Introduction to Simple Harmonic Motion

Restoring forces, period, frequency, and Hooke's Law

🌊 Introduction to Simple Harmonic Motion

What is Simple Harmonic Motion (SHM)?

Simple Harmonic Motion is periodic oscillatory motion where the restoring force is proportional to displacement:

$F = -kx$

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

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

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

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:

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

For Pendulum Problems:

Identify: length $L$
Check: small angle approximation valid? ( $\theta < 15°$ )
Calculate period: $T = 2\pi\sqrt{\frac{L}{g}}$
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

| Quantity | Symbol | Mass-Spring | Pendulum | |----------|--------|-------------|----------| | Restoring Force | $F$ | $-kx$ | $-mg\sin\theta \approx -mg\theta$ | | Period | $T$ | $2\pi\sqrt{\frac{m}{k}}$ | $2\pi\sqrt{\frac{L}{g}}$ | | Frequency | $f$ | $\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ | $\frac{1}{2\pi}\sqrt{\frac{g}{L}}$ | | Angular Frequency | $\omega$ | $\sqrt{\frac{k}{m}}$ | $\sqrt{\frac{g}{L}}$ | | Max Speed | $v_{max}$ | $A\omega$ | $A\omega$ | | Max Acceleration | $a_{max}$ | $A\omega^2$ | $A\omega^2$ |

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

📚 Practice Problems

1Problem 1easy

❓ Question:

A 0.5 kg mass is attached to a spring with spring constant k = 200 N/m. The mass is pulled 0.1 m from equilibrium and released. Find: (a) the period of oscillation, (b) the frequency, and (c) the maximum speed.

💡 Show Solution

Given Information:

Mass: $m = 0.5$ kg
Spring constant: $k = 200$ N/m
Amplitude: $A = 0.1$ m (pulled 0.1 m from equilibrium)

(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.

2Problem 2medium

❓ Question:

A mass on a spring oscillates with amplitude 0.10 m and frequency 2.0 Hz. (a) What is the period? (b) What is the maximum speed? (c) What is the maximum acceleration?

💡 Show Solution

Solution:

Given: A = 0.10 m, f = 2.0 Hz

(a) Period: T = 1/f = 1/2.0 = 0.50 s

(b) Maximum speed: Angular frequency: ω = 2πf = 2π(2.0) = 4π rad/s v_max = Aω = 0.10(4π) = 1.26 m/s or 0.4π m/s

Note: Maximum speed occurs at equilibrium (x = 0), maximum acceleration at maximum displacement (x = ±A).

3Problem 3medium

❓ Question:

A simple pendulum has a length of 2 m. (a) What is its period on Earth? (b) If the same pendulum were on the Moon (where g = 1.6 m/s²), what would its period be?

💡 Show Solution

Given Information:

Length: $L = 2$ m
Earth: $g_{Earth} = 9.8$ m/s²
Moon: $g_{Moon} = 1.6$ m/s²

(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!

4Problem 4hard

❓ Question:

A 0.50 kg mass on a spring oscillates with amplitude 0.15 m. At a displacement of 0.10 m from equilibrium, the speed is 0.80 m/s. (a) Find the angular frequency. (b) Find the spring constant. (c) Find the period.

💡 Show Solution

Solution:

Given: m = 0.50 kg, A = 0.15 m, at x = 0.10 m, v = 0.80 m/s

(a) Angular frequency: For SHM: v² = ω²(A² - x²) (0.80)² = ω²[(0.15)² - (0.10)²] 0.64 = ω²[0.0225 - 0.0100] 0.64 = ω²(0.0125) ω² = 51.2 ω = 7.16 rad/s or 7.2 rad/s

(b) Spring constant: ω = √(k/m) 7.16 = √(k/0.50) 51.2 = k/0.50 k = 25.6 N/m or 26 N/m

Or: T = 2π√(m/k) = 2π√(0.50/25.6) = 0.88 s ✓

5Problem 5hard

❓ Question:

A 2 kg mass on a spring oscillates with amplitude 0.15 m and period 1.5 s. Find: (a) the spring constant, (b) the maximum speed, (c) the maximum acceleration, and (d) the speed when the displacement is 0.1 m from equilibrium.

💡 Show Solution

Given Information:

Mass: $m = 2$ kg
Amplitude: $A = 0.15$ m
Period: $T = 1.5$ s

(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!

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Introduction to Simple Harmonic Motion

🌊 Introduction to Simple Harmonic Motion

What is Simple Harmonic Motion (SHM)?

Examples of SHM

Hooke's Law

Characteristics of SHM

Amplitude AAA

Period TTT

Frequency fff

Angular Frequency ω\omegaω

Position, Velocity, and Acceleration in SHM

Position

Velocity

Acceleration

Mass-Spring System

Simple Pendulum

⚠️ Common Mistakes

Mistake 1: Confusing Period and Frequency

Mistake 2: Thinking Amplitude Affects Period

Mistake 3: Sign of Restoring Force

Mistake 4: Pendulum Period and Mass

Restoring Force

Problem-Solving Strategy

For Mass-Spring Problems:

For Pendulum Problems:

Applications

Clocks and Timekeeping

Seismographs

Musical Instruments

Molecular Vibrations

Key Formulas Summary

📚 Practice Problems

1Problem 1easy

❓ Question:

2Problem 2medium

❓ Question:

3Problem 3medium

❓ Question:

4Problem 4hard

❓ Question:

5Problem 5hard

❓ Question:

Practice with Flashcards

Browse All Topics

Amplitude $A$

Period $T$

Frequency $f$

Angular Frequency $\omega$