💡 Key Idea: SHM occurs when an object experiences a restoring force proportional to its displacement from equilibrium. The negative sign means the force always tries to bring the object back to equilibrium.

Examples of SHM

Mass-Spring System:

• Mass attached to spring
• Pull and release → oscillates
• Spring force: $F = -kx$ (Hooke's Law)

Simple Pendulum:

• Mass on string
• Pull aside and release → swings
• For small angles: acts like SHM
• Restoring force: component of gravity

Other Examples:

• Vibrating guitar string
• Atoms in a solid (vibrate about equilibrium)
• Molecular bonds
• Earthquake oscillations (seismograph)

Hooke's Law

For an ideal spring, the force is proportional to displacement:

$F = -kx$

Spring Constant $k$:

• Measures spring "stiffness"
• Units: N/m
• Large $k$: stiff spring (hard to stretch)
• Small $k$: soft spring (easy to stretch)

Equilibrium Position:

• Where spring is unstretched ($x = 0$)
• No net force
• Object naturally returns here

Characteristics of SHM

Amplitude $A$

Maximum displacement from equilibrium:

• If object goes from $-A$ to $+A$, amplitude is $A$
• Always positive
• Units: meters
• Independent of mass or spring constant

Period $T$

Time for one complete oscillation:

• From max displacement, through equilibrium, to other max, and back
• Units: seconds (s)
• Does NOT depend on amplitude!

For mass-spring system: $T = 2\pi\sqrt{\frac{m}{k}}$

For simple pendulum (small angles): $T = 2\pi\sqrt{\frac{L}{g}}$

where $L$ is length, $g$ is gravitational acceleration.

Frequency $f$

Number of oscillations per second: $f = \frac{1}{T}$

Units: Hz (hertz) = 1/s = cycles/second

For mass-spring: $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Angular Frequency $\omega$

$\omega = 2\pi f = \frac{2\pi}{T}$

Units: rad/s

For mass-spring: $\omega = \sqrt{\frac{k}{m}}$

Position, Velocity, and Acceleration in SHM

Position

$x(t) = A\cos(\omega t + \phi)$

or

$x(t) = A\sin(\omega t + \phi)$

where:

• $A$ = amplitude
• $\omega$ = angular frequency
• $\phi$ = phase constant (depends on initial conditions)
• $t$ = time

Velocity

$v(t) = -A\omega\sin(\omega t + \phi)$

Maximum speed (at equilibrium, $x = 0$): $v_{max} = A\omega$

Acceleration

$a(t) = -A\omega^2\cos(\omega t + \phi) = -\omega^2 x$

Maximum acceleration (at amplitude, $x = \pm A$): $a_{max} = A\omega^2$

Key relationship: $a = -\omega^2 x$

Acceleration is always proportional to (and opposite in direction to) displacement!

Mass-Spring System

For a mass $m$ attached to spring with constant $k$:

Period: $T = 2\pi\sqrt{\frac{m}{k}}$

Frequency: $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Key Observations:

• Heavier mass → longer period (slower oscillation)
• Stiffer spring (larger $k$) → shorter period (faster oscillation)
• Independent of amplitude! (as long as spring obeys Hooke's Law)

Simple Pendulum

For a pendulum with length $L$ and small amplitude:

Period: $T = 2\pi\sqrt{\frac{L}{g}}$

Key Observations:

• Longer pendulum → longer period
• Independent of mass!
• Independent of amplitude! (for small angles < 15°)
• Only depends on length and gravity

Note: For large angles, period does depend on amplitude (not SHM anymore).

⚠️ Common Mistakes

Mistake 1: Confusing Period and Frequency

• Period $T$: time for ONE oscillation (seconds)
• Frequency $f$: oscillations per second (Hz)
• They are reciprocals: $f = 1/T$

Mistake 2: Thinking Amplitude Affects Period

In ideal SHM, period is independent of amplitude!

• Large swing or small swing, same period
• As long as Hooke's Law holds

Mistake 3: Sign of Restoring Force

Force is ALWAYS opposite to displacement: $F = -kx$

• If $x > 0$ (stretched right), then $F < 0$ (force left)
• If $x < 0$ (compressed left), then $F > 0$ (force right)

Mistake 4: Pendulum Period and Mass

Pendulum period does NOT depend on mass!

• Only length $L$ and gravity $g$ matter
• Heavy or light bob, same period (if same length)

Restoring Force

Definition: Force that always acts to return object to equilibrium

Characteristics:

• Always points toward equilibrium
• Proportional to displacement (for SHM)
• Creates oscillatory motion

For Spring: $F = -kx$

For Pendulum (small angle $\theta$):

• Restoring force = component of weight tangent to arc
• $F = -mg\sin\theta \approx -mg\theta$ (for small $\theta$)
• For small angles: $\theta \approx \frac{x}{L}$
• So: $F \approx -\frac{mg}{L}x$ (proportional to $x$!)

Problem-Solving Strategy

For Mass-Spring Problems:

1. Identify: mass $m$, spring constant $k$
2. Calculate period: $T = 2\pi\sqrt{\frac{m}{k}}$
3. Calculate frequency: $f = 1/T$
4. Find amplitude: from initial conditions
5. Maximum speed: $v_{max} = A\omega$
6. Maximum acceleration: $a_{max} = A\omega^2$

For Pendulum Problems:

1. Identify: length $L$
2. Check: small angle approximation valid? ($\theta < 15°$)
3. Calculate period: $T = 2\pi\sqrt{\frac{L}{g}}$
4. Note: independent of mass and (small) amplitude

Applications

Clocks and Timekeeping

• Pendulum clocks: invented by Christiaan Huygens (1656)
• Period independent of amplitude (for small swings)
• Reliable timekeeping

Seismographs

• Detect earthquake vibrations
• Record ground oscillations
• Mass-spring system measures ground motion

Musical Instruments

• Vibrating strings (guitars, violins)
• Vibrating air columns (flutes, organ pipes)
• Vibrating membranes (drums)

Molecular Vibrations

• Atoms in molecules vibrate about equilibrium
• Absorption spectra reveal molecular structure
• Chemical bond strength ~ spring constant

Key Formulas Summary

Universal: $f = 1/T$, $\omega = 2\pi f$, $a = -\omega^2 x$

(a) Find period of oscillation

Step 1: Calculate period

$T = 2\pi\sqrt{\frac{m}{k}}$

$T = 2\pi\sqrt{\frac{0.5}{200}}$

$T = 2\pi\sqrt{0.0025}$

$T = 2\pi(0.05)$

$T = 0.314 \text{ s}$

Answer (a): Period = 0.314 s (about 0.31 seconds per oscillation)

(b) Find frequency

Step 2: Calculate frequency

$f = \frac{1}{T}$

$f = \frac{1}{0.314}$

$f = 3.18 \text{ Hz}$

Answer (b): Frequency = 3.18 Hz (about 3.2 oscillations per second)

(c) Find maximum speed

Step 3: Calculate angular frequency

$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.314} = 20 \text{ rad/s}$

Or directly: $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}} = \sqrt{400} = 20 \text{ rad/s}$

Step 4: Calculate maximum speed

Maximum speed occurs at equilibrium position:

$v_{max} = A\omega$

$v_{max} = (0.1)(20)$

$v_{max} = 2.0 \text{ m/s}$

Answer (c): Maximum speed = 2.0 m/s

Summary: The mass oscillates with period 0.31 s, completing about 3.2 cycles per second, reaching a maximum speed of 2 m/s as it passes through equilibrium.