$F = -kx$

where:

• $F$ = restoring force (N)
• $k$ = spring constant or force constant (N/m)
• $x$ = displacement from equilibrium (m)
• Negative sign: force always points toward equilibrium

💡 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 on Earth

Step 1: Apply pendulum period formula

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

$T_{Earth} = 2\pi\sqrt{\frac{2}{9.8}}$

$T_{Earth} = 2\pi\sqrt{0.204}$

$T_{Earth} = 2\pi(0.452)$

$T_{Earth} = 2.84 \text{ s}$

Answer (a): Period on Earth = 2.84 s

(b) Find period on Moon

Step 2: Apply formula with Moon's gravity

$T_{Moon} = 2\pi\sqrt{\frac{L}{g_{Moon}}}$

$T_{Moon} = 2\pi\sqrt{\frac{2}{1.6}}$

$T_{Moon} = 2\pi\sqrt{1.25}$

$T_{Moon} = 2\pi(1.118)$

$T_{Moon} = 7.02 \text{ s}$

Step 3: Compare the periods

$\frac{T_{Moon}}{T_{Earth}} = \frac{7.02}{2.84} \approx 2.47$

The Moon pendulum has a period about 2.5 times longer than on Earth!

Answer (b): Period on Moon = 7.02 s

Explanation: Lower gravity means weaker restoring force, so the pendulum oscillates more slowly.

Ratio check: $\frac{T_{Moon}}{T_{Earth}} = \sqrt{\frac{g_{Earth}}{g_{Moon}}} = \sqrt{\frac{9.8}{1.6}} = \sqrt{6.125} \approx 2.47 \quad ✓$

Note: This is why pendulum clocks would run slower on the Moon!

(a) Find spring constant

Step 1: Use period formula and solve for k

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

Square both sides: $T^2 = 4\pi^2\frac{m}{k}$

Solve for $k$: $k = \frac{4\pi^2 m}{T^2}$

$k = \frac{4\pi^2 (2)}{(1.5)^2}$

$k = \frac{4(9.87)(2)}{2.25}$

$k = \frac{78.96}{2.25}$

$k = 35.1 \text{ N/m}$

Answer (a): Spring constant = 35.1 N/m

(b) Find maximum speed

Step 2: Calculate angular frequency

$\omega = \frac{2\pi}{T} = \frac{2\pi}{1.5} = 4.19 \text{ rad/s}$

Step 3: Calculate maximum speed

$v_{max} = A\omega$

$v_{max} = (0.15)(4.19)$

$v_{max} = 0.629 \text{ m/s}$

Answer (b): Maximum speed = 0.63 m/s

(c) Find maximum acceleration

Step 4: Calculate maximum acceleration

$a_{max} = A\omega^2$

$a_{max} = (0.15)(4.19)^2$

$a_{max} = (0.15)(17.6)$

$a_{max} = 2.64 \text{ m/s}^2$

Answer (c): Maximum acceleration = 2.64 m/s²

(d) Find speed at x = 0.1 m

Step 5: Use energy conservation

Total energy (constant): $E = \frac{1}{2}kA^2$

At position $x$: $E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Step 6: Set equations equal and solve for v

$\frac{1}{2}kA^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Multiply by 2 and cancel: $kA^2 = kx^2 + mv^2$

$k(A^2 - x^2) = mv^2$

$v^2 = \frac{k}{m}(A^2 - x^2)$

Note: $\frac{k}{m} = \omega^2 = (4.19)^2 = 17.6$ rad²/s²

$v^2 = 17.6[(0.15)^2 - (0.1)^2]$

$v^2 = 17.6[0.0225 - 0.01]$

$v^2 = 17.6(0.0125)$

$v^2 = 0.22$

$v = 0.47 \text{ m/s}$

Answer (d): Speed at $x = 0.1$ m is 0.47 m/s

Check: At $x = 0$: $v = \sqrt{17.6(0.0225)} = 0.63$ m/s = $v_{max}$ ✓

At $x = A = 0.15$ m: $v = \sqrt{17.6(0)} = 0$ m/s ✓

Makes sense: speed is maximum at equilibrium, zero at amplitude!