Oscillations and Simple Harmonic Motion

Misconceptions

Key Concepts

Simple Harmonic Motion (SHM)

Motion where the restoring force is proportional to displacement from equilibrium:

• •F = -kx (Hooke's Law for springs)
• •Results in sinusoidal motion
• •Acceleration always points toward equilibrium
• •a = -omega^2 * x

Key characteristics:

• •Motion repeats with constant period T
• •Amplitude A is the maximum displacement from equilibrium
• •Angular frequency omega = 2pi/T = 2pi*f

Period and Frequency

• •Period (T): Time for one complete oscillation (seconds)
• •Frequency (f): Number of oscillations per second (Hz = 1/s)
• •Angular frequency (omega): omega = 2pif = 2*pi/T (rad/s)

Relationships:

• •f = 1/T
• •omega = 2pif = 2*pi/T

Amplitude

• •Maximum displacement from equilibrium position
• •Does not affect period in ideal SHM
• •Determines maximum velocity and acceleration

Mass-Spring System

A mass m attached to a spring with spring constant k:

• •Period: T = 2pisqrt(m/k)
• •Frequency: f = (1/2*pi)*sqrt(k/m)
• •Angular frequency: omega = sqrt(k/m)

Key insights:

• •Period increases with mass (heavier = slower)
• •Period decreases with stiffer spring (larger k = faster)
• •Period is independent of amplitude and gravity

Simple Pendulum

A mass on a string of length L (for small angles < 15 degrees):

• •Period: T = 2pisqrt(L/g)
• •Frequency: f = (1/2*pi)*sqrt(g/L)

Key insights:

• •Period depends only on length and gravitational acceleration
• •Period is independent of mass and amplitude (for small angles)
• •Longer pendulum = longer period

Position, Velocity, and Acceleration in SHM

Starting from maximum displacement (x = A at t = 0):

• •Position: x(t) = Acos(omegat)
• •Velocity: v(t) = -Aomegasin(omega*t)
• •Acceleration: a(t) = -Aomega^2cos(omegat) = -omega^2x

Maximum values:

• •Maximum displacement: |x_max| = A
• •Maximum velocity: |v_max| = A*omega
• •Maximum acceleration: |a_max| = A*omega^2

Phase relationships:

• •Velocity leads position by 90 degrees (pi/2 radians)
• •Acceleration leads velocity by 90 degrees
• •Acceleration is 180 degrees out of phase with position

Energy in Simple Harmonic Motion

Total mechanical energy is conserved:

• •Potential Energy: PE = (1/2)kx^2
• •Kinetic Energy: KE = (1/2)mv^2
• •Total Energy: E = (1/2)kA^2 = (1/2)mv_max^2

At any position:

• •E = KE + PE = (1/2)mv^2 + (1/2)kx^2 = (1/2)kA^2

Energy exchange:

• •At x = 0 (equilibrium): KE = max, PE = 0
• •At x = +/- A (endpoints): KE = 0, PE = max

Damped Oscillations

Real oscillations lose energy due to friction or air resistance:

• •Amplitude decreases over time
• •Three regimes:
  • •Underdamped: Oscillates with decreasing amplitude
  • •Critically damped: Returns to equilibrium fastest without oscillating
  • •Overdamped: Returns slowly without oscillating

For underdamped systems:

• •x(t) = A*e^(-bt/(2m))*cos(omega't)
• •omega' < omega_0 (damped frequency is less than natural frequency)

Equations

[1] x = A*cos(omega*t) or x = A*sin(omega*t + phi) (position in SHM)
[2] v = -A*omega*sin(omega*t) (velocity in SHM)
[3] a = -A*omega^2*cos(omega*t) = -omega^2*x (acceleration in SHM)
[4] T = 2*pi*sqrt(m/k) (period of mass-spring system)
[5] T = 2*pi*sqrt(L/g) (period of simple pendulum)
[6] f = 1/T, omega = 2*pi*f = 2*pi/T (frequency relationships)
[7] v_max = A*omega (maximum velocity)
[8] a_max = A*omega^2 (maximum acceleration)
[9] E = (1/2)*k*A^2 = (1/2)*m*v_max^2 (total energy)
[10] (1/2)*m*v^2 + (1/2)*k*x^2 = (1/2)*k*A^2 (energy conservation)

Worked Examples

Example 1: Mass-Spring Period

Problem: A 0.5 kg mass is attached to a spring with k = 200 N/m. What is the period of oscillation?

Solution:

1. •Use the mass-spring period formula: T = 2pisqrt(m/k)
2. •T = 2pisqrt(0.5/200) = 2pisqrt(0.0025)
3. •T = 2*pi * 0.05 = 0.314 s

Example 2: Simple Pendulum

Problem: What length pendulum has a period of 2.0 s on Earth (g = 9.8 m/s^2)?

Solution:

1. •Use T = 2pisqrt(L/g), solve for L
2. •T^2 = 4*pi^2 * L/g
3. •L = gT^2 / (4pi^2) = 9.8 * 4 / (4*pi^2)
4. •L = 39.2 / 39.48 = 0.993 m (approximately 1 m)

Example 3: Energy in SHM

Problem: A 2 kg mass on a spring (k = 50 N/m) oscillates with amplitude 0.1 m. What is the maximum speed?

Solution:

1. •Total energy: E = (1/2)kA^2 = (1/2)(50)(0.1)^2 = 0.25 J
2. •At equilibrium, all energy is kinetic: E = (1/2)mv_max^2
3. •v_max = sqrt(2E/m) = sqrt(2*0.25/2) = sqrt(0.25) = 0.5 m/s

Alternative: v_max = Aomega = Asqrt(k/m) = 0.1sqrt(50/2) = 0.15 = 0.5 m/s

Example 4: Position and Velocity

Problem: A spring system has omega = 4 rad/s and A = 0.2 m. If x = A at t = 0, what is the position and velocity at t = 0.5 s?

Solution:

1. •Position: x = Acos(omegat) = 0.2cos(40.5) = 0.2cos(2) = 0.2(-0.416) = -0.083 m
2. •Velocity: v = -Aomegasin(omegat) = -0.24sin(2) = -0.8(0.909) = -0.727 m/s

Explanation Patterns

1. •Identify the type of oscillator: Is it a spring, pendulum, or other system?
2. •Determine given quantities: What do you know (m, k, L, A, T, f)?
3. •Choose the appropriate period formula:
   • •Spring: T = 2pisqrt(m/k)
   • •Pendulum: T = 2pisqrt(L/g)
4. •For kinematics problems: Write x(t), then differentiate for v(t) and a(t)
5. •For energy problems: Use E = (1/2)kA^2 and energy conservation
6. •Check your answer: Does the period make sense? Are units correct?

Common Problem Types

1. •Period/frequency calculations: Find T or f given system parameters
2. •Mass-spring problems: Apply T = 2pisqrt(m/k) or solve for m or k
3. •Pendulum problems: Apply T = 2pisqrt(L/g) or solve for L; compare periods on different planets
4. •Maximum velocity/acceleration: Use v_max = Aomega and a_max = Aomega^2
5. •Position at a given time: Use x = Acos(omegat) with correct initial conditions
6. •Energy problems: Apply conservation of energy between positions
7. •Velocity at a given position: Use energy conservation: (1/2)mv^2 + (1/2)kx^2 = (1/2)kA^2
8. •Damping effects: Qualitative understanding of amplitude decay