Simple Harmonic Motion Notes

SHM and UCM

There is a deep connection between Simple Harmonic Motion (SHM) and Uniform Circular Motion (UCM).
Simple Harmonic Motion can be thought of as a one-dimensional projection of Uniform Circular Motion.
All the ideas we learned for UCM, can be applied to SHM…we don't have to reinvent them.

UCM and Period

Uniform Circular Motion is when an object moves in a circle subject to a force that is pointed towards the center of the circular motion.
The object moves with a constant speed, but a changing velocity as its direction continually changes.
The time it takes for an object to complete one trip around a circular path is called its period.
- The symbol for period is "T."
- Periods are measured in units of time; we will usually use seconds (s).
Often we are given the time $(t)$ it takes for an object to make a number of trips $(n)$ around a circular path. In that case,
$T = \frac{t}{n}$

Frequency

The number of revolutions that an object completes in a given amount of time is called the frequency of its motion.
- The symbol for frequency is "f".
Periods are measured in units of revolutions per unit time; we will usually use 1/seconds $(s^{-1})$ .
- Another name for $s^{-1}$ is Hertz (Hz).
- Frequency can also be measured in revolutions per minute (rpm), etc.
Often we are given the time $(t)$ it takes for an object to make a number of revolutions $(n)$ . In that case,
$f = \frac{n}{t}$

Period and Frequency

Since $T = \frac{t}{n}$ and $f = \frac{n}{t}$ then
$T = \frac{1}{f}$ and $f = \frac{1}{T}$

Velocity

Recall from Uniform Circular Motion:
$v = \frac{2\pi r}{T}$
$v = 2\pi rf$
The magnitude of the velocity is constant.

SHM and Circular Motion

In UCM, an object completes one circle, or cycle, in every $T$ seconds.
- That means it returns to its starting position after $T$ seconds.
In Simple Harmonic Motion, the object does not go in a circle, but it also returns to its starting position in $T$ seconds.
Any motion that repeats over and over again, always returning to the same position is called "periodic."

SHM and Circular Motion

The green ball moved in UCM - circular motion with a constant speed.
The red block moved in SHM - up and down, with a constantly changing speed.
But, they were always at the same height!
The ball was moving in two dimensions and the block was moving in one dimension.
Simple Harmonic Motion can be thought of as a one-dimensional projection of Uniform Circular Motion.

Simple Harmonic Motion

There are two conditions for a system to be undergoing simple harmonic motion:
- The object that is oscillating must be subject to a nonconstant force, proportional to and opposite its displacement from an equilibrium position.
  - The force is a restorative force and is explained by Hooke's Law:
    $F = -kx$
- The object is described by two forms of energy, such as kinetic energy and elastic potential energy.
  - Energy is continually being shifted between the two types, but the total energy is constant.

Example 1: Period and Frequency

A mass-spring system makes 40 complete oscillations in 8.0 seconds. What is the period and frequency of the oscillations?
Given: $n = 40$ , $t = 8.0 s$

Example 2: Period and Frequency

A simple pendulum oscillates with a period of 2.0 s. What is the frequency?
Given: $T = 2.0 s$

Example 3: Period and Frequency

A simple pendulum oscillates with a frequency of 25.0 Hz. What is the period?
Given: $f = 25.0 Hz$

Mass - Spring System Motion

A mass - spring system consists of a mass attached to a spring which is attached to a support.
Illustrated is a horizontal mass - spring system, where the spring is first compressed or stretched, and it oscillates between the maximum compressed/stretched point.
Energy is constantly being transferred between kinetic energy and elastic potential energy.

Mass - Spring System

A cycle is a full to-and-fro motion where the mass returns to its starting point (same as one trip around a circle in UCM).
The time it takes to complete one cycle is the period.
Frequency is the number of cycles completed per second.

Mass - Spring System

Displacement $(x)$ is measured from the equilibrium point $(x = 0)$ .
- The spring is neither compressed nor stretched.
- Hooke's Law states that the force at the equilibrium point is zero.
Amplitude $(A)$ is the maximum displacement (corresponds to the radius in UCM).

Example 1: Amplitude and Period

The period of a mass-spring system is 2.0 s and the amplitude of its motion is 0.40 m. How far does the mass travel in 4.0 s?
Stretch the spring until the mass is at $x(t = 0) = +A$ , and release it.
In one period, $T = 2.0 s$ , the mass will travel to $x = -A$ and then return to $x = +A$ , covering a distance of $4A = 4 (0.40m) = 1.6 m$ .
In 4.0 s, the mass oscillates for two periods.
The distance covered is $2 (1.6 m) = 3.2 m$ .
Given: $T = 2.0 s$ , $A = 0.40 m$ , $t = 4.0 s$

Example 1: Spring Force

What is the spring force in a spring with the spring constant of $k = 100.0 N/m$ that is stretched by 10.0 cm?
Given: $k = 100 N/m$ , $x = 10.0 cm = 0.10 m$
Use Hooke's Law
$F = -kx$
$F = -(100 N/m)(0.10 m)$
$F = -10 N$
Substitute in givens
The negative sign indicates that the force acts to move the mass back towards the equilibrium point.

Simple Harmonic Motion

The restoring force is described by: $F = -kx$
- The minus sign indicates that it is a restoring force – it is directed to restore the mass to its equilibrium position.
- $k$ is the spring constant.
The force is not constant, so the acceleration is not constant either.

Simple Harmonic Motion

The maximum force exerted on the mass is when the spring is most stretched or compressed ( $x = -A$ or $+A$ ):
$F = -kA$ (when $x = -A$ or $+A$ )
The minimum force exerted on the mass is when the spring is not stretched at all ( $x = 0$ )
$F = 0$ (when $x = 0$ )

Simple Harmonic Motion

When the spring is all the way compressed:
- The displacement is at the negative amplitude.
- The force of the spring is in the positive direction.
- The acceleration is in the positive direction.
- The velocity is zero.

Simple Harmonic Motion

When the spring is at equilibrium and heading in the positive direction:
- The displacement is zero.
- The force of the spring is zero.
- The acceleration is zero.
- The velocity is positive and at a maximum.

Simple Harmonic Motion

When the spring is all the way stretched:
- The displacement is at the positive amplitude.
- The force of the spring is in the negative direction.
- The acceleration is in the negative direction.
- The velocity is zero.

Simple Harmonic Motion

When the spring is at equilibrium and heading in the negative direction:
- The displacement is zero.
- The force of the spring is zero.
- The acceleration is zero.
- The velocity is negative and at a maximum.

Mass-Spring System

If the spring is hung vertically, the only change is in the equilibrium position, which is at the point where the spring force equals the gravitational force.
The effect of gravity is cancelled out by changing to this new equilibrium position.

Example 1: Vertical Spring

A spring stretches 5.0 cm when a 1.00 kg mass is suspended from it. What is the spring constant?
Given: $m = 1.00 kg$ , $x = 5.0 cm = 0.050 m$
Draw a FBD showing that the spring force acts opposite the stretching force due to gravity.
Write Newton's second law.
The mass is at rest at the equilibrium position
$F{spring} - Fg = 0$

Example 1: Vertical Spring

A spring stretches 5.0 cm when a 1.00 kg mass is suspended from it. What is the spring constant?
Given: $m = 1.00 kg$ , $x = 5.0 cm = 0.050 m$
Use Hooke's Law
$F_{spring} = kx$
Solve for k
$k = F_{spring}/x$
Substitute in givens, noting that x is negative as the displacement is in the down direction
$k = (1.00 kg * 9.8 m/s^2)/0.05 m$
$k = 196 N/m$

Mass - Spring System Energy

Any vibrating system where the restoring force is proportional to the negative of the displacement is in simple harmonic motion (SHM), and is often called a simple harmonic oscillator.
Also, SHM requires that a system has two forms of energy and a method that allows the energy to go back and forth between those forms.

Energy in the Mass-Spring System

There are two types of energy in a mass-spring system.
- The energy stored in the spring because it is stretched or compressed:
  $EPE = \frac{1}{2} kx^2$
- AND The kinetic energy of the mass:
  $KE = \frac{1}{2} mv^2$

Energy in the Mass-Spring System

The total mechanical energy is constant.
At any moment, the total mechanical energy of the system is constant and comprised of those two forms.
$E = KE + EPE$
$E = \frac{1}{2} mv^2 + \frac{1}{2} kx^2$

Energy in the Mass-Spring System

When the mass is at the limits of its motion ( $x = A$ or $x = -A$ ), the energy is all potential:
$E = \frac{1}{2} kA^2$
When the mass is at equilibrium ( $x = 0$ ) the spring is not stretched and all the energy is kinetic:
$E = \frac{1}{2} mv_{max}^2$
The total energy is constant:
$E = \frac{1}{2} mv^2 + \frac{1}{2} kx^2$

Energy in the Mass-Spring System

When the spring is all the way compressed….
- EPE is at a maximum.
- KE is zero.
- Total energy is constant.
  $v = 0$
  E = {\frac{1}{2}kA^2

Energy in the Mass-Spring System

When the spring is passing through the equilibrium….
- EPE is zero.
- KE is at a maximum.
- Total energy is constant.
  $E = \frac{1}{2}mv_{max}^2$

Energy in the Mass-Spring System

When the spring is all the way stretched….
- EPE is at a maximum.
- KE is zero.
- Total energy is constant.
  $v = 0$
  $E = \frac{1}{2}kA^2$

Problem Solving using Energy

Since the energy is constant, and the work done on the system is zero, you can always find the velocity of the mass at any location by using:
$E0 = Ef$
The most general equation becomes:
$\frac{1}{2} mv0^2 + \frac{1}{2} kx0^2 = \frac{1}{2} mvf^2 + \frac{1}{2} kxf^2$
This is simplified by being given the energy at some point where it is all EPE ( $x = A$ or $-A$ ) or when it is all KE ( $x = 0$ ).

Period and Frequency of Mass - Spring System

We will build on the relationship between UCM and SHM to calculate the period and frequency of a mass - spring system.
The concept that SHM is a one dimensional projection of the two dimensional UCM is the basis for these calculations.

SHM relationship to UCM

Picture a particle moving in a circle of radius r, and its projection on the x axis below the circle.
- Displacement is measured from the equilibrium point
- Amplitude is the maximum displacement (equivalent to the radius, r, in UCM).

SHM relationship to UCM

A cycle is a full back and forth motion (the same as one trip around the circle in UCM).
The Period is the time required to complete one cycle (the same as period in UCM).
The Frequency is the number of cycles completed per second (the same as frequency in UCM).

The Period and Frequency of a Mass-Spring System

We now use the period and frequency of a particle moving in a circle to find the period and frequency of a mass moving in SHM.
The particle is moving in a circle of radius r, with constant speed, v.
The mass is oscillating on a spring with a maximum amplitude of A.
The time it takes for a mass to move back and forth along the x axis is the time it takes for a particle to make one complete revolution of the circle.
The period of each is the same.

The Period and Frequency of a Mass-Spring System

The velocity of the mass on the x axis is a maximum at x = 0. At this point, $KE{max} = E{total} = \frac{1}{2} kA^2$ .
The particle's speed anywhere on the circle is the same, $v = \frac{2\pi r}{T}$ , and at the top of the circle is equal to the speed of the mass on the x axis.
This is only true at the top and bottom of the circle.
At all other points, the speed of the mass is less than the speed of the particle.

The Period and Frequency of a Mass-Spring System

When the particle on the circle is at the points shown below, its velocity is in the ±y direction.
There is zero projection of the particle's velocity on the line where the mass is oscillating.
Thus, the velocity of the mass is zero when $x = ±A$ ; just like we learned earlier.
Back to the calculation of period and frequency……

The Period and Frequency of a Mass-Spring System

Substitute in the particle on a circle speed
$v_{max} = \sqrt{\frac{k}{m}} A = \frac{2\pi A}{T}$
$A = r$ substitution
Solving for Period
$T = 2\pi \sqrt{\frac{m}{k}}$
Solving for Frequency
$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Example 1: Energy of a Mass- Spring System

A mass of 1.4 kg is attached to a horizontal spring with a spring constant of 75.0 N/m. The spring is stretched from equilibrium position by 5.0 cm and released.
What is the maximum elastic potential energy?
Given: $m = 1.4 kg$ , $k = 75.0 N/m$ , $x_{max} = A = 5.0 cm = 0.050 m$
$EPE = \frac{1}{2} kx^2$
$EPE = \frac{1}{2} (75.0 N/m) (0.050 m)^2$
$EPE = 0.094 J$
EPE Equation ( $x_{max} = A$ )
Substitute in givens

Example 2: Energy of a Mass- Spring System

A mass of 1.4 kg is attached to a horizontal spring with a spring constant of 75.0 N/m. The spring is stretched from equilibrium position by 5.0 cm and released.
What is the maximum kinetic energy?
Given: $m = 1.4 kg$ , $k = 75.0 N/m$ , $x_{max} = A = 5.0 cm = 0.050 m$
The total mechanical energy in SHM (mass-spring system) is conserved.
The maximum kinetic energy is equal to the maximum potential energy that was found in the previous question:
$KE_{max} = 0.094 J$

Example 3: Energy of a Mass- Spring System

A mass of 1.4 kg is attached to a horizontal spring with a spring constant of 75.0 N/m. The spring is stretched from equilibrium position by 5.0 cm and released.
What is the maximum speed of the mass?
Given: $m = 1.4 kg$ , $k = 75.0 N/m$ , $x_{max} = A = 5.0 cm = 0.050 m$
We will use the KE equation to find the maximum speed:
$KE = \frac{1}{2} mv^2$
$0.094 J = \frac{1}{2} (1.4 kg) v{max}^2$ $v{max} = \sqrt{\frac{2 * 0.094 J}{1.4 kg}}$
$v_{max} = 0.37 m/s$
Solve for $v_{max}$ .

Example 4: Period of a Mass- Spring System

What is the period of a mass-spring oscillation system with a spring constant of 120.0 N/m and mass of 0.5 kg?
Given: $m = 0.50 kg$ , $k = 120.0 N/m$
Use the period equation
$T = 2\pi \sqrt{\frac{m}{k}}$
$T = 2\pi \sqrt{\frac{0.50 kg}{120.0 N/m}}$
$T = 0.41 s$
Substitute in givens

Example 5: Frequency of a Mass- Spring System

What is the frequency of a mass-spring oscillation system with a spring constant of 125.0 N/m and mass of 2.00 kg?
Given: $m = 2.00 kg$ , $k = 125.0 N/m$
Use the frequency equation
$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$
$f = \frac{1}{2\pi} \sqrt{\frac{125.0 N/m}{2.00 kg}}$
$f = 1.26 Hz$
Substitute in givens

Simple Pendulum

A simple pendulum consists of a mass at the end of a lightweight cord.
We assume that the cord does not stretch, and that its mass is negligible.

The Simple Pendulum

The restoring force is proportional to $\sin \theta$ and not to $\theta$ itself.
We don't really need to worry about this because for small angles (less than 15 degrees or so), $\sin \theta ≈ \theta$ and $x = L\theta$ .
So we can replace $\sin \theta$ with $x/L$ .
In order to be in SHM, the restoring force must be proportional to the negative of the displacement. Here we have:
$F = -mg \sin \theta$

The Simple Pendulum

has the form of
$F = -kx$
if
$k = mg/L$
But we learned before that
$T = 2\pi \sqrt{\frac{m}{k}}$
Substituting for k
$T = 2\pi \sqrt{\frac{m}{mg/L}}$
$T = 2\pi \sqrt{\frac{L}{g}}$
Notice the "m" canceled out, the mass doesn't matter.

Example 1: Period of a Simple Pendulum

A simple pendulum with a length of 2.00 m oscillates on the earth’s surface. What is the period of oscillations?
Given: $L = 2.00 m$ , $g = 9.8 m/s^2$
Use the period equation
$T = 2\pi \sqrt{\frac{L}{g}}$
$T = 2\pi \sqrt{\frac{2.00 m}{9.8 m/s^2}}$
$T = 2.84 s$
Substitute in givens

Example 1: Frequency of a Simple Pendulum

A simple pendulum with a length of 2.60 m oscillates on the Earth’s surface. What is the frequency of oscillations?
Given: L = 2.60 m, g = 9.8 m/s^2
$f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$
Use the frequency equation
Substitute in givens

Energy in the Pendulum

The two types of energy in a pendulum are:
- Gravitational Potential Energy
  $U_g = mgh$
- AND The kinetic energy of the mass:
  $KE = \frac{1}{2} mv^2$

Energy in the Pendulum

The total mechanical energy is constant.
At any moment in time the total energy of the system is constant and comprised of those two forms.

Example 1: Energy of a Simple Pendulum

A mass of 0.50 kg oscillates on a simple pendulum with a length of 1.50 m that reaches a maximum height of 0.080 m when it is in SHM.
What is the maximum gravitational potential energy?
The gravitational potential energy is a maximum when the pendulum reaches its maximum height.
Given: $m = 0.50 kg$ , $g = 9.8 m/s^2$ , $h{max} = 0.080 m$ , $L = 1.50 m$ $Ug = mgh{max}$ $Ug = (0.50 kg)(9.8 m/s^2)(0.080 m)$
$U_g = 0.39 J$
Substitute in givens

Example 2: Energy of a Simple Pendulum

A mass of 0.50 kg oscillates on a simple pendulum with a length of 1.50 m that reaches a maximum height of 0.080 m when it is in SHM.
What is the maximum kinetic energy?
The total mechanical energy (TME) in SHM is conserved.
The maximum kinetic energy is equal to the maximum gravitational potential energy that was found in the previous example problem.
Given: $m = 0.50 kg$ , $g = 9.8 m/s^2$ , $h{max} = 0.080 m$ , $L = 1.50 m$ $KE{max} = 0.39 J$

Example 3: Energy of a Simple Pendulum

A mass of 0.50 kg oscillates on a simple pendulum with a length of 1.50 m that reaches a maximum height of 0.080 m when it is in SHM.
What is the maximum speed of the mass?
Use the KE equation to find the maximum speed
$KE = \frac{1}{2} mv^2$
$0.39 J= \frac{1}{2} (0.50 kg) v_{max}^2$
Given: $m = 0.50 kg$ , $g = 9.8 m/s^2$ , $h{max} = 0.080 m$ , $L = 1.50 m$ $v{max} = \sqrt{\frac{2 * 0.39 J}{0.50 kg}}$
$v_{max} = 1.3 m/s$
Solve for v max.
Use KEmax from previous slide

Sinusoidal Nature of SHM

The position as a function of time for an object in simple harmonic motion can be derived from the equation:
$x = A \cos \theta$
Where A is the amplitude of oscillations.
Take note that it doesn't really matter if you are using sine or cosine since that only depends on when you start your clock.
For our purposes lets assume that you are looking at the motion of a mass-spring system and that you start the clock when the mass is at the positive amplitude.

Position as a function of time

Now we can derive the equation for position as a function of time.
Since we can replace $\theta$ with $\omega t$ .
And we can also replace $\omega$ with $2\pi f$ or $\frac{2\pi}{T}$ .
$x = A \cos (\omega t)$
$x = A \cos (2\pi f t)$
$x = A \cos (\frac{2\pi}{T} t)$
Where A is amplitude, T is period, and t is time.

Position as a function of time

We can also derive the equation for velocity as a function of time.
Since $v = \omega r$ can replace $v_0$ with $\omega A$ as well as $\theta$ with $\omega t$ .
And again we can also replace $\omega$ with $2\pi f$ or ${2\pi}{T}$ .
$v = -v_0 \sin (\omega t)$
$v = -\omega A \sin (\omega t)$
Where A is amplitude, T is period, and t is time.

Velocity as a function of time

We can also derive the equation for acceleration as a function of time.
Since $a = r\omega^2$ , we can replace $a0$ with $A\omega^2$ as well as $\theta$ with $\omega t$ . $a = - a0 \cos (\omega t)$
$a = -\omega^2 A \cos (\omega t)$
And again we can also replace $\omega$ with $2\pi f$ or ${2\pi}{T}$ .
Where A is amplitude, T is period, and t is time.

The Sinusoidal Nature of SHM

Now you can see all of the graphs together.
Take note that when the position is at the positive amplitude, the acceleration is negative and the velocity is zero.
Or when the velocity is at a maximum both the position and acceleration are zero.