Simple Harmonic Motion, Circular Motion, and Phasors

Learning Intentions and Success Criteria

• Learning Intention: We are learning to explain simple harmonic motion (SHM).

• Success Criteria:

  • Ability to use phasors to solve displacement, velocity, and acceleration problems.

  • Ability to determine the displacement, velocity, and acceleration of a reference particle during simple harmonic motion.

Fundamental Physical Quantities in Simple Harmonic Motion

Simple Harmonic Motion (SHM) describes repetitive back-and-forth oscillation through a central equilibrium position. The key physical variables involved in describing SHM at any instant are displacement ($y$ or $x$), acceleration ($a$), speed/velocity ($v$), kinetic energy ($E_k$), and potential energy ($E_p$).

Below is the summary of physical state parameters across key positions in an oscillating system:

• Displacement $y = -A$ (Negative Amplitude / Lower Extreme Position):

  • Acceleration ($a$): Maximum magnitude (directed towards the equilibrium position, $0$)

  • Speed ($v$): $0$

  • Kinetic Energy ($E_k$): $0$

  • Potential Energy ($E_p$): Maximum

• Displacement $y = 0$ (Equilibrium Position / Position at Rest):

  • Acceleration ($a$): $0$

  • Speed ($v$): Maximum

  • Kinetic Energy ($E_k$): Maximum

  • Potential Energy ($E_p$): $0\,\text{J}$ (minimum)

• Displacement $y = +A$ (Positive Amplitude / Upper Extreme Position):

  • Acceleration ($a$): Maximum magnitude (directed towards the equilibrium position, $0$)

  • Speed ($v$): $0$

  • Kinetic Energy ($E_k$): $0$

  • Potential Energy ($E_p$): Maximum

Simple Harmonic Motion, Circular Motion, and Reference Circles

When an object undergoing uniform circular motion is observed sideways (in the plane of rotation), the motion appears as linear simple harmonic motion along a straight line.

A reference circle is defined as a circle whose radius is equal to the amplitude ($A$) of the simple harmonic motion, and whose projection along a diameter gives the linear simple harmonic motion.

Equations of Motion for Simple Harmonic Motion

Important Calculation Rule: For all SHM trigonometric equations, the scientific calculator must strictly be set to radian mode.

Case 1: Motion Starting at Maximum Amplitude ($t = 0$ at $y = A$ or $y = -A$)

When timing starts as the object is at its maximum displacement position:

• Displacement ($y$):   $y = A \cos(\omega t)$

• Velocity ($v$):   $v = -A\omega \sin(\omega t)$

• Acceleration ($a$):   $a = -A\omega^2 \cos(\omega t)$

Case 2: Motion Starting at Equilibrium ($t = 0$ at $y = 0$)

When timing starts as the object passes through the equilibrium position moving towards positive displacement:

• Displacement ($y$):   $y = A \sin(\omega t)$

• Velocity ($v$):   $v = A\omega \cos(\omega t)$

• Acceleration ($a$):   $a = -A\omega^2 \sin(\omega t)$

Core Variables and Constants

• $A$: Amplitude (measured in meters, $\text{m}$)

• $\omega$: Angular frequency (measured in radians per second, $\text{rad s}^{-1}$)

• $t$: Time elapsed (measured in seconds, $\text{s}$)

• $T$: Time period for one complete oscillation (measured in seconds, $\text{s}$)

• $f$: Frequency of oscillation (measured in Hertz, $\text{Hz}$)

• $\theta$: Angle through which displacement phasor turns, where:   $\theta = \omega t$

• Angular frequency formula:   $\omega = \frac{2\pi}{T} = 2\pi f$

• Maximum Velocity ($v_{\text{max}}$):   $v_{\text{max}} = A\omega$

• Maximum Acceleration ($a_{\text{max}}$):   $a_{\text{max}} = -A\omega^2$

Practical Application Problems and Solutions

Example 1: Playground Swing Dynamics (Serena)

Serena sits on a rigid swing that is $3.00\,\text{m}$ long. She swings from end A to end B with an amplitude of $1.50\,\text{m}$ and a period of $3.50\,\text{s}$.

• a) Calculate the angular frequency of the SHM of the swing.

  • Formula:     $\omega = \frac{2\pi}{T}$

  • Calculation:     $\omega = \frac{2\pi}{3.50} = 1.795\,\text{rad s}^{-1}$

  • Rounded to 3 significant figures:     $\omega = 1.80\,\text{rad s}^{-1}$

• b) Serena swings $1.00\,\text{m}$ from the equilibrium position. Using the reference circle or otherwise, calculate the angle through which the displacement phasor turns.

  • Formula:     $\theta = \sin^{-1}\left(\frac{1.00}{1.50}\right)$

  • Calculation (in radian mode):     $\theta = 0.7297\,\text{rad}$

  • Rounded to 3 significant figures:     $\theta = 0.730\,\text{rad}$

• c) Calculate the time it takes for Serena to swing to this position.

  • Rearranging $\theta = \omega t$:     $t = \frac{\theta}{\omega}$

  • Substitution:     $t = \frac{0.7297}{1.795} = 0.407\,\text{s}$

• d) Calculate the velocity of Serena and the swing at this position.

  • Velocity formula starting from equilibrium:     $v = A\omega \cos(\omega t)$

  • Substitution:     $v = (1.50 \times 1.795) \cos(1.795 \times 0.407)$

  • Calculation:     $v = 2.01\,\text{m s}^{-1}$

Question 1: Ball Bearing on a Watch Glass

A ball bearing is released on a watch glass, and rolls back and forth with simple harmonic motion. The watch glass is a shallow, semi-circular glass bowl with a radius of curvature, $R$.

The ball bearing is released $0.0400\,\text{m}$ from the right of the equilibrium position, and oscillates with a time period of $0.882\,\text{s}$. Using reference circles or otherwise, calculate the displacement of the ball bearing after $1.20\,\text{s}$.

• Solution:

  • Calculate angular frequency $\omega$:     $\omega = 2\pi f = \frac{2\pi}{0.882} = 7.1238\,\text{rad s}^{-1}$

  • Since the particle is released from its maximum displacement, use the cosine function:     $y = A \cos(\omega t)$

  • Substitution:     $y = 0.0400 \cos(7.1238 \times 1.20)$

  • Calculation:     $y = -0.025602\,\text{m}$

  • Final value rounded to 3 significant figures:     $y = -0.0256\,\text{m}$

Question 2: Astronaut Landing Seat Spring System

When astronauts return to Earth, a spring under their seat reduces the force during the landing. The astronaut's kinetic energy is converted to spring potential energy as the spring is compressed. If friction is negligible, this will set the astronaut into simple harmonic motion.

• a) During a landing, an astronaut and seat had a combined mass of $80.0\,\text{kg}$ and were set into simple harmonic motion with an amplitude of $0.150\,\text{m}$ and a period of $0.940\,\text{s}$. Determine:

  • i) The spring constant of the spring:

  • Formula for time period of a mass-spring system:       $T = 2\pi \sqrt{\frac{m}{k}}$

  • Substitution:       $0.940 = 2\pi \sqrt{\frac{80.0}{k}}$

  • Solving for $k$:       $k = 3574\,\text{N m}^{-1}$

  • Rounded to 3 significant figures:       $k = 3570\,\text{N m}^{-1}$

  • ii) The amount of energy stored in the spring at maximum displacement:

  • Potential energy formula:       $E = \frac{1}{2} k x^2$

  • Substitution:       $E = 0.5 \times 3574 \times (0.150)^2 = 40.2\,\text{J}$

• b) Using a reference circle or otherwise, determine the velocity of the astronaut when the astronaut is $0.100\,\text{m}$ above the equilibrium position.

  • Calculate angular frequency $\omega$:     $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.940} = 6.684\,\text{rad s}^{-1}$

  • Calculate phasor angle $\phi$:     $\phi = \sin^{-1}\left(\frac{0.100}{0.150}\right) = 0.7297\,\text{radians}$

  • Calculate time $t$:     $t = \frac{\phi}{\omega} = \frac{0.7297}{6.684} = 0.109\,\text{s}$

  • Calculate velocity $v$:     $v = A\omega \cos(\omega t)$     $v = 0.150 \times 6.684 \times \cos(6.684 \times 0.109) = 0.748\,\text{m s}^{-1}$

Question 3: Simple Pendulum Calibration on Mars

Some space explorers on Mars want to check that their electronic timers are functioning correctly. They make a simple pendulum, using a large rock, mass $2.30\,\text{kg}$, tied to a wire.

• a) The distance from the centre of mass of the rock to the fixing point is $1.83\,\text{m}$. On Mars, the gravitational field strength is $3.72\,\text{N kg}^{-1}$. Show that the time period of the pendulum is $4.41\,\text{s}$.

  • Formula for simple pendulum period:     $T = 2\pi \sqrt{\frac{l}{g}}$

  • Substitution:     $T = 2\pi \sqrt{\frac{1.83}{3.72}} = 4.4069\,\text{s}$

  • Rounding confirms:     $T \approx 4.41\,\text{s}$

• b) They set the pendulum oscillating by releasing the pendulum bob $0.300\,\text{m}$ away from its rest position, and at the same moment they start a timer. Determine the position of the pendulum bob from its release point $2.00\,\text{s}$ after it is released.

  • Angular frequency $\omega$:     $\omega = \frac{2\pi}{T} = \frac{2\pi}{4.407} = 1.43\,\text{s}^{-1}$

  • Angle turned in $t = 2.00\,\text{s}$:     $\theta = \omega t = 1.43 \times 2.00 = 2.86\,\text{rad}$

  • Displacement from rest position $x$ (since released at maximum displacement $A = 0.300\,\text{m}$):     $x = 0.300 \cos(2.86) = -0.288\,\text{m}$

  • Distance from initial release point:     $\text{Distance} = 0.300 + 0.288 = 0.588\,\text{m}$

  • (Note: If a calculator is mistakenly set to degree mode instead of radian mode, the calculated displacement yields $x = 0.299\,\text{m}$).