitude small for a pendulum and within the linear range for a spring.
  6. Vary only one independent variable at a time.
  7. Use a linearized graph and uncertainty bars to evaluate the model.

19. A Reliable Problem-Solving Workflow

  1. Identify one complete repeated state and determine whether timing or cycle count is given.
  2. Use \(T=\Delta t/N\) and \(f=N/\Delta t\) before selecting a system-specific model.
  3. For an ideal mass–spring system, use \(T=2\pi\sqrt{m/k_{\mathrm{eff}}}\).
  4. For a small-angle simple pendulum, use \(T=2\pi\sqrt{L/g}\).
  5. Check that all quantities use SI units.
  6. For comparisons, form a ratio and apply the square root.
  7. For experiments, square the period and use the appropriate linear graph.
  8. Check whether the result changes correctly when mass, stiffness, length, or gravity changes.

20. Common Traps and Key Takeaways

  • One cycle returns to the same position and velocity direction.
  • Period and frequency are reciprocals, not equal quantities.
  • Use seconds for period and hertz for frequency.
  • Do not confuse \(f\) with \(\omega=2\pi f\).
  • Spring period grows as \(\sqrt m\) and decreases as \(1/\sqrt k\).
  • Pendulum period grows as \(\sqrt L\) and decreases as \(1/\sqrt g\).
  • Ideal spring period does not depend on amplitude or \(g\).
  • Small-angle pendulum period does not depend on bob mass.
  • The pendulum formula requires the small-angle approximation.
  • Plot \(T^2\), not \(T\), to obtain a linear relationship with \(m\) or \(L\).
Checkpoint

An ideal \(0.60\,\mathrm{kg}\) mass oscillates on a spring with \(k=38\,\mathrm{N/m}\). Calculate \(T\), \(f\), and \(\omega\); predict each value if the mass becomes \(2.40\,\mathrm{kg}\); and state the effect of doubling amplitude. A small-angle pendulum at the same location has the same original period. Calculate its required length using \(g=9.8\,\mathrm{m/s^2}\), then predict its period if moved to a location where \(g=2.45\,\mathrm{m/s^2}\). Finally, describe the axes and expected slopes of two experiments—one using \(T^2\) versus spring mass to determine \(k\), and one using \(T^2\) versus pendulum length to determine \(g\)—including a timing procedure that reduces reaction-time uncertainty.

Concept scope cross-checked with the College Board AP Physics 1 Course and Exam Description and the open educational references OpenStax University Physics: Simple Harmonic Motion and OpenStax University Physics: Pendulums. Explanations, diagram, derivations, examples, tables, graph analysis, checkpoint, and investigation are independently written for Insomnia Studyroom.

Official curriculum reference: College Board AP Physics 1 course page. The explanation and worked example are independently written for this study site.