AP Physics C Mechanics- 7.2 Frequency and Period of SHM- Study Notes- New Syllabus
AP Physics C Mechanics- 7.2 Frequency and Period of SHM – Study Notes
AP Physics C Mechanics- 7.2 Frequency and Period of SHM – Study Notes – per latest Syllabus.
Key Concepts:
- Relationship Between Period, Frequency, and Angular Frequency
- Period of an Ideal Spring–Mass Oscillator
- Period of a Simple Pendulum (Small-Angle Approximation)
Relationship Between Period, Frequency, and Angular Frequency
In simple harmonic motion, the angular frequency \( \mathrm{\omega} \), frequency \( \mathrm{f} \), and period \( \mathrm{T} \) are related through the following equations:
\( \mathrm{\omega = 2\pi f = \dfrac{2\pi}{T}} \)
or equivalently,
\( \mathrm{T = \dfrac{2\pi}{\omega} = \dfrac{1}{f}} \)
- \( \mathrm{\omega} \): angular frequency (radians per second)
- \( \mathrm{f} \): frequency (oscillations per second)
- \( \mathrm{T} \): period (time for one complete oscillation)
Key Idea: The faster an oscillator moves (larger \( \mathrm{\omega} \) or \( \mathrm{f} \)), the shorter its period \( \mathrm{T} \). Angular frequency measures how rapidly the object completes each cycle of oscillation in radians per second.
Example
An oscillator completes \( \mathrm{5} \) full oscillations every second. Find its angular frequency and period.
▶️ Answer / Explanation
Step 1: The frequency is \( \mathrm{f = 5\,Hz} \).
Step 2: Find the angular frequency:
\( \mathrm{\omega = 2\pi f = 2\pi(5) = 31.4\,rad/s} \)
Step 3: Find the period:
\( \mathrm{T = \dfrac{1}{f} = \dfrac{1}{5} = 0.20\,s} \)
Result: Angular frequency \( \mathrm{\omega = 31.4\,rad/s} \); Period \( \mathrm{T = 0.20\,s} \).
Period of an Ideal Spring–Mass Oscillator
For a mass–spring system undergoing simple harmonic motion, the period depends on the mass of the object and the spring constant \( \mathrm{k} \).
\( \mathrm{T_s = 2\pi \sqrt{\dfrac{m}{k}}} \)
- \( \mathrm{T_s} \): period of oscillation
- \( \mathrm{m} \): mass attached to the spring
- \( \mathrm{k} \): spring constant (stiffness)
Derivation:
From \( \mathrm{F = -kx = ma} \Rightarrow a = -\dfrac{k}{m}x \).
This is the defining equation of SHM with angular frequency \( \mathrm{\omega = \sqrt{\dfrac{k}{m}}} \).
Thus, \( \mathrm{T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{m}{k}}} \).
Key Idea:
- The period increases with greater mass (heavier objects oscillate slower).
- The period decreases with stiffer springs (larger \( \mathrm{k} \) → faster oscillation).
Example
A \( \mathrm{0.25\,kg} \) mass is attached to a spring with spring constant \( \mathrm{k = 100\,N/m} \). Find the period of oscillation.
▶️ Answer / Explanation
Step 1: Use the spring–mass period formula:
\( \mathrm{T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{0.25}{100}}} \)
Step 2: Simplify:
\( \mathrm{T = 2\pi(0.05) = 0.314\,s} \)
Result: The oscillator has a period of \( \mathrm{0.31\,s} \) (oscillates roughly 3 times per second).
Period of a Simple Pendulum (Small-Angle Approximation)
For a simple pendulum swinging at small angles (\( \mathrm{\theta < 15°} \)), the restoring force is approximately proportional to displacement, producing SHM.
The period depends on the length of the pendulum and the acceleration due to gravity:
\( \mathrm{T_p = 2\pi\sqrt{\dfrac{L}{g}}} \)
- \( \mathrm{T_p} \): period of the pendulum
- \( \mathrm{L} \): length of the pendulum
- \( \mathrm{g} \): acceleration due to gravity
Derivation:
For small angles, the tangential restoring force is \( \mathrm{F_t = -mg\sin\theta \approx -mg\theta = -mg\dfrac{x}{L}} \).
Using \( \mathrm{F = ma = -\dfrac{g}{L}x} \), we identify \( \mathrm{\omega = \sqrt{\dfrac{g}{L}}} \),
so \( \mathrm{T = 2\pi\sqrt{\dfrac{L}{g}}} \).
Key Idea:
- The period depends only on \( \mathrm{L} \) and \( \mathrm{g} \), not on the mass or amplitude (for small displacements).
- Longer pendulums swing slower; larger \( \mathrm{g} \) results in faster oscillations.
Example
A simple pendulum of length \( \mathrm{0.80\,m} \) is displaced by a small angle and released. Find the period of oscillation. (Take \( \mathrm{g = 9.8\,m/s^2} \)).
▶️ Answer / Explanation
Step 1: Use the pendulum period formula:
\( \mathrm{T = 2\pi\sqrt{\dfrac{L}{g}} = 2\pi\sqrt{\dfrac{0.80}{9.8}}} \)
Step 2: Simplify:
\( \mathrm{T = 2\pi(0.285) = 1.79\,s} \)
Result: The pendulum completes one full swing every \( \mathrm{1.79\,s} \).