AP Physics C Mechanics- 7.1 Defining Simple Harmonic Motion (SHM)- Study Notes- New Syllabus
AP Physics C Mechanics- 7.1 Defining Simple Harmonic Motion (SHM) – Study Notes
AP Physics C Mechanics- 7.1 Defining Simple Harmonic Motion (SHM) – Study Notes – per latest Syllabus.
Key Concepts:
- Simple Harmonic Motion (SHM)
- Restoring Force and the Definition of SHM
Simple Harmonic Motion (SHM)
Simple Harmonic Motion (SHM) is a special case of periodic motion , motion that repeats itself in equal intervals of time.
In SHM, the motion of an object is such that it oscillates back and forth about an equilibrium position with a regular, repeating pattern. The motion can be described by sinusoidal functions (sine or cosine).
Key Idea:
- All SHM is periodic, but not all periodic motion is SHM.
- Examples of SHM: a mass on a spring, a simple pendulum (for small angles), vibrations of molecules.
- The motion is predictable and symmetric about the equilibrium point.
Example
A block attached to a horizontal spring oscillates back and forth on a frictionless surface. The spring constant is \( \mathrm{k = 100\,N/m} \), and the block has a mass of \( \mathrm{0.50\,kg} \). The block is pulled \( \mathrm{0.10\,m} \) from its equilibrium position and released from rest.
Show that this motion is an example of simple harmonic motion and find the period of oscillation.
▶️ Answer / Explanation
Step 1: Determine if the restoring force satisfies Hooke’s Law.
The spring exerts a restoring force proportional to displacement:
\( \mathrm{F = -k\,x = -100\,x} \)
The force is directly proportional to displacement (\( \mathrm{x} \)) and opposite in direction → this satisfies the condition for SHM.
Step 2: Write the equation of motion:
\( \mathrm{m\,a = -k\,x \Rightarrow a = -\dfrac{k}{m}x} \)
Since acceleration is directly proportional to \( \mathrm{-x} \), the block undergoes simple harmonic motion.
Step 3: Find the angular frequency:
\( \mathrm{\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{100}{0.50}} = \sqrt{200} = 14.14\,rad/s} \)
Step 4: Find the period of oscillation:
\( \mathrm{T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{14.14} = 0.444\,s} \)
Result: The block performs simple harmonic motion with a period of \( \mathrm{0.44\,s} \). The restoring force \( \mathrm{F = -k\,x} \) confirms that this is a special case of periodic motion known as SHM.
Restoring Force and the Definition of SHM
Simple harmonic motion occurs when the magnitude of the restoring force is directly proportional to the displacement from the equilibrium position and acts in the opposite direction of that displacement.
Derived Equation:
\( \mathrm{F_{net} = -k\,\Delta x} \)
Applying Newton’s second law:
\( \mathrm{m a = -k\,\Delta x} \)
or equivalently,
\( \mathrm{a = -\dfrac{k}{m}\,\Delta x} \)
- \( \mathrm{k} \): spring constant (stiffness of the system)
- \( \mathrm{\Delta x} \): displacement from equilibrium
- \( \mathrm{a} \): acceleration directed toward the equilibrium position
The negative sign indicates that the restoring acceleration (and force) always points opposite to the displacement, pulling the object back toward equilibrium.
Nature of the Restoring Force
A restoring force is a force that always acts toward the equilibrium position, opposing displacement.
In SHM systems:
- When the object is displaced in one direction, the restoring force acts in the opposite direction.
- The greater the displacement, the larger the restoring force this linear relationship ensures harmonic motion.
Mathematically:
\( \mathrm{F_{restoring} = -k\,\Delta x} \)
Examples:
- Mass–spring system: spring pulls back toward equilibrium when stretched or compressed.
- Pendulum (for small angles): gravitational component acts to restore the pendulum toward its central position.
Equilibrium Position
The equilibrium position is the point where the net force on the object is zero. At this position, the object would remain at rest if undisturbed.
For SHM systems:
- Displacement \( \mathrm{\Delta x = 0} \).
- Net restoring force \( \mathrm{F_{net} = 0} \).
- Acceleration \( \mathrm{a = 0} \).
- Velocity \( \mathrm{v} \) is maximum as the object passes through equilibrium.
Physical Meaning: At the equilibrium position, potential energy is minimum, and kinetic energy is maximum this continuous interchange gives rise to oscillatory motion.
Example
A block of mass \( \mathrm{0.50\,kg} \) is attached to a spring with \( \mathrm{k = 200\,N/m} \) and displaced \( \mathrm{0.10\,m} \) from equilibrium. Find the restoring force and acceleration at that position.
▶️ Answer / Explanation
Step 1: Apply Hooke’s Law for restoring force:
\( \mathrm{F = -k\,\Delta x = -(200)(0.10) = -20\,N} \)
The negative sign indicates that the force acts opposite the displacement.
Step 2: Find acceleration using Newton’s second law:
\( \mathrm{a = \dfrac{F}{m} = \dfrac{-20}{0.50} = -40\,m/s^2} \)
Result: At a displacement of \( \mathrm{0.10\,m} \), the spring exerts a restoring force of \( \mathrm{20\,N} \) toward equilibrium, causing an acceleration of \( \mathrm{40\,m/s^2} \) toward the center.