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Introduction to Simple Harmonic Motion

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

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🌊 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=−kxF = -kx

where:

  • FF = restoring force (N)
  • kk = spring constant or force constant (N/m)
  • xx = 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=−kxF = -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=−kxF = -kx

Spring Constant kk:

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

Equilibrium Position:

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

Characteristics of SHM

Amplitude AA

Maximum displacement from equilibrium:

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

Period TT

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=2πmkT = 2\pi\sqrt{\frac{m}{k}}

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

where LL is length, gg is gravitational acceleration.

Frequency ff

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

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

For mass-spring: f=12πkmf = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

Angular Frequency ω\omega

ω=2πf=2πT\omega = 2\pi f = \frac{2\pi}{T}

Units: rad/s

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


Position, Velocity, and Acceleration in SHM

Position

x(t)=Acos⁡(ωt+ϕ)x(t) = A\cos(\omega t + \phi)

or

x(t)=Asin⁡(ωt+ϕ)x(t) = A\sin(\omega t + \phi)

where:

  • AA = amplitude
  • ω\omega = angular frequency
  • ϕ\phi = phase constant (depends on initial conditions)
  • tt = time

Velocity

v(t)=−Aωsin⁡(ωt+ϕ)v(t) = -A\omega\sin(\omega t + \phi)

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

Acceleration

a(t)=−Aω2cos⁡(ωt+ϕ)=−ω2xa(t) = -A\omega^2\cos(\omega t + \phi) = -\omega^2 x

Maximum acceleration (at amplitude, x=±Ax = \pm A): amax=Aω2a_{max} = A\omega^2

Key relationship: a=−ω2xa = -\omega^2 x

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


Mass-Spring System

For a mass mm attached to spring with constant kk:

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

Frequency: f=12πkmf = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

Key Observations:

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

Simple Pendulum

For a pendulum with length LL and small amplitude:

Period: T=2πLgT = 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 TT: time for ONE oscillation (seconds)
  • Frequency ff: oscillations per second (Hz)
  • They are reciprocals: f=1/Tf = 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=−kxF = -kx

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

Mistake 4: Pendulum Period and Mass

Pendulum period does NOT depend on mass!

  • Only length LL and gravity gg 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=−kxF = -kx

For Pendulum (small angle θ\theta):

  • Restoring force = component of weight tangent to arc
  • F=−mgsin⁡θ≈−mgθF = -mg\sin\theta \approx -mg\theta (for small θ\theta)
  • For small angles: θ≈xL\theta \approx \frac{x}{L}
  • So: F≈−mgLxF \approx -\frac{mg}{L}x (proportional to xx!)

Problem-Solving Strategy

For Mass-Spring Problems:

  1. Identify: mass mm, spring constant kk
  2. Calculate period: T=2πmkT = 2\pi\sqrt{\frac{m}{k}}
  3. Calculate frequency: f=1/Tf = 1/T
  4. Find amplitude: from initial conditions
  5. Maximum speed: vmax=Aωv_{max} = A\omega
  6. Maximum acceleration: amax=Aω2a_{max} = A\omega^2

For Pendulum Problems:

  1. Identify: length LL
  2. Check: small angle approximation valid? (θ<15°\theta < 15°)
  3. Calculate period: T=2πLgT = 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

QuantitySymbolMass-SpringPendulum
Restoring ForceFF−kx-kx−mgsin⁡θ≈−mgθ-mg\sin\theta \approx -mg\theta
PeriodTT2πmk2\pi\sqrt{\frac{m}{k}}2πLg2\pi\sqrt{\frac{L}{g}}
Frequencyff12πkm\frac{1}{2\pi}\sqrt{\frac{k}{m}}12πgL\frac{1}{2\pi}\sqrt{\frac{g}{L}}
Angular Frequencyω\omegakm\sqrt{\frac{k}{m}}gL\sqrt{\frac{g}{L}}
Max Speedvmaxv_{max}AωA\omegaAωA\omega
Max Accelerationamaxa_{max}Aω2A\omega^2Aω2A\omega^2

Universal: f=1/Tf = 1/T, ω=2πf\omega = 2\pi f, a=−ω2xa = -\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=0.5m = 0.5 kg
  • Spring constant: k=200k = 200 N/m
  • Amplitude: A=0.1A = 0.1 m (pulled 0.1 m from equilibrium)

(a) Find period of oscillation


Step 1: Calculate period

T=2πmkT = 2\pi\sqrt{\frac{m}{k}}

T=2π0.5200T = 2\pi\sqrt{\frac{0.5}{200}}

T=2π0.0025T = 2\pi\sqrt{0.0025}

T=2π(0.05)T = 2\pi(0.05)

T=0.314 sT = 0.314 \text{ s}


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


(b) Find frequency


Step 2: Calculate frequency

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

f=10.314f = \frac{1}{0.314}

f=3.18 Hzf = 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

ω=2πT=2π0.314=20 rad/s\omega = \frac{2\pi}{T} = \frac{2\pi}{0.314} = 20 \text{ rad/s}

Or directly: ω=km=2000.5=400=20 rad/s\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:

vmax=Aωv_{max} = A\omega

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

vmax=2.0 m/sv_{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

(c) Maximum acceleration: a_max = Aω² = 0.10(4π)² = 0.10(16π²) a_max = 15.8 m/s² or 1.6π² m/s²

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=2L = 2 m
  • Earth: gEarth=9.8g_{Earth} = 9.8 m/s²
  • Moon: gMoon=1.6g_{Moon} = 1.6 m/s²

(a) Find period on Earth


Step 1: Apply pendulum period formula

T=2πLgT = 2\pi\sqrt{\frac{L}{g}}

TEarth=2π29.8T_{Earth} = 2\pi\sqrt{\frac{2}{9.8}}

TEarth=2π0.204T_{Earth} = 2\pi\sqrt{0.204}

TEarth=2π(0.452)T_{Earth} = 2\pi(0.452)

TEarth=2.84 sT_{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

TMoon=2πLgMoonT_{Moon} = 2\pi\sqrt{\frac{L}{g_{Moon}}}

TMoon=2π21.6T_{Moon} = 2\pi\sqrt{\frac{2}{1.6}}

TMoon=2π1.25T_{Moon} = 2\pi\sqrt{1.25}

TMoon=2π(1.118)T_{Moon} = 2\pi(1.118)

TMoon=7.02 sT_{Moon} = 7.02 \text{ s}


Step 3: Compare the periods

TMoonTEarth=7.022.84≈2.47\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: TMoonTEarth=gEarthgMoon=9.81.6=6.125≈2.47✓\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

(c) Period: T = 2π/ω = 2π/7.16 = 0.88 s

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

(a) Find spring constant


Step 1: Use period formula and solve for k

T=2πmkT = 2\pi\sqrt{\frac{m}{k}}

Square both sides: T2=4π2mkT^2 = 4\pi^2\frac{m}{k}

Solve for kk: k=4π2mT2k = \frac{4\pi^2 m}{T^2}

k=4π2(2)(1.5)2k = \frac{4\pi^2 (2)}{(1.5)^2}

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

k=78.962.25k = \frac{78.96}{2.25}

k=35.1 N/mk = 35.1 \text{ N/m}


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


(b) Find maximum speed


Step 2: Calculate angular frequency

ω=2πT=2π1.5=4.19 rad/s\omega = \frac{2\pi}{T} = \frac{2\pi}{1.5} = 4.19 \text{ rad/s}


Step 3: Calculate maximum speed

vmax=Aωv_{max} = A\omega

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

vmax=0.629 m/sv_{max} = 0.629 \text{ m/s}


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


(c) Find maximum acceleration


Step 4: Calculate maximum acceleration

amax=Aω2a_{max} = A\omega^2

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

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

amax=2.64 m/s2a_{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=12kA2E = \frac{1}{2}kA^2

At position xx: E=12kx2+12mv2E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2


Step 6: Set equations equal and solve for v

12kA2=12kx2+12mv2\frac{1}{2}kA^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2

Multiply by 2 and cancel: kA2=kx2+mv2kA^2 = kx^2 + mv^2

k(A2−x2)=mv2k(A^2 - x^2) = mv^2

v2=km(A2−x2)v^2 = \frac{k}{m}(A^2 - x^2)

Note: km=ω2=(4.19)2=17.6\frac{k}{m} = \omega^2 = (4.19)^2 = 17.6 rad²/s²

v2=17.6[(0.15)2−(0.1)2]v^2 = 17.6[(0.15)^2 - (0.1)^2]

v2=17.6[0.0225−0.01]v^2 = 17.6[0.0225 - 0.01]

v2=17.6(0.0125)v^2 = 17.6(0.0125)

v2=0.22v^2 = 0.22

v=0.47 m/sv = 0.47 \text{ m/s}


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

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

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

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

Explain using:

📌 Related Topics in Oscillations

❓ Frequently Asked Questions

What is Introduction to Simple Harmonic Motion?▾
Restoring forces, period, frequency, and Hooke's Law
How can I study Introduction to Simple Harmonic Motion effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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What course covers Introduction to Simple Harmonic Motion?▾
Introduction to Simple Harmonic Motion is part of the AP Physics 1 course on Study Mondo, specifically in the Oscillations section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Introduction to Simple Harmonic Motion?▾
Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.