Periodic Motion

Periodic motion is a type of motion that repeats itself at regular time intervals. Examples include the swinging of a pendulum, the motion of a mass on a spring, and the orbit of planets around the sun.

The key characteristics of periodic motion are amplitude, period, and frequency. Amplitude is the maximum displacement from the equilibrium position. Period ($T$) is the time taken for one complete cycle of the motion. Frequency ($f$) is the number of cycles per unit time.

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

Where $T$ is the period and $f$ is the frequency. The units of period are seconds ($s$), and the units of frequency are in hertz ($Hz$), where $1 Hz = \frac {1} {s}$.

Example

What is the freqency of a heart beating every $0.75$ seconds?

Answer

$T = \frac {1} {f} \ \Rightarrow \ f = \frac {1} {T} = \frac {1} {(0.75)} = 1.3 Hz$

$\therefore$ The heart beats $1.3$ times a second.

Springs

Springs are elastic objects that, when a deforming force is applied, a reaction to get it back to its original state occurs.

They have an oscillating behaviour that can be described by simple harmonic motion. Its back and forth movement resembles the graph of $\sin$ graph:

Hooke's Law

Hooke's Law describes the behavior of springs and other elastic materials. It states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.

$F_s = -kx$

Where $F_s$ is the restoring force, $k$ is the spring constant (a measure of the stiffness of the spring), and $x$ is the displacement from the equilibrium position. The negative sign indicates that the force is in the opposite direction of the displacement.

Here what this relation looks like on a graph:

Example

A spring with a spring constant of $k = 7.50 \frac {N} {m}$ is stretched $4.30m$. What is the resisting force it applies due to the deformation?

Answer

$\vec {F}_s = -kx = -(7.50 \frac {N} {m})(4.30m) = -32.3N$

$\therefore$ The resistive force is $32.3N$.

Simple Harmonic Motion

Energy of Simple Harmonic Motion

In simple harmonic motion (SHM), the energy of the system oscillates between kinetic energy ($E_k$) and potential energy ($E_p$). At maximum displacement (amplitude), the energy is entirely potential, and at the equilibrium position, the energy is entirely kinetic.

The total mechanical energy in SHM is constant and is given by:

$E_{ \text {total} } = E_k + E_p$

Where $E_{ \text {total} }$ is the total mechanical energy in the system.

For a mass-spring system, the kinetic potential energy formulas are:

$E_k = \frac {1} {2} mv^2 \quad \text {and} \quad E_p = \frac {1} {2} F_s x = \frac {1} {2} kx^2$

Period and Frequency of Simple Harmonic Motion

The period and frequency of a simple harmonic oscillator depends on the properties of the system. For a mass-spring system, the period and frequency are given by:

$T = 2 \pi \sqrt{ \frac {m} {k} } \quad \text {and} \quad f = \frac {1} {T} = \frac {1} { 2 \pi } \sqrt { \frac {k} {m} }$

Where $m$ is the mass and $k$ is the spring constant.

Example

If the maximum distance of a spring with a spring constant $52.5 \frac {N} {m}$ is $0.500m$, what is the maximum speed at which a $2.00kg$ mass on its end can move.

Answer

First, let's determine the maximum energy that can be stored in the spring:

$E_{ \text {max} } = \frac {1} {2} k x_{ \text {max} }^2 = \frac {1} {2} (52.5 \frac {N} {m})(0.500m) = 13.1J$

Now we need to determine the maximum speed by using the kinetic energy of a spring equation:

$E_{ \text {max} } = \frac {1} {2} mv_{ \text {max} }^2 \ \Rightarrow \ v_{ \text {max} } = \sqrt { \frac { E_{ \text {max} } } { m } } = \sqrt { \frac { (13.1J) } { (2.00kg) } } = 2.56 \frac {m} {s}$

$\therefore$ The maximum speed the mass will experience is $2.56 \frac {m} {s}$.

Example

A $0.64kg$ mass is on a spring with a spring constant of $k = 30 \frac {N} {m}$. What is the period of the vibration of the spring?

Answer

$T = 2 \pi \sqrt { \frac {m} {k} } = 2 \pi \sqrt { \frac { (0.64kg) } { (30 \frac {N} {m}) } } = 0.918s$

Pendulums

A pendulum is a weight suspended from a pivot so that it can swing freely. The restoring force that brings the pendulum back to its equilibrium position is gravity.

For small angles that are less than $15^{\circ}$, the period and frequency of the pendulum can be approximated by the following equation:

$T = 2 \pi \sqrt { \frac {L} {g} } \quad \text {and} \quad f = \frac {1} {T} = \frac {1} { 2 \pi } \sqrt { \frac {g} {L} }$

Where $L$ is the length of the pendulum, and $g$ is the acceleration due to gravity.

Example

What is the period of a pendulum of length $0.24m$?

Answer

$T = 2 \pi \sqrt { \frac {L} {g} } = 2 \pi \sqrt { \frac { (0.24m) } { (9.81 \frac {m} {s^2} }) } = 0.983s$