Simple harmonic motion (SHM) is a back-and-forth movement around an equilibrium point. It's characterized by displacement, velocity, and acceleration, which can be described using equations and graphs.

Understanding SHM involves analyzing its key components: amplitude, frequency, and period. These elements help predict an object's position, speed, and direction of motion at any given time during its oscillation.

Displacement, velocity, and acceleration in SHM

When an object undergoes SHM, its displacement, velocity, and acceleration follow predictable patterns that can be described mathematically and graphically.

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Equations for Displacement in SHM

The position of an object in SHM can be described using sine or cosine functions, depending on the initial conditions.

$x = A \cos(2 \pi ft)$ or $x = A \sin(2 \pi ft)$

• $A$ represents the amplitude, the maximum displacement from equilibrium
• $f$ is the frequency, the number of oscillations per second (measured in Hz)
• $t$ is time elapsed since the motion started

The choice between sine and cosine depends on the object's initial position:

• Use cosine if the object starts at maximum displacement
• Use sine if the object starts at equilibrium

Understanding the Relationship Between Position, Velocity, and Acceleration

In SHM, the displacement, velocity, and acceleration are interconnected but reach their maximum and minimum values at different times.

• At the equilibrium position, displacement and acceleration are zero, while velocity is at its maximum
• At the maximum displacement (amplitude), velocity is zero and acceleration is at its maximum in the opposite direction of displacement

This relationship creates a continuous cycle where energy transforms between potential and kinetic forms without loss (in an ideal system).

• When the object is at maximum displacement, it has maximum potential energy and zero kinetic energy
• When the object passes through equilibrium, it has maximum kinetic energy and zero potential energy

Amplitude and Period in SHM

The period of oscillation is independent of the amplitude in SHM, which is one of its defining characteristics.

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

• The amplitude ($A$) can change without affecting the period ($T$)
• For a mass-spring system, the period depends only on the mass and spring constant: $T = 2\pi\sqrt{\frac{m}{k}}$
• For a simple pendulum, the period depends primarily on length (for small angles): $T = 2\pi\sqrt{\frac{L}{g}}$

This independence means that whether you pull a pendulum back a small amount or a large amount, it will take the same time to complete one full swing (assuming the angle remains small).

Graphical Analysis of SHM

The periodic nature of SHM creates distinctive graphical patterns that help visualize the motion.

Displacement-time graphs show a sinusoidal pattern where:

• The amplitude equals the maximum displacement from equilibrium
• The period equals the time for one complete oscillation
• The frequency equals the number of complete oscillations per second

Velocity is the rate of change of displacement, so:

• Velocity-time graphs are also sinusoidal but shifted by $\frac{1}{4}$ period relative to displacement
• Maximum velocity occurs at equilibrium (zero displacement)
• Velocity is zero at maximum displacement
• Velocity can be found by taking the derivative of displacement: $v = -A \omega \sin(\omega t)$ for cosine displacement or $v = A \omega \cos(\omega t)$ for sine displacement, where $\omega = 2\pi f$

Acceleration is the rate of change of velocity:

• Acceleration-time graphs are sinusoidal and shifted by $\frac{1}{2}$ period relative to displacement
• Maximum acceleration occurs at maximum displacement and always points toward equilibrium
• Acceleration equals zero when passing through equilibrium
• Acceleration can be found from: $a = -A \omega^2 \cos(\omega t)$ for cosine displacement or $a = -A \omega^2 \sin(\omega t)$ for sine displacement

Practice Problem 1: Spring-Mass System

Solution:

(a) The amplitude is the initial displacement: $A = 0.10$ m

(b) The period of oscillation: $T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.5 \text{ kg}}{20 \text{ N/m}}} = 2\pi\sqrt{0.025} = 0.99$ s

(c) Maximum velocity occurs at equilibrium: $v_{max} = A\omega = A \cdot \frac{2\pi}{T} = 0.10 \text{ m} \cdot \frac{2\pi}{0.99 \text{ s}} = 0.635$ m/s

(d) Maximum acceleration occurs at maximum displacement: $a_{max} = A\omega^2 = A \cdot \left(\frac{2\pi}{T}\right)^2 = 0.10 \text{ m} \cdot \left(\frac{2\pi}{0.99 \text{ s}}\right)^2 = 4.02$ m/s²

Practice Problem 2: Pendulum Motion

Solution:

(a) The period of a simple pendulum: $T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{0.8 \text{ m}}{9.8 \text{ m/s²}}} = 2\pi\sqrt{0.082} = 1.79$ s

(b) For small angles, the maximum displacement can be calculated as: $s = L\theta$ where $\theta$ is in radians $s = 0.8 \text{ m} \cdot (5° \cdot \frac{\pi}{180°}) = 0.8 \text{ m} \cdot 0.087 \text{ rad} = 0.070$ m

Maximum speed (at the lowest point): $v_{max} = \omega A = \frac{2\pi}{T} \cdot s = \frac{2\pi}{1.79 \text{ s}} \cdot 0.070 \text{ m} = 0.245$ m/s

(c) On the Moon, where $g_{moon} \approx \frac{g_{earth}}{6} \approx 1.63$ m/s²: $T_{moon} = 2\pi\sqrt{\frac{L}{g_{moon}}} = 2\pi\sqrt{\frac{0.8 \text{ m}}{1.63 \text{ m/s²}}} = 2\pi\sqrt{0.491} = 4.39$ s

This shows the period would increase by a factor of $\sqrt{6} \approx 2.45$ times on the Moon.

Practice Problem 3: Graphical Analysis

Solution:

(a) Since the particle starts at equilibrium (x = 0) at t = 0, we use the sine function: $x = A\sin(2\pi ft) = 0.05 \text{ m} \cdot \sin\left(2\pi \cdot \frac{1}{2} \text{ s}^{-1} \cdot t\right) = 0.05\sin(\pi t)$ meters

(b) Maximum displacement first occurs at t = T/4: $t = \frac{T}{4} = \frac{2 \text{ s}}{4} = 0.5$ seconds

(c) At t = 0.75 seconds:

Position: $x = 0.05\sin(\pi \cdot 0.75) = 0.05\sin(0.75\pi) = 0.05\sin(135°) = 0.05 \cdot 0.707 = 0.035$ meters

Velocity: $v = 0.05\pi\cos(\pi \cdot 0.75) = 0.05\pi\cos(135°) = 0.05\pi \cdot (-0.707) = -0.111$ m/s

Acceleration: $a = -0.05\pi^2\sin(\pi \cdot 0.75) = -0.05\pi^2\sin(135°) = -0.05\pi^2 \cdot 0.707 = -0.348$ m/s²