Simple Harmonic Motion

Simple harmonic motion can be understood mathematically by comparing it to uniform circular motion.

Reference Circle Model

• Imagine an object moving in a circle of radius $A$ at constant angular velocity $\omega$.
• Viewed from above, the object moves in a circle.
• Viewed from the side (edge of the table), the projection of its motion on the x-axis appears as back-and-forth oscillation—simple harmonic motion.

From circular motion:

$x = A \cos(\theta)$

If $\theta = \omega t$, then:

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

Where:

• $A$ is the amplitude (maximum displacement),
• $\omega$ is the angular velocity in radians per second,
• $t$ is time.

Since $\omega = 2\pi f$, where $f$ is the frequency in hertz, we can also write:

$x(t) = A \cos(2\pi f t)$

This equation describes how the position changes over time in SHM.

Visual Representation

If a pen is attached to an oscillating mass and a sheet of paper is moved steadily beneath it, the resulting trace is a sinusoidal wave—a physical representation of SHM.

Example: Identifying SHM Forces

Which of the following forces can cause SHM?

1. $F = -0.5x$
2. $F = -2.5x$
3. $F = 9.8x$
4. $F = -5x$

Correct Answers: (b) and (d)

These forces follow the form $F = -kx$, indicating SHM because:

• The force is proportional to displacement.
• The negative sign shows it is a restoring force, directed toward the equilibrium.

Options (a) and (c) do not produce SHM:

• (a) lacks proper proportionality for SHM.
• (c) has a force in the same direction as displacement, not opposite.

To explore further sinusoidal characteristics of simple harmonic motion, Figure 3.17 will be used to establish its displacement, period, velocity, and acceleration. This figure is a combination of ideas presented in Figure 3.15 and 3.16.

Displacement

This is the linear distance of the particle from the equilibrium position of the motion, i.e., distance ON (Figure 3.17). It is given by:

$T = \frac{2\pi r}{v}$

Since $v = \omega r$, therefore:

$T = \frac{2\pi}{\omega}$

Linear Velocity

Velocity is the rate of change of displacement, that is $v = \frac{dy}{dt}$, but:

$y = A \sin(\omega t)$

Therefore:

$\frac{dy}{dt} = A\omega \cos(\omega t)$

Thus:

$v = A\omega \cos(\omega t) \quad (3.25)$

Equation Derivation

Alternatively, squaring both sides of equation (3.24) and (3.25) gives:

$y^2 = A^2 \sin^2(\omega t)$

and

$v^2 = A^2 \omega^2 \cos^2(\omega t) \quad (3.27)$

Adding equations (3.26) and (3.27) and solving for $v$ gives:

$v = \pm \omega \sqrt{A^2 - y^2} \quad (3.28)$

Period

The period $T$ is the time taken by an oscillating object to complete one oscillation or cycle:

$T = \frac{2\pi}{\omega}$

Example 3.9

A particle executing simple harmonic motion has a period of 4 seconds and an amplitude of 0.02 m. Find:

(a) Maximum velocity:

Maximum velocity is obtained when $y = 0$. Hence:

$v_{\text{max}} = \omega A = \frac{2\pi}{T} \times A$

$v_{\text{max}} = \frac{2\pi}{4\,\text{s}} \times 0.02\,\text{m} = 3.14 \times 10^{-2} \, \text{ms}^{-1}$

(b) The velocity of the particle when $y = \frac{1}{2}A$:

$v = \frac{2\pi}{T} \times A \sqrt{1 - \left(\frac{1}{2}\right)^2} = 2.72 \times 10^{-2} \, \text{ms}^{-1}$

Acceleration

Acceleration is defined as the time rate of change of velocity, i.e., $a = \frac{dv}{dt}$. Therefore:

$a = -\omega^2 A \sin(\omega t) \quad (3.29)$

Example 3.10

Calculate the period for a particle executing simple harmonic motion with an acceleration of $16 \, \text{cm/s}^2$ at a distance of 4 cm from the equilibrium position.

Solution:

$a = -\omega^2 y$

$16\,\text{cm/s}^2 = \omega^2 \times 4\,\text{cm}$

Thus, the period is:

$T = 3.14\,\text{s}$

Problem Statement:

The velocity of a particle executing simple harmonic motion is $16 \, \text{cm/s}$ at distances of 8 cm and 12 cm from the mean position. Determine the amplitude of the motion.

Solution

In simple harmonic motion (SHM), the velocity of a particle is given by the relation:

$v = \omega \sqrt{A^2 - x^2}$

Where:

• $v$ is the velocity at a given position,
• $\omega$ is the angular frequency,
• $A$ is the amplitude,
• $x$ is the displacement from the mean position.

We are given that the velocity is $16 \, \text{cm/s}$ when the particle is at distances of 8 cm and 12 cm from the mean position. We can use this information to form two equations:

1. For $x = 8 \, \text{cm}$

$16 = \omega \sqrt{A^2 - 8^2}$

$16 = \omega \sqrt{A^2 - 64} \quad \text{(Equation 1)}$

2. For $x = 12 \, \text{cm}$

$16 = \omega \sqrt{A^2 - 12^2}$

$16 = \omega \sqrt{A^2 - 144} \quad \text{(Equation 2)}$

Step 1: Solve for $\omega$ in both equations

From Equation 1:

$\omega = \frac{16}{\sqrt{A^2 - 64}}$

Substitute this into Equation 2:

$16 = \frac{16}{\sqrt{A^2 - 64}} \sqrt{A^2 - 144}$

Step 2: Simplify the equation

Divide both sides by 16:

$1 = \frac{\sqrt{A^2 - 144}}{\sqrt{A^2 - 64}}$

Square both sides:

$1 = \frac{A^2 - 144}{A^2 - 64}$

Step 3: Solve for $A^2$

Multiply both sides by $A^2 - 64$:

$A^2 - 144 = A^2 - 64$

Simplifying:

$-144 = -64$

This leads to a contradiction, indicating there might be a mistake in our initial setup or assumptions.