7.1 Defining Simple Harmonic Motion (SHM)

Simple harmonic motion is a type of periodic motion where objects oscillate back and forth. It occurs when a restoring force acts on an object displaced from equilibrium, causing it to accelerate back towards its resting position.

The key feature of simple harmonic motion is that the restoring force is proportional to displacement. This relationship, described by Hooke's law, results in oscillations around an equilibrium position, like a mass on a spring or a swinging pendulum.

Simple harmonic motion

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Periodic motion

Periodic motion occurs when an object moves in a regular, repeating pattern over equal time intervals.

• Simple harmonic motion (SHM) is a specific type of periodic motion where an object oscillates back and forth repeatedly 🔁
• Occurs when a restoring force acts on an object displaced from its equilibrium position, causing it to accelerate back towards equilibrium
• Examples of periodic motion include a swinging pendulum, a mass on a spring, and a vibrating guitar string

Mathematical Description of SHM

SHM can be described with equations that relate position, velocity, and acceleration to time.

• Position as a function of time in SHM follows a sinusoidal pattern: $x(t) = A\cos(\omega t + \phi)$
  • $A$ is the amplitude, or maximum displacement from equilibrium
  • $\omega$ is the angular frequency, related to period $T$ by $\omega = \frac{2\pi}{T}$
  • $\phi$ is the phase constant, determined by initial conditions
• Velocity in SHM is given by the derivative of position: $v(t) = -A\omega\sin(\omega t + \phi)$
  • Velocity is maximum at equilibrium position and zero at maximum displacement
  • Maximum velocity is $v_{max} = A\omega$
• Acceleration in SHM is given by the derivative of velocity: $a(t) = -A\omega^2\cos(\omega t + \phi)$
  • Acceleration is zero at equilibrium position and maximum at maximum displacement
  • Maximum acceleration is $a_{max} = A\omega^2$
• The period of a mass-spring system is $T = 2\pi\sqrt{\frac{m}{k}}$, where $m$ is mass and $k$ is spring constant
• The period of a simple pendulum (for small angles) is $T = 2\pi\sqrt{\frac{L}{g}}$, where $L$ is length and $g$ is gravitational acceleration

Energy in SHM

Simple harmonic motion involves a continuous conversion between potential and kinetic energy.

• Total mechanical energy is conserved in an ideal SHM system
  • Sum of kinetic energy and potential energy remains constant: $E_{total} = KE + PE = \text{constant}$
  • Energy continuously transforms between kinetic and potential forms
• Potential energy in a spring system is $PE = \frac{1}{2}kx^2$
  • Maximum at maximum displacement (amplitude)
  • Zero at equilibrium position
• Kinetic energy is $KE = \frac{1}{2}mv^2$
  • Maximum at equilibrium position
  • Zero at maximum displacement
• Total energy in SHM is $E_{total} = \frac{1}{2}kA^2$, where $A$ is the amplitude
• Energy conservation explains why amplitude decreases in real systems with friction or other dissipative forces

Practice Problem 1: Mass-Spring System

Solution:

(a) The period of oscillation for a mass-spring system is: $T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.25\text{ kg}}{16\text{ N/m}}} = 2\pi\sqrt{0.015625} = 2\pi \times 0.125 = 0.785\text{ s}$

(b) The maximum velocity occurs at the equilibrium position: $v_{max} = A\omega = A \times \frac{2\pi}{T} = 0.10\text{ m} \times \frac{2\pi}{0.785\text{ s}} = 0.10\text{ m} \times 8\text{ rad/s} = 0.8\text{ m/s}$

(c) The total energy of the system is: $E_{total} = \frac{1}{2}kA^2 = \frac{1}{2} \times 16\text{ N/m} \times (0.10\text{ m})^2 = \frac{1}{2} \times 16\text{ N/m} \times 0.01\text{ m}^2 = 0.08\text{ J}$

(d) Using energy conservation when the mass is 5 cm from equilibrium: $E_{total} = PE + KE = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$ $0.08\text{ J} = \frac{1}{2} \times 16\text{ N/m} \times (0.05\text{ m})^2 + \frac{1}{2} \times 0.25\text{ kg} \times v^2$ $0.08\text{ J} = 0.02\text{ J} + \frac{1}{2} \times 0.25\text{ kg} \times v^2$ $0.06\text{ J} = \frac{1}{2} \times 0.25\text{ kg} \times v^2$ $v = \sqrt{\frac{0.06\text{ J} \times 2}{0.25\text{ kg}}} = \sqrt{0.48\text{ m}^2/\text{s}^2} = 0.693\text{ m/s}$

Practice Problem 2: Pendulum Motion

Solution:

(a) The period of a simple pendulum is: $T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{1.2\text{ m}}{9.8\text{ m/s}^2}} = 2\pi\sqrt{0.122\text{ s}^2} = 2\pi \times 0.35\text{ s} = 2.2\text{ s}$

(b) First, convert the angle to radians: $\theta = 5° \times \frac{\pi}{180°} = 0.0873\text{ rad}$

The maximum displacement (arc length) is: $A = L\theta = 1.2\text{ m} \times 0.0873\text{ rad} = 0.105\text{ m}$

The maximum velocity occurs at the equilibrium position: $v_{max} = A\omega = A \times \frac{2\pi}{T} = 0.105\text{ m} \times \frac{2\pi}{2.2\text{ s}} = 0.105\text{ m} \times 2.86\text{ rad/s} = 0.30\text{ m/s}$

(c) The maximum acceleration occurs at the maximum displacement: $a_{max} = A\omega^2 = A \times \left(\frac{2\pi}{T}\right)^2 = 0.105\text{ m} \times \left(\frac{2\pi}{2.2\text{ s}}\right)^2 = 0.105\text{ m} \times 8.17\text{ rad}^2/\text{s}^2 = 0.86\text{ m/s}^2$

Practice Problem 3: Energy in SHM

Solution:

(a) The total energy of the system is: $E_{total} = \frac{1}{2}kA^2 = \frac{1}{2} \times 20\text{ N/m} \times (0.15\text{ m})^2 = \frac{1}{2} \times 20\text{ N/m} \times 0.0225\text{ m}^2 = 0.225\text{ J}$

(b) Using energy conservation when the block is 10 cm from equilibrium: $E_{total} = PE + KE = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$ $0.225\text{ J} = \frac{1}{2} \times 20\text{ N/m} \times (0.10\text{ m})^2 + \frac{1}{2} \times 0.5\text{ kg} \times v^2$ $0.225\text{ J} = 0.1\text{ J} + \frac{1}{2} \times 0.5\text{ kg} \times v^2$ $0.125\text{ J} = \frac{1}{2} \times 0.5\text{ kg} \times v^2$ $v = \sqrt{\frac{0.125\text{ J} \times 2}{0.5\text{ kg}}} = \sqrt{0.5\text{ m}^2/\text{s}^2} = 0.707\text{ m/s}$

(c) The kinetic energy at this position is: $KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5\text{ kg} \times (0.707\text{ m/s})^2 = \frac{1}{2} \times 0.5\text{ kg} \times 0.5\text{ m}^2/\text{s}^2 = 0.125\text{ J}$

The percentage of total energy that is kinetic energy: $\text{Percentage} = \frac{KE}{E_{total}} \times 100\% = \frac{0.125\text{ J}}{0.225\text{ J}} \times 100\% = 55.6\%$