AP Physics C Mechanics- 7.3 Representing and Analyzing SHM- Study Notes- New Syllabus
AP Physics C Mechanics- 7.3 Representing and Analyzing SHM – Study Notes
AP Physics C Mechanics- 7.3 Representing and Analyzing SHM – Study Notes – per latest Syllabus.
Key Concepts:
- Mathematical and Physical Characteristics of Simple Harmonic Motion (SHM)
- Differential Equation of SHM
- Kinematic Relationships in SHM
- Resonance in Simple Harmonic Motion
- Amplitude and Period Independence
- Graphical Analysis of Simple Harmonic Motion (SHM)
Displacement Representation in SHM
For an object undergoing simple harmonic motion, the displacement \( \mathrm{x} \) from its equilibrium position at time \( \mathrm{t} \) can be described by a sinusoidal function:
\( \mathrm{x = A\cos(2\pi f t)} \quad \text{or} \quad \mathrm{x = A\sin(2\pi f t)} \)
- \( \mathrm{A} \): amplitude (maximum displacement)
- \( \mathrm{f} \): frequency of oscillation
- \( \mathrm{t} \): time
- \( \mathrm{x} \): instantaneous displacement
The motion repeats itself periodically and can be modeled using either a cosine or sine function depending on the initial phase.
Features of Harmonic Motion
During SHM, the displacement, velocity, and acceleration vary continuously and exhibit distinct patterns of maxima, minima, and zeros:
- Displacement is maximum at the endpoints of motion.
- Velocity is maximum at the equilibrium position.
- Acceleration is maximum when displacement is maximum (opposite in direction).
Qualitative Analysis Using Extrema
Recognizing the times or positions at which displacement, velocity, and acceleration reach their maximum or zero values allows for qualitative understanding of oscillatory motion.
For example:
- At equilibrium (\( \mathrm{x=0} \)): \( \mathrm{v} \) is maximum, \( \mathrm{a=0} \).
- At maximum displacement (\( \mathrm{x=\pm A} \)): \( \mathrm{v=0} \), \( \mathrm{a} \) is maximum and opposite to \( \mathrm{x} \).
Example
An oscillator follows \( \mathrm{x = 0.05\cos(4\pi t)} \). Determine its amplitude, frequency, and period.
▶️ Answer / Explanation
Step 1: Compare with \( \mathrm{x = A\cos(2\pi f t)} \):
\( \mathrm{A = 0.05\,m},\ \ 2\pi f = 4\pi \Rightarrow f = 2\,Hz} \)
Step 2: Period \( \mathrm{T = \dfrac{1}{f} = 0.5\,s} \)
Result: Amplitude = \( \mathrm{0.05\,m} \), Frequency = \( \mathrm{2\,Hz} \), Period = \( \mathrm{0.5\,s} \).
Differential Equation of SHM
The motion of an SHM system satisfies a second-order differential equation derived from Newton’s second law. The acceleration is proportional to and opposite the displacement:
\( \mathrm{\dfrac{d^2x}{dt^2} = -\omega^2 x} \)
Its general solution is the sinusoidal function:
\( \mathrm{x = A\cos(\omega t + \phi)} \)
- \( \mathrm{A} \): amplitude
- \( \mathrm{\omega} \): angular frequency (\( \mathrm{2\pi f} \))
- \( \mathrm{\phi} \): phase constant, determined by initial conditions
Key Idea: This equation defines all kinematic quantities (velocity, acceleration) for SHM and shows that the motion is harmonic and bounded.
Example
A mass attached to a spring obeys the differential equation \( \mathrm{\dfrac{d^2x}{dt^2} = -25x} \). Find the angular frequency, frequency, and the general equation of motion for the system.
▶️ Answer / Explanation
Step 1: Compare with the SHM form:
\( \mathrm{\dfrac{d^2x}{dt^2} = -\omega^2 x} \)
Thus, \( \mathrm{\omega^2 = 25 \Rightarrow \omega = 5\,rad/s.} \)
Step 2: Find the frequency:
\( \mathrm{f = \dfrac{\omega}{2\pi} = \dfrac{5}{2\pi} = 0.796\,Hz} \)
Step 3: Write the general solution:
\( \mathrm{x = A\cos(\omega t + \phi) = A\cos(5t + \phi)} \)
Interpretation:
- The system oscillates with an angular frequency of \( \mathrm{5\,rad/s} \).
- The solution \( \mathrm{x = A\cos(5t + \phi)} \) describes all possible motions depending on initial displacement and velocity (encoded in \( \mathrm{\phi} \)).
Result: \( \mathrm{\omega = 5\,rad/s,\ f = 0.796\,Hz,\ T = 1.26\,s.} \)
Acceleration–Position Relationship
The acceleration of an object in SHM is related to its displacement and angular frequency:
\( \mathrm{a = -\omega^2 x} \)
The acceleration always points toward the equilibrium position and increases with displacement magnitude.
Maximum Velocity and Acceleration
The maximum velocity and acceleration occur when the sine or cosine terms reach ±1:
\( \mathrm{v_{max} = A\omega} \quad \text{and} \quad \mathrm{a_{max} = A\omega^2} \)
Key Idea:
- Velocity is maximum at equilibrium (\( \mathrm{x=0} \)).
- Acceleration is maximum at the extremes of motion (\( \mathrm{x=\pm A} \)).
Example
A mass oscillates according to \( \mathrm{x = 0.10\cos(5t)} \). Find its \( \mathrm{v_{max}} \) and \( \mathrm{a_{max}} \).
▶️ Answer / Explanation
Given: \( \mathrm{A = 0.10\,m,\ \omega = 5\,rad/s} \)
Step 1: \( \mathrm{v_{max} = A\omega = 0.10(5) = 0.5\,m/s} \)
Step 2: \( \mathrm{a_{max} = A\omega^2 = 0.10(25) = 2.5\,m/s^2} \)
Result: Maximum velocity = \( \mathrm{0.5\,m/s} \), Maximum acceleration = \( \mathrm{2.5\,m/s^2} \).
Resonance in Simple Harmonic Motion
When a sinusoidal external force is applied to a system capable of oscillation, the system may exhibit resonance. Resonance occurs when the frequency of the external driving force matches the system’s natural frequency.
Derived Relationships:
- Natural frequency: \( \mathrm{f_0 = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}} \)
- Angular frequency: \( \mathrm{\omega_0 = \sqrt{\dfrac{k}{m}}} \)
Key Idea: At resonance, the system absorbs energy efficiently from the external force, resulting in a large amplitude of oscillation.
Condition for Resonance
Resonance occurs when the driving frequency \( \mathrm{f_d} \) equals the natural frequency \( \mathrm{f_0} \):
\( \mathrm{f_d = f_0 \quad \text{or} \quad \omega_d = \omega_0} \)
At this condition, every push from the external force arrives in sync with the system’s motion, reinforcing it on each cycle.
Effect of Resonance
At resonance:
- The amplitude of oscillation increases dramatically.
- The system’s energy grows with each cycle (until damping or energy loss limits it).
- Even small periodic forces can cause large oscillations over time.
Natural Frequency
The natural frequency of a system is the frequency at which it oscillates freely when displaced from equilibrium, without any external driving force:
\( \mathrm{f_0 = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}} \)
It depends on the system’s inertia (mass) and stiffness (spring constant or restoring force). Each oscillating system (bridge, pendulum, molecule, etc.) has its own natural frequency.
Key Relationships:
| Parameter | Symbol | Relationship | Description |
|---|---|---|---|
| Natural Frequency | \( \mathrm{f_0} \) | \( \mathrm{f_0 = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}} \) | Frequency of free oscillation |
| Driving Frequency | \( \mathrm{f_d} \) | Applied periodic frequency | External force oscillation rate |
| Resonance Condition | \( \mathrm{f_d = f_0} \) | Maximum amplitude occurs | Energy transfer most efficient |
Example
A bridge vibrates at its natural frequency of \( \mathrm{2\,Hz} \) due to wind gusts matching that frequency. Explain why the vibration amplitude increases dramatically.
▶️ Answer / Explanation
Step 1: The external wind force oscillates at \( \mathrm{f_{drive} = 2\,Hz} \), the same as the bridge’s \( \mathrm{f_0} \).
Step 2: Resonance occurs because the driving frequency matches the natural frequency.
Step 3: The bridge continually absorbs energy each cycle, increasing oscillation amplitude.
Result: The amplitude increases due to resonance — a dangerous effect that can lead to structural failure.
Amplitude and Period Independence
Changing the amplitude of a system in SHM does not change its period. This means that the period of oscillation depends only on system parameters (mass, stiffness, length, etc.), not on how far it is displaced provided the motion remains harmonic.
Key Relationships:
- For a spring: \( \mathrm{T = 2\pi\sqrt{\dfrac{m}{k}}} \)
- For a pendulum: \( \mathrm{T = 2\pi\sqrt{\dfrac{L}{g}}} \)
- Amplitude \( \mathrm{A} \) affects energy, not period.
Physical Meaning: In SHM, the restoring force is linear (\( \mathrm{F = -kx} \)). As long as this linearity holds, the time for one oscillation remains the same regardless of amplitude.
Example
A spring–mass oscillator with \( \mathrm{k = 80\,N/m} \) and \( \mathrm{m = 0.20\,kg} \) oscillates with amplitude \( \mathrm{A = 0.05\,m} \). If the amplitude is doubled to \( \mathrm{0.10\,m} \), find how the period changes.
▶️ Answer / Explanation
Step 1: Formula for period:
\( \mathrm{T = 2\pi\sqrt{\dfrac{m}{k}}} \)
Step 2: Substitute values:
\( \mathrm{T = 2\pi\sqrt{\dfrac{0.20}{80}} = 2\pi\sqrt{0.0025} = 2\pi(0.05) = 0.314\,s} \)
Step 3: When amplitude doubles, \( \mathrm{A’ = 2A = 0.10\,m} \), but \( \mathrm{T} \) depends only on \( \mathrm{m} \) and \( \mathrm{k} \).
Therefore:
\( \mathrm{T’ = T = 0.314\,s.} \)
Result: The period does not change when amplitude changes. Amplitude affects energy, not timing.
Physical Explanation:
- In SHM, \( \mathrm{F = -kx} \) remains linear.
- Larger amplitude → more distance and higher velocity, but each cancel proportionally, keeping \( \mathrm{T} \) constant.
Graphical Analysis of Simple Harmonic Motion (SHM)
Simple Harmonic Motion (SHM) can be represented and analyzed graphically by plotting displacement, velocity, and acceleration as functions of time. These graphs reveal how these quantities vary sinusoidally and how they are related in phase and magnitude.
Each graph—displacement vs. time, velocity vs. time, and acceleration vs. time—provides insight into how the motion evolves through a full cycle.
Mathematical Relationships:
\( \mathrm{x = A\cos(\omega t)} \)
\( \mathrm{v = -A\omega\sin(\omega t)} \)
\( \mathrm{a = -A\omega^2\cos(\omega t)} \)
- Displacement, velocity, and acceleration are all sinusoidal.
- Velocity and acceleration can be found by taking the first and second derivatives of displacement.
- They differ in phase by \( \mathrm{90°} \) (or \( \mathrm{\pi/2} \) radians).
Graphical Features:
| Quantity | Equation | Phase Relationship | Maxima / Zeros |
|---|---|---|---|
| Displacement | \( \mathrm{x = A\cos(\omega t)} \) | Reference curve | Max at \( \mathrm{x = \pm A} \); zero at equilibrium |
| Velocity | \( \mathrm{v = -A\omega\sin(\omega t)} \) | Lags \( \mathrm{x} \) by \( \mathrm{90°} \) | Zero at extremes; max at equilibrium |
| Acceleration | \( \mathrm{a = -A\omega^2\cos(\omega t)} \) | Opposite phase (180°) to \( \mathrm{x} \) | Max at extremes; zero at equilibrium |
Qualitative Interpretation of Graphs:
- When \( \mathrm{x} \) is maximum, \( \mathrm{v = 0} \) and \( \mathrm{a} \) is maximum (opposite direction).
- When \( \mathrm{x = 0} \), \( \mathrm{v} \) is maximum and \( \mathrm{a = 0} \).
- Acceleration is always directed opposite to displacement, confirming the restoring nature of SHM.
Example
An object performs SHM with displacement \( \mathrm{x = 0.10\cos(4\pi t)} \). Sketch or describe how displacement, velocity, and acceleration vary with time over one period.
▶️ Answer / Explanation
Step 1: Identify amplitude and angular frequency:
\( \mathrm{A = 0.10\,m,\ \omega = 4\pi\,rad/s,\ f = 2\,Hz,\ T = 0.5\,s.} \)
Step 2: Write the velocity and acceleration equations:
\( \mathrm{v = -A\omega\sin(\omega t) = -0.10(4\pi)\sin(4\pi t) = -1.26\sin(4\pi t)} \)
\( \mathrm{a = -A\omega^2\cos(\omega t) = -0.10(4\pi)^2\cos(4\pi t) = -15.8\cos(4\pi t)} \)
Step 3: Describe phase relationships:
- Displacement (\( \mathrm{x} \)) and acceleration (\( \mathrm{a} \)) are **in opposite phase** — when one is positive, the other is negative.
- Velocity (\( \mathrm{v} \)) is **90° (or \( \mathrm{\pi/2} \) radians)** out of phase with displacement.
- Velocity reaches zero when displacement is at maximum and vice versa.
Step 4: Describe behavior over one cycle (0 → T = 0.5 s):
| Time | Displacement \( \mathrm{x} \) | Velocity \( \mathrm{v} \) | Acceleration \( \mathrm{a} \) |
|---|---|---|---|
| 0 | +A (max) | 0 | Maximum negative |
| \( \mathrm{T/4 = 0.125\,s} \) | 0 | Maximum negative | 0 |
| \( \mathrm{T/2 = 0.25\,s} \) | −A (min) | 0 | Maximum positive |
| \( \mathrm{3T/4 = 0.375\,s} \) | 0 | Maximum positive | 0 |
| \( \mathrm{T = 0.5\,s} \) | +A (max) | 0 | Maximum negative |
Result: The SHM motion is fully cyclic: displacement, velocity, and acceleration are sinusoidal and phase-shifted by \( \mathrm{90°} \) intervals.