SIMPLE HARMONIC MOTION

In  this Lesson we will consider the step‐to‐step explanation on Simple Harmonic Motion (SHM) that explains the key concepts, parameters, and formulas. Included are diagram descriptions (which you may later convert into drawn or digital images) and six worked examples styled in a way similar to JAMB exam questions.

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1. Introduction and Definition

Simple Harmonic Motion (SHM) is a type of periodic motion in which an object oscillates about an equilibrium position under a restoring force that is directly proportional to its displacement and acts in the opposite direction. In mathematical form, if an object at position $x$ is displaced from equilibrium, then the restoring force is given by:

$$F = -kx$$

where

• $k$ is the force (or spring) constant, and
• the minus sign indicates that the force acts opposite to the displacement.

Examples of SHM include:

• A mass attached to a spring (mass–spring system)
• A simple pendulum (for small angles)

(See also citeturn0search0 and citeturn0search2)

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2. Key Parameters and Equations

A. Parameters

1. Amplitude ($A$)
   – The maximum displacement from the equilibrium position.

2. Angular Frequency ($\omega$)
   – For a mass–spring system:

   $$\omega = \sqrt{\frac{k}{m}}$$

   – For a simple pendulum (small-angle approximation):

   $$\omega = \sqrt{\frac{g}{l}}$$

   where:

   • $m$ = mass,
   • $l$ = length of the pendulum,
   • $g$ = acceleration due to gravity.

3. Period ($T$)
   – The time taken for one complete oscillation.

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

4. Frequency ($f$)
   – The number of oscillations per unit time:

   $$f = \frac{1}{T} = \frac{\omega}{2\pi}$$

5. Phase Constant ($\phi$)
   – Determines the starting position (or “phase”) of the oscillation at $t=0$.

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B. Equations of Motion

For an undamped SHM the general solution for the displacement $x(t)$ is written as:

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

Alternatively, it may be written as:

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

depending on the initial conditions.

From this, we derive:

• Velocity:

  $$v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t + \phi)$$

  (The maximum speed is $v_{\text{max}} = A\omega$ when $x=0$.)

• Acceleration:

  $$a(t) = \frac{d^2x}{dt^2} = -A\omega^2\cos(\omega t + \phi) = -\omega^2 x(t)$$

  (Notice the acceleration is always directed toward the equilibrium position.)

(For further derivations, see citeturn0search2 and the detailed treatment in citeturn0search16.)

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C. Energy in SHM

In a mass–spring system the energy alternates between kinetic energy (KE) and potential energy (PE):

• Potential Energy (Spring Energy): $$U = \frac{1}{2} k x^2$$
• Kinetic Energy: $$K = \frac{1}{2} m v^2 = \frac{1}{2} m (A\omega)^2\sin^2(\omega t + \phi)$$
• Total Energy (Conserved): $$E = \frac{1}{2} k A^2$$

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3. Diagram Descriptions

Diagram 1: Mass–Spring System

Imagine a horizontal frictionless table with a wall on the left. A spring is fixed at the wall; its free end is attached to a mass.

• Equilibrium position: The point where the spring is neither stretched nor compressed.
• Amplitude ($A$) is shown as the maximum distance the mass moves from equilibrium.
• Arrows indicate the restoring force pointing toward equilibrium.

Diagram 2: Sinusoidal Graph of SHM

A typical graph of $x(t) = A\cos(\omega t + \phi)$:

• The vertical axis represents displacement.
• The horizontal axis represents time.
• The peaks and troughs (at $+A$ and $-A$) mark the amplitude.
• A full cycle (period $T$) is the distance on the time axis between two successive peaks.
• A horizontal shift indicates the phase constant $\phi$.

(These ideas are also illustrated in online tutorials such as citeturn0search17.)

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4. Worked Examples (JAMB Exam Style)

Below are six worked examples that mirror the style of typical JAMB exam questions.

Example 1: Mass–Spring System Equation of Motion

Question:
A 0.5 kg mass is attached to a spring with a constant $k = 200$ N/m. If the mass is pulled 0.1 m from equilibrium and released from rest, find the equation of motion and the period of oscillation.

Solution:

• Amplitude: $A = 0.1$ m
• Angular frequency: $$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}} = \sqrt{400} = 20\, \text{rad/s}$$
• Period: $$T = \frac{2\pi}{\omega} = \frac{2\pi}{20} = 0.314\, \text{s}$$
• Since released from rest from the maximum displacement, $\phi = 0$ (if using cosine form).
• Equation of Motion: $$x(t) = 0.1\cos(20t)$$

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Example 2: Simple Pendulum

Question:
A simple pendulum of length 0.64 m is displaced slightly. Using the small-angle approximation, calculate the period of oscillation.

Solution:
For a simple pendulum,

$$T = 2\pi \sqrt{\frac{l}{g}}$$

Using $l = 0.64$ m and $g = 9.8$ m/s$^2$:

$$T = 2\pi \sqrt{\frac{0.64}{9.8}} \approx 2\pi \sqrt{0.0653} \approx 2\pi(0.2555) \approx 1.606\, \text{s}$$

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Example 3: Maximum Velocity and Acceleration

Question:
For the mass–spring system in Example 1, determine the maximum velocity and the maximum acceleration.

Solution:

• Maximum Velocity: $$v_{\text{max}} = A\omega = 0.1 \times 20 = 2\, \text{m/s}$$
• Maximum Acceleration: $$a_{\text{max}} = A\omega^2 = 0.1 \times (20)^2 = 0.1 \times 400 = 40\, \text{m/s}^2$$

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Example 4: Energy Calculation

Question:
A 0.5 kg mass attached to a spring with $k = 200$ N/m oscillates with amplitude 0.1 m. Calculate the total mechanical energy stored in the system.

Solution:
Total energy (all stored as potential energy at the extremes):

$$E = \frac{1}{2}kA^2 = \frac{1}{2} \times 200 \times (0.1)^2 = 100 \times 0.01 = 1\, \text{J}$$

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Example 5: Determining Amplitude and Phase

Question:
An oscillator’s displacement is given by:

$$x(t) = 0.05\cos(15t) - 0.0866\sin(15t)$$

Express this in the form $x(t) = A\cos(15t + \phi)$ and find $A$ and $\phi$.

Solution:
Write the given expression in the cosine with phase-shift form. Compare:

$$A\cos(15t + \phi) = A\cos\phi\cos(15t) - A\sin\phi\sin(15t)$$

Equate coefficients:

$$A\cos\phi = 0.05 \quad \text{and} \quad A\sin\phi = 0.0866$$

Find amplitude:

$$A = \sqrt{(0.05)^2 + (0.0866)^2} \approx \sqrt{0.0025 + 0.0075} = \sqrt{0.01} = 0.1\, \text{m}$$

Determine phase angle:

$$\tan\phi = \frac{0.0866}{0.05} = 1.732 \quad \Longrightarrow \quad \phi \approx 60^\circ \text{ or } 1.047\, \text{rad}$$

Thus,

$$x(t) = 0.1\cos(15t + 1.047)$$

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Example 6: Frequency from a Given Equation

Question:
The displacement of a particle executing SHM is given by:

$$x(t) = 0.2\sin(10t - \frac{\pi}{6})$$

Find the frequency of the oscillation and the maximum acceleration.

Solution:

• Angular frequency: $\omega = 10\, \text{rad/s}$
• Frequency: $$f = \frac{\omega}{2\pi} = \frac{10}{2\pi} \approx 1.59\, \text{Hz}$$
• Maximum Acceleration: $$a_{\text{max}} = A\omega^2 = 0.2 \times 10^2 = 0.2 \times 100 = 20\, \text{m/s}^2$$

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5. Summary

In summary, SHM is characterized by sinusoidal motion described by the displacement equation

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

with velocity and acceleration given by its derivatives. The period $T$, frequency $f$, and energy of the system depend on parameters such as mass, spring constant, length (for a pendulum), and amplitude. These relationships are used extensively in solving problems in physics—including exam-style questions like those in JAMB.

(For more detailed derivations and additional examples, see citeturn0search0, citeturn0search2, and citeturn0search16.)