Simple Harmonic Motion

To fully specify the initial conditions of a particle in simple harmonic motion, we need:

1. The initial displacement (x₀) from the equilibrium position
2. The initial velocity (v₀) at that position

These two parameters completely determine the subsequent motion of the particle, including:

• The amplitude of oscillation
• The phase constant
• The exact position and velocity at any future time

The equation of motion can then be written as: x = A·sin(ωt + δ)

where A and δ are determined from x₀ and v₀.

Each phase value indicates a specific position and motion state in the oscillation cycle:

• φ = 0: Particle is at the mean position (x = 0) moving toward positive direction with maximum velocity • φ = π/2: Particle is at the positive extreme position (x = A) with zero velocity • φ = π: Particle is at the mean position (x = 0) moving toward negative direction with maximum velocity • φ = 3π/2: Particle is at the negative extreme position (x = -A) with zero velocity • φ = 2π: Particle returns to initial state at mean position moving toward positive direction

The full cycle of 2π radians represents one complete oscillation.

In a pendulum:

• The gravitational force (mg) resolves into two components:

  1. Tangential component: mg·sinθ (parallel to motion)
  2. Radial component: mg·cosθ (countered by tension)

• The tangential component mg·sinθ acts as the restoring force

• For small angles: sinθ ≈ θ (in radians)

• Therefore: restoring force ≈ -mg·θ

• Since θ = x/l (where x is arc displacement and l is length), this gives: F = -mg·(x/l) = -(mg/l)·x

• This follows Hooke's Law form F = -kx where k = mg/l

• The negative sign indicates the force opposes displacement

This tangential component is what creates the characteristic simple harmonic motion of a pendulum with period T = 2π√(l/g).

Derivation of pendulum as linear SHM:

1. Starting with Newton's Second Law applied to tangential motion:

   • Force = mass × tangential acceleration
   • -mg·sinθ = m·(d²x/dt²), where x = lθ

2. Rearranging to get the differential equation:

   • d²x/dt² = -g·sinθ

3. The critical approximation: For small angles θ (measured in radians)

   • sinθ ≈ θ = x/l

4. Substituting this approximation:

   • d²x/dt² = -g·(x/l) = -(g/l)·x
   • d²x/dt² + (g/l)·x = 0

5. This is precisely the form of simple harmonic motion: d²x/dt² + ω²x = 0

   • Where ω² = g/l
   • ω = √(g/l) is the angular frequency

This small-angle approximation is what allows us to treat pendulum motion as simple harmonic, with period T = 2π√(l/g), valid when the amplitude is small (generally <15°).

The time period of a physical pendulum is:

$T = 2\pi\sqrt{\frac{I}{mgl}}$

Where:

• $I$ = moment of inertia about the axis of rotation
• $m$ = mass of the pendulum
• $g$ = acceleration due to gravity
• $l$ = distance from pivot point to center of mass

Key insights: • Unlike a simple pendulum, the time period depends on the mass distribution (via moment of inertia) • For small oscillations, the motion is approximately simple harmonic • The formula applies to any rigid body suspended from a fixed support • As $I$ increases, the time period increases

Example: A uniform rod suspended from one end has $I = \frac{mL^2}{3}$ and $l = \frac{L}{2}$, giving $T = 2\pi\sqrt{\frac{2L}{3g}}$

Torque equation for a physical pendulum:

$\tau = -mgl\sin\theta$

Where:

• $m$ = mass of the pendulum
• $g$ = acceleration due to gravity
• $l$ = distance from pivot to center of mass
• $\theta$ = angular displacement

This creates angular acceleration: $\alpha = \frac{\tau}{I} = -\frac{mgl}{I}\sin\theta$

Simple harmonic motion condition: • When θ is small, $\sin\theta \approx \theta$ • This approximation gives: $\alpha = -\frac{mgl}{I}\theta = -\omega^2\theta$ • Where $\omega^2 = \frac{mgl}{I}$

The motion becomes simple harmonic only for small oscillations because only then is the restoring torque proportional to the angular displacement.

For large angles, the motion is not strictly simple harmonic due to the nonlinear nature of the $\sin\theta$ term.

Analysis of forces on a physical pendulum:

1. Weight ($mg$): • Acts vertically downward through the center of mass • Magnitude = $mg$ • Creates torque of magnitude $mgl\sin\theta$ around pivot • This is the restoring torque that drives oscillation

2. Normal/contact force ($N$): • Acts at the pivot point • Balances other forces to maintain constraint • Contributes zero torque about the pivot point • Reason: force line of action passes through the axis of rotation

Key principle: Only forces with a moment arm relative to the pivot contribute to torque.

The torque equation becomes: $\tau = -mgl\sin\theta$

This analysis applies to any rigid body suspended from a fixed support, such as a circular ring on a nail or a rod suspended at a hole.

Note: While the normal force doesn't affect the rotational dynamics directly, it's essential for maintaining the pendulum's constraint.

Equation of motion: $I\alpha = -k\theta$, which becomes $\alpha = -\frac{k}{I}\theta$

Where: • $I$ = moment of inertia of suspended body • $\alpha$ = angular acceleration • $k$ = torsional constant of wire • $\theta$ = angular displacement

This is simple harmonic motion because: • Angular acceleration is directly proportional to angular displacement • Acceleration is directed opposite to displacement • The system oscillates around an equilibrium position ($\theta = 0$)

Required conditions: • Small amplitude oscillations only • No significant damping forces (air resistance, internal friction) • Torsional constant must remain linear (Hooke's Law applies) • Wire must remain vertical throughout oscillation

Angular frequency: $\omega = \sqrt{\frac{k}{I}}$

Example: Using a torsional pendulum to determine moment of inertia of irregularly shaped objects by measuring oscillation period

Special cases in combining SHMs of same frequency:

1. In-phase SHMs (δ = 0°):

   • Resultant amplitude: A = A₁ + A₂ + A₃ + ... (direct sum)
   • Resultant phase: Same as component phases
   • This gives maximum possible amplitude

2. Opposite-phase SHMs (δ = 180°):

   • For two SHMs: A = |A₁ - A₂| (absolute difference)
   • If A₁ = A₂: complete cancellation (A = 0)
   • Resultant phase: Phase of the larger amplitude component

3. Quadrature SHMs (δ = 90°):

   • Resultant amplitude: A = √(A₁² + A₂²)
   • Resultant phase: tanε = A₂/A₁
   • If A₁ = A₂: resultant phase is 45° relative to first component

4. Multiple equal amplitudes with equal phase differences:

   • Creates special patterns like circles or complex Lissajous figures when components are in perpendicular directions

Damping effects on resonance width:

• Larger damping creates broader, flatter resonance curves
• Smaller damping produces narrower, sharper resonance peaks
• The width of the resonance curve is proportional to the damping coefficient

Practical implications:

• Systems with low damping (high Q-factor) are highly selective frequency filters
• Highly damped systems respond to a wider range of driving frequencies
• Engineering applications must balance response amplitude vs. frequency selectivity
• Precision instruments (tuning forks, quartz oscillators) use low damping for frequency stability
• Vehicle suspension systems use higher damping to absorb vibrations across multiple frequencies