# Oscillations and Wave Identities
Waves have two modes - transverse waves, which oscillate perpendicular to the direction of the particle's movement, as well as longitudinal waves, which oscillate parallel to the direction of the wave.
%%all-inclusive diagram pls for both transverse and longitudinal! %%
**Transverse waves** include EM waves, which constitute light. Light is actually made up of different subclasses, ranging from gamma rays to radio light.
Below is a list of the objects or events that make or use each wavelength:
- **Gamma rays** --> supernovae, gamma ray bursts, pulsar beams
- **X-rays** --> CT scans, high-energy explosions (supernovae), x-ray binaries, brown dwarf x-ray flares, medical x-rays,
- **UV** --> bee UV light, fluorescence detectors
- **IR** --> JWST, TV remote
- **Radio** --> black holes, FRBs (fast radio bursts), quasars
**Longitudinal waves** include sound (acoustic) waves, which make up sound - they're just vibrations of molecules. This is why you can't have sound in space (for the most part)!
%%make a diagram comparing the two.%%
>[!Success]- Useful Identities
>There are a bunch of identities we commonly use when it comes to waves, a couple of them listed as:
>
>1. The speed of light, $3\times 10^8$ m/s
>2. The speed of sound in air, $330$ m/s
>3. The speed of sound in water, $1500$ m/s
>MORE TO BE ADDED!
## Frequency, Wave-speed, Wavelength
One of the most important wave equations is shown below. This is:
$ v = f\lambda$
Where:
- $v$ is the velocity of the wave,
- $f$ is the frequency of the wave, how many times it oscillates in a length of $2 \pi$
- $\lambda$ is the wavelength
Look, you're not getting past the GCSE if you don't have this down. It's simple, it's elegant, and it's just v = d/t transmuted into wave terms.
## More Transverse Waves
As a transverse wave is given by the definition where it oscillates perpendicular to the direction of the wave's movement, we're going to need a [[Sinusoidal Curves (Maths)|sinusoidal]] - where the $y$ axis defines the *transverse* displacement of the wave as a function of the time displacement $t$ and the change in the linear displacement $x$:
$y(x,t) = h(x,t)$
Since this is an oscillation the motion of a particle on the wave can be described with a sinusoidal - so depending on where the oscillation is a $\cos$ or $\sin$ operator can be used. However, let's imagine that all of the graphs have a y-intercept of 0, making it a sinusoidal. This gives us:
$y(x,t) = y_{m}\sin(kx-\omega t)$
Where $y_m$ is the amplitude of the sinusoidal that can be created. Some of you may be asking: "What's the bit inside the brackets?" Well, that's the *phase argument* of the wave - which defines where the particle is on its oscillation as it varies by $t$.
Letting the time be 0 allows us to rewrite the transverse wave equation as just $y(x,t) = y_{m}\sin(kx)$
The total displacement $y$ is going to be the same after a displacement of one wavelength - giving us an equation:
$y_{m}\sin(kx) = ym\sin(kx+k\lambda)$
With our knowledge of the sinusoidal wave as having a wavelength of $2\pi$, we get an expression for the value $k$ - which is also the *wave number density*:
$k = \frac{2\pi}{\lambda}$
This quantity $k$ has a unit $rad/m$. However, there's still one more unresolved component for us to find - the $\omega$ before the time $t$! At the point $x = 0$, the equation is:
$y(x,t) = y_{m}\sin(-\omega t) = -y_{m}\sin \omega t$
This $\omega$ is the [[Simple Harmonic and Pendulum Motion (Waves)#More on Angular Frequency|angular frequency]] of the wave. To link this with the period of the wave, note that after one period $T$ the transverse wave equation will reach the same transverse value again. This means we can create the following identity:
$-y_{m}\sin \omega t = -y_{m}\sin \omega (t+T) = -y_{m}\sin \omega t + \omega T$
This means that $\omega T$ must be equal to $2 \pi$, giving us another expression for the angular frequency:
$\omega = \frac{2\pi}{T}$
Since the angular frequency is also equal to $2\pi f$ (as we found previously in SHM) we can write the frequency in terms of the period:
$f = \frac{1}{T}$
Sometimes, if there's a vertical displacement we're also going to be adding an extra variable - the **phase constant**. Some of you may remember this from simple harmonic motion - this basically deals with cases where the y-intercept is not 0 and we want to keep using our $\sin$ operator!
>[!Abstract]- Deriving $v = f\lambda$ from the Transverse Wave Equation
>For our final act, let's use what we've learned here to derive the most fundamental equation in waves!
>
>As the wave moves, each point on it retains its nominal $y$ axis (transverse) displacement. As the phase $kx - \omega t$ gives it this transverse displacement, this value must remain constant as the shape of the sinusoidal does not change and we are measuring a horizontal shift in the sinusoidal as $x$. Since wavespeed is also given by the equation $\frac{dx}{dt}$, we can change the equation by [[Calculus Basics (Maths)#Differentiation|taking the derivative]] with respect to $dt$ to get:
>$k \frac{dx}{dt} - \omega = 0$
>$\frac{dx}{dt} = v = \frac{\omega}{k}$
>Time to plug in the identities we found earlier! Remember, $\omega = \frac{2\pi}{k}$ and $k = \frac{2\pi}{\lambda}$!
>$v = \frac{\lambda}{T} = f\lambda$
# Waves on Strings
## Wavespeeds of Strings
Ever seen a single 'wave pulse'? No? Well, you can imagine it as just a little bump in a really, really long rope - but if that bump were to be moving.
Before we begin, let's learn some new terms! The **Linear Density** of a wave is given by the mass of the string divided by its length - much like the [[Newton's Laws and SUVAT#Density|density]] we all know and love. This quantity is given by the greek letter $\mu$ - as seen:
$\mu = \frac{m}{L}$
There are also a couple boundary conditions we need to keep in mind when doing this, namely:
1. **The pulse must be symmetrical**.
2. **The pulse must be in an inertial reference frame**, meaning that to us, it shouldn't be moving!
Now that's out of the way, let's model the little "bump" as it's own circle:
![[Pasted image 20230902121355.png]]
Now let's consider a small wave element towards the top of the circle, where the string touches the circle. If we measure the angular displacement from the midpoint in both directions (symmetrically), we can define a small angle $\theta$ as being the displacement in direction, giving us a total angle of $2\theta$.
This angle directly translates to a small arclength $dx$. As the rope is taut, there is a tension force $F$ acting on both sides of the rope. We can draw these new terms onto the diagram as shown:
![[Pasted image 20230902124433.png]]
The way the two tension forces are arranged means that although their horizontal components cancel out their radial (vertical) components do not. As a result, the rope experiences centripetal acceleration. We can add up the vertical components of the tension forces to get:
$F = 2\tau \sin \theta = \tau(2\theta) = \tau\frac{ dx}{R}$
Where $R$ is the radius of the circle. As we're taking the forces as close as possible together, we can use a small angle approximation for the function. Going back to our definition of the linear density (where we let the small length $dx$ be considered one unit) we can isolate the equation to get the mass of $dx$:
$m = \mu dx$
The forces on the string must be in equilibrium. Since there is a centripetal acceleration, we can create an equivalence equation with [[Newton's Laws and SUVAT#Newton's Second Law|Newton's Second Law]].
However, we have to multiply everything by the unit length multiple of $dx$ - let's give it a variable, $\ell$. As a result, we get this equation:
$T\frac{ \ell}{R} = \mu \ell \frac{v^2}{R}$
The $\ell$ is representative of the length $dx$ and any multiples associated with it! Now all we have to do is solve for $v$ to get our expression for the velocity of *any* wave pulse!
$\frac{T\ell R}{\mu \ell R} = v^2$
$\frac{T}{\mu} = v^2$
$v = \sqrt{ \frac{T}{\mu} }$
We're done! Give yourself a pat on the back.
## Energies of a String
Strictly speaking, a wave on a string is going to have two types of intrinsic energy - its kinetic energy, from its movement, and its elastic potential energy, from the back-and-forth motion of the string.
![[Pasted image 20230902133822.png]]
*The wave diagram - reiterated!*
Using the information given in the diagram above, the oscillating string (the sine wave in this case) can be split into arbitrary lengths of $dx$, each with an arbitrary mass $dm$. Note that a string element moves fastest when the gradient of the curve is highest - when the $y$ value of the equation is at 0. The elastic potential energy is also highest at $y = 0$, as for every linear length element $dx$ the string is most stretched when the line rushes down towards $y = 0$.
This means that at a displacement of $y_m$ (the positive amplitude of this first graph) the energy is 0. Yet the sinusoidal still oscillates - this means the tension forces in the rope are required to transfer energy over to the string.
Mind you, this isn't limited to just $y=0$ - any corresponding $y$ value to wherever the sinusoidal has its highest rate of change will work!
For my final act, let's let the vector quantity expressed as the "direction of oscillation" in the diagram above be equal to the value $\overrightarrow{u}$.
I'll split the derivations of the elastic potential energy and the kinetic energy into their respective boxes. Happy reading!
Mind you, I have no idea if the proof for Elastic Potential Energy holds. I have no idea what it is itself, besides.
>[!Danger]- Kinetic Energy Derivation
>Let's remember some things, shall we? If the mass of each unit length of the string changes, the kinetic energy is also going to be changing - allowing us to write an equation:
>$dK = \frac{1}{2} dm u^2$
>The general equation for a transverse wave is given by $y(x,t) = y_{m}\sin(kx-\omega t)$. This is the *transverse (vertical) displacement* - meaning we'll have to differentiate it to get the transverse velocity, $\overrightarrow{u}$. Notice that there's an $x$ component to the equation - so we'll pretend that its a constant while we differentiate $y$ - a [[Partial Derivatives (Maths)|partial derivative!]] This gives us:
>$u = \frac{\delta y}{dt} = -\omega y_{m}\cos(kx-\omega t)$
>Substitute this into the kinetic energy equation:
>$dK = \frac{1}{2} dm (-\omega y_{m})^2\cos(kx-\omega t)^2$
>Using the concept of [[Waves Foundations#Wavespeeds of Strings|linear density]] we can rewrite the component $dm$ as $\mu dx$. Substituting into the equation gives us:
>$dK = \frac{1}{2} (\mu dx) (-\omega y_{m})^2\cos(kx-\omega t)^2$
>We can divide the entire equation by the term $dt$ to get the rate of change of the kinetic energy, $dt$. Dividing by $dt$ on the right yields a term $\frac{dx}{dt}$, which is just the wavespeed $v$:
>$\frac{dK}{dt} = \frac{1}{2}\mu v \omega^2 y_{m}^2 \cos^2(kx-\omega t)$
>
>$\frac{dK}{dt_{avg}} = \frac{1}{4}\mu v \omega^2 y_{m}^2$
>This formula isn't too pretty but it works! We should probably find a way to simplify it so that there aren't as many variables as there are now...
>[!Danger]- Proving the EPE Equation (HRW NOT INCLUDED!)
>Diagram for reference!
>
>![[Pasted image 20230903202140.png]]
>
>Notice that we can make a right angled triangle relating the extended length of the string $ds$ and the unit of length $dx$. This gives us an expression for $ds$:
>$ds = \frac{dx}{\cos \theta} = dx\sec \theta$
>Use the trig identity $sec^2 \theta = 1 + \tan^2\theta$ to further break the expression down!
>$ds = dx\sqrt{ 1+\tan^2\theta }$
>Since a right triangle can be created we can also generalise the $\tan^2 \theta$ term we have created into a ratio between the $y$ and $x$ axes of the stretching - giving us a partial derivative (as we only want to measure the stretching of the string, given by changes in the $x$ variable):
>$ds = dx\sqrt{ 1 + (\frac{\delta y}{\delta x})^2}$
>
>An alternative equation for the elastic potential energy is given by the tensional force (which we will keep as constant for now) times the net horizontal (on the x axis) extension of the string relative to the unit length $dx$, as given by $(ds - dx)$. As this is the EPE of a small section of the rope we must define it with the prefix $dU$ as shown:
>$dU = T \cdot (ds - dx) = T \cdot (dx\sqrt{ 1 + (\frac{\delta y}{\delta x})^2} - dx)$
>$dU = T dx \cdot (\sqrt{ 1 + (\frac{\delta y}{\delta x})^2} - 1)$
>The %%add link and info to this pls it'll take 15 mins%% transverse wave equation means that we can model the $y$ value of the equation as the expression $y_m \sin (kx - \omega t)$, giving us an expression to differentiate:
>$(\frac{\delta y}{\delta x})^2 = y_{m}^2k^2\cos^2(kx-\omega t)$
>Substitute this into the equation along with the identity[[Waves Foundations#Waves on Strings| proven here]] $T = v^2\mu$, giving us:
>$dU = \frac{1}{2}v^2\mu y_{m}^2k^2\cos(kx-\omega t)$
>More identities! Remember the original equation $v = \nu \lambda$, and our purpose - to equate this identity with the kinetic energy equation, which contains an $\omega$ term. This $\omega$ is the angular frequency of the wave - which is given by the identity $2\pi\nu$, allowing us to create the following substitution for the velocity:
>$v = \frac{2\pi \nu}{\frac{2\pi}{\lambda}} = \frac{\omega}{k}$
>Where $k$ in this context denotes the number of wavelengths in a length of $2\pi$. Substituting this into the main equation gives us:
>$dU = \frac{1}{2}\left( \frac{\omega^2}{k^2} \right)\mu y_{m}^2k^2\cos(kx-\omega t) = \frac{1}{2}\omega^2 \mu y_{m}^2\cos(kx-\omega t)$
>This is the same as the term for the kinetic energy, and averaging it will also yield the same:
>$dU = \frac{1}{4}\omega^2 \mu y_{m}^2$
>Just a couple hours.. :(
>[!Abstract]- Power of a Wave on a String
>The elastic potential energy of the string follows the same $\frac{dU}{dt}$ relationship as the kinetic energy. This is as the potential energy can also be interpreted as being dependant on the change in each unit length.
>
>Check the two proofs above to see why! We can express the power of the system as a function of two times the kinetic energy:
>$\frac{dP}{dt} = 2\left( \frac{dK}{dt} \right)$
>$\frac{dP}{dt} = \frac{1}{2}\mu v\omega^2 y_{m}^2$
>[!Tip]- The use of the "$d
quot; prefix before Variables
>Note that here the usage of the $d$ prefix doesn't mean a change in the quantity - it just shows us that the quantity is always changing, though we can standardise it.
>[!Tip]- Standing Waves?
>Why the string? Surely this would've been easier without leaving the string stationary, right? Well, you've been 'played' - this is the precursor to a standing wave. If you hadn't realised, the inertial frame gave it all away! Mwhahahaha!!!
%%
## History (this goes into an EM dedicated one)
--> gneeral relativity
--> louis debroglie waveparticle duality / how he found it, surface level layman's discovery synopsis
Finding the energy of the wave can be given
The Energy of a wave
$E = h\nu$
Where $h$ is the **Planck Constant**, a constant value that can be found from the Blackbody Equation.
Using $v = \nu\lambda$ we can substitute $\nu$ for $\frac{c}{\lambda}$
$E = \frac{hc}{\lambda}$
Rearranging to give the wavelength yields an expression in terms of $E$:
$\lambda = \frac{hc}{E}$
This is the **de Broglie** wavelength - the standardised method to find the wavelength of an object given it's energy.
There are many other ways to express this! We can take the momentum exerted on an object by the wave, $p$, as a function of the Planck Constant as shown:
$p = \frac{h}{\lambda}$
>[!Abstract]- Proving the momentum relationship
>This has its roots in quantum mechanics, meaning that we'll have to construct several wavefunctions... oh, don't be so scared! By virtue of some higher power a simple proof of this through algebra is possible!
>First, we have to remember that the energy of a particle like a photon can also be described with another equation - Einstein's own mass-energy equivalence equation. The very one.
>As a result, we can create a relationship as shown:
>$mc^2 = \frac{hc}{\lambda}$
>Remember that momentum is given by the relationship $p = mv$. We can therefore let $c$, the speed of light, be the speed of the photon, giving us:
>$pc = \frac{hc}{\lambda}$
>Simplify then isolate $\lambda$ This gives us our final equation in terms of momentum:
>$\lambda = \frac{h}{p}$
This means we can also write the de Broglie wavelength as:
$\lambda = \frac{h}{mv}$
//going further into this proof requires quantum mechanics. i won't even try unless i have to.
>[!Tip]- Units!
>Most of the time the energy of a photon will be given in **electron volts**, or eV, as the energies given in the form of joules are so small we need to convert it into a much smaller, electron-based unit.
//add derivations...later (after you learn general relativity). this is very difficult, ok?
// For the benefit of Astronomy people
//notes --> haven't done this, will figure out what this does eventually!
$\frac{v}{c}$ is the **lorentz factor** - the factor which denotes how the time, length or other physical quantities of an object change over time.
$E = \frac{v}{c} = \frac{1}{n}$
%%
# Example Problems
Examples. Enter the Dragon's Lair if you dare...
>[!Example]- Module 16.1 Problem 1
>If a wave $y(x, t) = 6.0 mm \sin(kx + (600 \ rad/s)t + \phi)$ travels along a string, how much time does any given point on the string take to move between displacements y = +2.0 mm and y = -2.0 mm?
>
>**Solution:**
>This means that the total transverse displacement is 4mm. All we have to do now is isolate the time $t$!
>$\frac{2}{3} = \sin (kx + (600\ rad/s)t) + \phi$
>$\arcsin \frac{2}{3} = kx + (600\ rad/s)t + \phi$
>Since the question asks us the time for any point to move that distance we can ignore any phase angle as the total distance travelled by a point on the wave will remain equal regardless. We can also let the point we take be $x = 0$, meaning that we can remove the term $kx$ from our equation as shown:
>$\arcsin \frac{2}{3} = (600\ rad/s)t$
>Make sure you give the $\arcsin$ answer in radians! Units always matter.
>$0.7297 = 600t$
>$t = 1.217 \times 10^{-3}s$
# What's Next?
For our next instalment in the waves series, go here: [[Standing Waves]]
For the electronvolt, go here:
[[The ElectronVolt (eV) (Nuclear & Quantum)]]
We'll be reiterating much of what we've said as foundational knowledge here:
[[Sinusoidal Curves (Maths)]]
For the next instalment on waves, go:
[[Simple Harmonic and Pendulum Motion (Waves)]]
For more on the Planck constant, go:
1. [[Spectra - A Technical Dive (Waves)]]
2. [[Nuclear & Quantum Contents (Nuclear & Quantum)]]
If you're interested in diving deeper into the energies of a wave, go here:
[[Blackbody Radiation (Thermodynamics)]]
To go back to navigating through physics topics, go:
[[The (Incomplete) Physics Almanac]]