# Preface Stuff like clocks, earthquakes - what do they have in common? That's right, they're *oscillations*, movements back and forth within a medium that occur pretty[^1] regularly! To model this motion, we have to use something known as *simple harmonic motion* - the transposition of the pendulum bob's motion as a wave, using displacement from the center as a means of doing so. It's a common science fair experiment, it's a common answer. Let's delve in and learn! Example problems included. ![[shm-comic.png]] [^1]: Check [[Simple Harmonic and Pendulum Motion (Waves)#Damped SHM|the notes on damped simple harmonic motion]] for more! Make sure you do everything before it first though. # Restoring Forces - Intro Before I get into the nitty-gritty about springs, a reminder that you **need** to be very, very confident in springs themselves, such as with topics like [[All About Elasticity (Mechanics)|Hooke's Law]] and [[Power, Work Done and Energy|Elastic Potential Energy]]. Brush up on these if you haven't already! We can express the **restoring force** of a spring on an object - the force it exerts on an object to attempt to bring it back through the following equation: $F = -kx$ How does Hooke's Law reappear here? Another way we can think about Hooke's law is through understanding that it is a function of the extension of an object - the negative sign appears as the force acts *opposite* to the direction of linear force. Notice how we can also model this force as a function of Newton's Second Law, $F=ma$ - the spring force is a direct consequence of an object's own stretching. We can therefore create the relationship: $-kx = ma$ >[!Danger]- Even More On Restoring Forces - Differentials >The acceleration is the second derivative of the displacement which is given by the extension $x$. >$m \frac{d^2x}{dt^2} = -kx$ >We can therefore isolate the differential term to get: >$-\frac{k}{m}x = \frac{d^2x}{dt^2}$ >This allows us to find the **net acceleration** of an object through details given by the spring constant. We can solve this differential equation %%add the maths once you can%%to get a sinusoidal wave, giving us the position (or displacement from the ground state) as a function of time: >$x(t) = x_{m} \cos (\omega t + \phi)$ >Now I don't know how to do this just yet but I'll figure it out eventually. # General Simple Harmonic Motion This section takes knowledge for the [[Waves Foundations|period]] of an oscillation as a given. Go brush up on it if you haven't already! We can generalise the equation for simple harmonic motion into a [[Sinusoidal Curves (Maths)|sinusoidal]]: $x(t) = x_{m} \cos (\omega t + \phi)$ **Where:** - $\omega$ is the angular frequency of the pendulum, which allows us to determine the frequency to which the same value for $x(t)$ repeats itself. - $x_m$ is the *amplitude* of the oscillation - the maximum displacement from the rest state. - $\phi$ is the *phase angle* of the oscillation, which allows us to set linear displacements for the object, allowing us to model the pendulum regardless of where it is during its oscillation. This gives us our displacement as a function of time - meaning we can always know where the object is if we know the time. >[!Success]- Deriving the Angular Frequency >Strictly, angular frequency refers to the **number of oscillations per unit displacement - or $2\pi$.** Since this is a cosine function, we can set the phase angle to $T$, one period, or $2\pi$: >$\omega(t + T) = \omega t + 2\pi$ >$\omega T = 2\pi$ >$\omega = \frac{2\pi}{T}$ Continuing onwards, we can take the **velocity** of a particle in Simple Harmonic Motion by differentiating the expression we have for the displacement, giving us: $v(t) = -\omega x_{m} \sin(\omega t + \phi)$ **Where:** - $\omega x_{m}$ is the *velocity amplitude* of the oscillation - $\phi$ is the phase constant This shows us that the particle is actually at its fastest when it gets to a displacement of $x(t) = 0$, the centre of the pendulum. Finally, a second derivative of the displacement function allows us to get an expression for the acceleration of a particle in SHM! $a(t) = -\omega^2 x_{m} \cos(\omega t + \phi)$ For a phase constant $\phi = 0$, we can create a simplified expression for the acceleration of the particle at time $t$: $a(t) = -\omega^2 x(t)$ How elegant! It's just the angular frequency times the displacement of the particle. Likewise, the term $\omega^2x_{m}$ is the *acceleration amplitude* of the simple harmonic oscillation. Most of the time, questions are going to ask you to express the angular frequency with the *spring constant*, of all things. To figure out why, we'll have to go deeper into angular frequency, so check out the box below! >[!Abstract]- Angular Frequency Extended! >Remember from restoring forces that the forces due to a spring can be modeled with both Newton's Second Law and Hooke's Law. Despite this being ironic as both of those people were actually rivals when they were alive, it allows us to write the following identity out: >$-kx = ma$ >Since this system is in simple harmonic motion (where the spring can be interpreted as a direct translation of the back-and-forth movement of the object), we can use the expression we get for acceleration: >$-kx = -m\omega^2 x$ >$kx = m\omega^2x$ >It's time to isolate the $\omega$! This is our angular frequency. >$k = m\omega^2$ >Therefore... >$\omega = \sqrt{ \frac{k}{m} }$ We can model the period of oscillation (the time it takes for a spring to return to its original state after 1 cycle) with the following equation: $T = 2\pi \sqrt{ \frac{m}{k} }$ **Where:** - $m$ is the load on the spring - $k$ is the spring constant >[!Abstract]- SHM - $\sin$ Version >We've been only taking a look at oscillations at max displacement - what would happen if we started the curve at 0? Wouldn't the displacement be 0? > >$x = x_{m}\sin(\omega t+\phi)$ >$v = \omega x_{m}\cos(\omega t+\phi)$ >$a = -\omega^2 x_{m}\sin(\omega t+\phi)$ > >As the curve is 0 when the angle is 0, we can therefore infer that instead of a cosine curve (where the curve is 1 when the angle is 0) there is a sine curve. It's basically the same though there are a couple differences: >$a = -\omega^2x_{m}$ >[!Example]- Problem Taster - Module 15-1 Problem 6 >A particle with a mass of $1.00 \times 10^{-20} kg$ is oscillating with simple harmonic motion with a period of $1.00 \times 10^{-5} /s$ and a maximum speed of $1.00 \times 10^3 m/s$. Calculate: > >(a) the angular frequency and >(b) the maximum displacement of the particle. > >**Solution:** >a) Angular frequency is the equivalent of angular velocity. Use the formula below and substitute numbers to get your answer! >$\omega = \frac{2\pi}{T}$ >$\omega = \frac{2\pi}{1.00 \times 10^{-5} /s} = 2 \times 10^5 \pi rad s^{-1}$ >b) The equation(s) for the velocity of the particle is: >$v(t) = -\omega x_{m}\sin(\omega t+\phi)$ >$v(t) = -\omega x_{m}\cos(\omega t+\phi)$ >For the purposes of this question we'll be using the first one. Let's substitute as many values as we can into the equation as possible: >$1.00 \times 10^3 m/s = 2 \times 10^5 \pi x_{m}\sin(2 \times 10^5 \pi (0.25 \times 10^{-5}))$ >We're taking a fourth of the period, as on a sinusoidal the first crest occurs after 1/4th of a period. Our goal now is to find $x_m$, the amplitude of the particle, or the largest displacement from the origin. All we have to do now is isolate it: >$\frac{1.00 \times 10^3 m/s}{2 \times 10^5 \pi }\sin(2 \times 10^5 \pi (0.25 \times 10^{-5})) = x_{m}$ >Throwing all the numbers together gives us: >$x_{m} = 0.0580600\dots \approx 0.058m$ ## The Energy of a SHM Particle Remember [[Power, Work Done and Energy|conservation of energy]]? If you don't, you better start remembering fast - not sure how you'll get through physics courses without mastering it. In [[Power, Work Done and Energy|the energy doc]], we went over some forms of potential energy, such as gravitational potential energy and chemical potential energy. Though elastic potential energy was also covered, I'm not too confident that what I've written would suffice for the concepts we'll cover here - so onwards with the SHM deep dive! In SHM, there are two forms of energy we need to take - the elastic potential energy (some textbooks call it tension), which is the energy a compressed spring has stored, and the kinetic energy, which is a measure of the energy due to how fast it (the spring) moves. An energy within an oscillator can be interpreted as having a total energy - a mechanical energy, or the sum of its potential and kinetic energies. Remember that the change in potential energy is equal to the negative value for work done. Since the initial potential energy in SHM can be written off as equivalent to the mechanical energy of the system (as the object doesn't move, meaning there is no kinetic energy), we can write the equation for potential energy as shown: $U(t) = \frac{1}{2}kx^2 = \frac{1}{2}kx_{m}^2\cos^2(\omega t+\phi)$ This allows us to find the potential energy at *any* given displacement value within the spring! Using this logic, we can find the potential energy using the velocity equation for SHM as shown in the above sections: $K(t) = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2x_{m}^2\sin^2(\omega t+\phi)$ We can substitute $\omega$ with $\sqrt{ \frac{k}{m} }$ for a better-looking (less variables making it more practical too!) equation as follows: $K(t) = \frac{1}{2}kx_{m}^2\sin^2(\omega t+\phi)$ Now it's finally time to find the total (mechanical) energy of the spring! Let's create an equation to describe the relationship shown: $E_{mec} = U+K$ $E_{mec} = \frac{1}{2}kx_{m}^2\cos^2(\omega t+\phi) + \frac{1}{2}kx_{m}^2\sin^2(\omega t+\phi)$ This equation's actually really, really nice for us - note how the equation is just the [[Trig Identities - Derivations DLC Ver. (Maths)|trigonometric identity]] $\sin^2x + \cos^2x = 1$, so the equation just becomes: $E_{mec}=\frac{1}{2}kx_{m}^2$ This shows us that it obeys Newton's First Law - by relating it with the [[Power, Work Done and Energy#Work Done|work-kinetic energy theorem]], it's obvious that there needs to be another force for the mechanical energy to increase! ## Simple Harmonic Oscillator - Angular Edition Yo-yos have always been great fun, though the "mobile" seems to have supplanted the role of toys, now of a simpler, bygone era. Sometimes, the string of a yo-yo gets all twisted and turned - yet when that happens, it seems to unravel itself while twisting in different directions ... eerily similar to simple harmonic motion, no? Well, that's because it's a type of simple harmonic motion, this time with a twist! Quite literally - it's the act of *twisting* the string that actually enables the Yo-Yo to undergo Simple Harmonic Motion. Note that in rotational circles (haha, get it?), the *torque*, or moment, of an object is the analog for force. As a result, we can create an expression for the restoring force in rotational motion as: $\tau=-\kappa\theta$ *Where:* - $\theta$ is the angular displacement of the object spinning - $\kappa$ is the **torsion constant**, which is dependant on the length, material and diameter of the wire (string) that is spinning Note the similarities between this and Hooke's Law. Generalising this similarity, we can therefore create an expression for the period of this oscillation: $T = 2\pi \sqrt{ \frac{I}{\kappa} }$ Where we replace the mass of the disc $m$ with its moment of inertia $I.$ >[!Danger]- More on Torsion >Simply put, torsion is a rotational interpretation of extending a spring. Kind of like those licorice or sour strips of candy they have which you can twist - it deforms, much like you'd expect a spring to deform when you stretch it. >Simply put, the formula for torsion (in terms of [[All About Elasticity (Mechanics)#Shear - Shearing|shear stresses]]) is given by: >$T = \frac{\kappa}{L}G \theta$ >Where $\theta$ is the angle the bar is twisted (angular displacement), $L$ is the length the shearing force acts on and $G$ is the [[All About Elasticity (Mechanics)#Shear - Shearing|shear modulus]]. > >This shear modulus shows us that torsion is actually a perpendicular deformation of an object! Now the next time you twist a metal bar with your bare hands, try to think about the forces acting on it... please? It had a loving family... # Pendulum Motion - Emotional Ups and Downs >[!Tip]- Pendulum Conundrum >It's difficult understanding Simple Harmonic Motion the first time you through it, and as somehow who did just that, I understand it! Make sure that you follow the rules listed out on the Homepage, where you could pick an unwitting 'subject' of sorts to explain abstract concepts that only exist within the human frame of mind. > >You got this! We believe in you. Remember - practice makes perfect. ## A Pendulum - Intro Doubling back to the yo-yo, you ever tried swinging it in front of someone to pretend you'd hypnotised them? While that may be a core memory for many, the back-and-forth motion of a pendulum can also be modelled with simple harmonic motion - maybe after this section you'll finally be able to hypnotise your parents! ![[Pasted image 20231112115153.png]] A pendulum can be thought of as a see-saw - where a pivot point (in the case of the yoyo, your hand holding the pivot) sits in the middle of an oscillating beam. This similarity with the see-saw allows us to create an equation for the *restoring torque* on the pendulum: $\tau = L F_{g}\sin \theta$ Where: - $\theta$ is the angle of displacement from the pivot at the instant when the object swings highest - $L$ is the length of the rod/string holding the pendulum - $F_g$ is the gravitational force felt by the object >[!Abstract]- Deriving the Equation >Note the similarities between the equation above and the original torque equation, given below: >$\tau = rF\sin \theta$ >The pendulum in motion draws out part of a circle - an arclength. The only force on the pendulum is the weight of the object that is suspended from the rod/string with a distance $L$. As a result, it is possible to just directly replace the variable $r$ with $L$ and the force with the weight $F_{g}$: >$\tau = LF_{g}\sin \theta$ Let's break the equation down into its base constituents to get a better intuition! $I\alpha = Lmg\sin \theta$ We can therefore get an equation for the angular acceleration using $\tau=I\alpha$: $\alpha = \frac{-L\ mg\sin\theta}{I}$ Notice that for small angles this equation becomes (thanks to approximations): $\alpha = -\frac{Lmg}{I}\theta$ Now use the formula for acceleration we have, $a = \omega^2x_{m}$, to get: $\omega = \sqrt{ \frac{mgL}{I} }$ Armed with this equation, the period just becomes a matter of dividing $\omega$, the angular frequency, by $2\pi$ - this gives us an equation for the period of a pendulum with $L$ as the linear length of the string: $T = 2\pi \sqrt{ \frac{I}{mgL} }$ For small swinging angles this is generalised into: $T = 2\pi \sqrt{ \frac{I}{mgh} }$ Where $h$ is the vertical length of the pendulum. If the mass of the pendulum is a point mass with a centre of mass at the tip of the rod and has a small swinging angle, we can further generalise this into: $T = 2\pi \sqrt{ \frac{L}{g} }$ This equation is also the base equation for a pendulum undergoing simple harmonic motion. ## Physical Pendulums Now for the "useful" bit! Physical Pendulums are pendulums we typically see in real life - not the idealised version where mass is distributed equally, but where mass and shape can be arbitrary at best. The main difference we'll have to look out for is with the moment arm - instead of using the length of the string $L$, we have to use an arbitrary perpendicular distance $h$ - the distance from the centre of mass of the object to the pivot arm. We can therefore rewrite the equation as shown: $T = 2\pi \sqrt{ \frac{I}{mgh} }$ >[!Danger]- Measuring Gravitational Acceleration with a Pendulum >We can measure $g$, the gravitational attraction of the Earth, with the parallel axis theorem: >$I = I_{com} + mh^2$ >Let's model the moment of inertia of the string as a rod, a really, really thin one at that. We use the moment of inertia of a rod as: >$I = \frac{1}{2}mL^2$ >Giving us: >$I_{com} = \frac{1}{12}mL^2 + m \frac{1}{2} L^2 = \frac{1}{3}mL^2$ >Since our period equation has the variable $g$, we can plug in $h = \frac{1}{2}L^2$ and $I = \frac{1}{3}mL^2$ to solve for it, giving us an expression: >$g = \frac{8\pi^2L}{3T^2}$ >Since the length $L$ and the gravitational acceleration $g$ can be used to find the period of a pendulum with a small angle we can take measurements to find $g$. ## Unifying SHM with Circular Motion If you've ever looked at Jupiter for a prolonged period of time, given that you have a telescope powerful enough you should be able to see it's four major moons, orbiting around their parent planet. Plotting a graph of the *angular separation* - that is, the angular distance in arcminutes relative to us, an observer on earth, follows a certain sinusoidal wave that we commonly see in SHM... Let's see the diagram for a reference of what we mean! ![[Pasted image 20230806125246.png]] (unintelligible diagram possible problem?) Remember that we can find the tangential velocity using the rotational identity $v = \omega r$ and that the radial (centripetal) acceleration can be found with the formula: $a = \frac{v^2}{r}$ A simple substitution of variables ($x_{m}$ being $r$, $\omega x_{m}$ being the velocity) yields a value for the radial acceleration as $\omega^2x_{m}$. Now that that's done and dusted with, we can finally, finally express the rest of the expressions as a function of uniform circular motion! $x(t) = x_{m}\cos(\omega t+\phi)$ Note that the object's velocity is anticlockwise. This means that the velocity amplitude is negative, creating another simple harmonic motion analog: $v(t)=-\omega x_{m}\sin(\omega t+\phi)$ This logic applies to the acceleration vectors: $a(t) = -\omega^2x_{m}\cos(\omega t+\phi)$ Note how these equations are identical to the SHM equations in the first section. ***As a result, we can map an object undergoing uniform circular motion with the simple harmonic equivalent for their motion.*** >[!Example]- Problem Refresher - Module 15-4 Problem 40 >A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance $d$ from the 50 cm mark. The period of oscillation is $2.5/s$. Find $d$. >[!Example]- Problem Refresher - Module 15-4 Problem 47 >A physical pendulum consists of a uniform solid disk (of radius $R = 2.35 cm$) supported in a vertical plane by a pivot located a distance $d = 1.75 cm$ from the centre of the disk. The disk is displaced by a small angle and released. What is the period of the resulting simple harmonic motion? # Damped SHM So far, the pendulums we've been talking about have all been modelled under ideal conditions - where the drag force on the pendulum is almost zero. That changes today. Unfortunately for us, it's time to see what happens if we model a pendulum under water, in oil, or in any medium where the drag force is significant. ![[Pasted image 20231006235055.png]] Nice displacement-time graph, when has anyone ever gone wrong with this? Note that a damping force varies from spring to spring - depending on the system there are different constants we have to use to describe the system! As the damping force acts opposite the direction of the pendulum's motion, we can express the force as: $F_{d} = -bv$ Where: - $b$ is the *damping constant*, which has a unit of kg/s (kilogram per second) - $v$ is the velocity of the object >[!Warning]- Damping Force Intuition >Remember that the drag force $\frac{1}{2}C\rho Av^2$ is dependant on the velocity of the object to change it, as the other attributes of an object do not change as easily. As a result, the damping force can be written as a function of just the velocity! To model this to oscillations, remember that the force on an object on a spring is given by Hooke's law. The force on the spring by an object is given by Newton's second law, which thanks to Newton's Third Law is equal to the Hooke's Law force. However, thanks to the damping force, the force a block exerts on a spring is actually decreasing with time - as a result, we can write the expression as: $-bv - kx = ma$ Rearrange to create an equation equal to zero: $ma + bv + kx = 0$ We can substitute $a$ and $v$ for their counterparts in terms of $x$, the displacement vector: $m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$ Finding $x$ becomes a matter of integrating the expression. Doing so yields the following: $x(t) = x_{m} e^\frac{-bt}{2m}\cos (\omega't + \phi)$ Where: - $x_me^{\frac{-bt}{2m}}$ is the amplitude of the oscillation - $w'$ is the angular frequency of the damped oscillation >[!Danger]- Deriving the Equation (unfinished) >Knowledge about [[Ordinary Differential Equations (Maths)|Ordinary Differential Equations]] and [[Calculus Basics (Maths)| integration]] here! >Note that the differential equation above is ac %%ftw sturm-liouville bs (include when found pls)%% %%what is life???%% We can express the angular frequency with the following expression: $w' = \sqrt{ \frac{k}{m} - \frac{b^2}{4m^2} }$ As the amplitude of the oscillations gets smaller with time, the mechanical energy of the system will also get smaller with time. This means we can substitute the $x_{m}$ in the original mechanical energy equation with the damped oscillation amplitude $x_{m}e^{\frac{-bt}{2m}}$, giving us: $E_{mec} = \frac{1}{2}kx_{m}e^{\frac{-bt}{m}}$ # Forced Oscillations and Resonance Escaping the damped oscillators, our weary heroes travel to their second-to-last destination - Asian Parent Simple Harmonic Motion! That's right, the oscillations in question are now being forced to, well, oscillate, one way or another! Hurray!!.... For this, we can imagine that a pendulum has been slightly "nudged" by its parents to take up piano lessons - where a driving force has been exerted on it. This force can be interpreted as translating into an angular frequency of $\omega_d$. The angular frequency of the pendulum will thus be equal to the angular frequency of the driving force $\omega_d$. This means we can rewrite the standing wave for SHM as: $x(t) = x_{m}\cos(\omega_{d}t + \phi)$ Notice that a swing follows the same principles of motion - as you push it, there seems to be an upper limit to how far you can push the swing (assuming that the person on the swing loses a significant amount of their kinetic energy). This is called **resonance** - where the velocity amplitude of the standing wave is highest and where the angular frequency of the driving force and the original standing wave are equal: $\omega_{d} = \omega$ Note that this can be interpreted as a continuation of[[Interference Basics| constructive interference]]. Resonant oscillations are very, very dangerous! Being the most powerful seismic waves, they usually cause significant damage upon reaching the surface. Earthquake-proofing is always a must. >[!Danger]- More Resonance %%(add 2ndorderDE working)%% >Note that resonant oscillations must be damped oscillations given their practical use (as a medium has to be stopping the object from increasing its frequency, or else it'd go to infinity!). We can therefore imagine a damping force $-bv$ that acts on the pendulum. Given that the tension of the rope is given by the force $-kx$ and that the force can be written as an oscillation with the Newton's second law therefore gives: >$-bv - kx + F_{m}\cos(\omega_{b}t) = ma_{x}$ >Rewrite everything in terms of $x$. This gives us a second-order differential equation: >$m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = F_{m}\cos \omega_{b}t$ >Note that the driving force and the tension in the rope must be constant. This means that $F_m\cos \omega_{b}t$ must decrease as $t$ increases, so $m \frac{d^2x}{dt^2}$ and $b \frac{dx}{dt}$ must remain the same. This gives us a solution for the differential equation: >$x(t) = \frac{F_{m}}{G}(\cos \omega_{b}t - \beta)$ >Where: >$G = \sqrt{ m^2(\omega_{b}^2 - \omega^2)^2 + b^2\omega^2 }$ >$\beta = \arccos \frac{b\omega^n}{G}$ ## More on Angular Frequency We can express the angular frequency $\omega$ of a pendulum as an equation as shown: $-\frac{g}{l} = -\omega^2 $ # Two-Body Oscillations (pain) In front of our heroes stand a colossus, undefeated by man, weathered by time. That's right, we're going to be looking at ways to model *two* pendulums now - so buckle up, because this is going to be painful. put on hold until i remember this, i'm very rusty! # Example Problems >[!Example]- Module 15.1 Problem 1 >An object undergoing simple harmonic motion takes 0.25 s to travel from one point of zero velocity to the next such point. The distance between those points is 36 cm. Calculate the >a) period >b) frequency >c) amplitude of the motion > >**Solution a)** >The next point with a velocity of 0 is just halfway through a single oscillation (or the next crest/trough if you're a visualisation maniac). Therefore, the period is just two times the time value given: >$T = 0.25 \times 2 = 0.5s$ >**Solution b)** >Remember the gold standard - that frequency is equal to the reciprocal of the period. We can plug in our above value to get: >$f = \frac{1}{T} = \frac{1}{0.5} = 2 Hz$ >**Solution c)** >As the oscillation starts from a crest, there won't be a stupid phase angle for us to deal with - so we can just write the displacement $x$ as its sinusoidal $x = x_m \cos (\omega t + phi)$, where our goal is to find $x_m$. Now all we do is to throw in our existing values for all the variables, letting $\omega$ be $2\pi f$: >$0.36 = x_m \cos (2\pi (2)(0.5)) = x_{m}$ >Therefore the amplitude is equal to the displacement, 0.36 metres. Remember to convert to SI Units! >[!Example] # What's Next? You're done! Sorry for the detour - oscillations are just imperative for waves and anything electromagnetic. We can continue with the waves' syllabus by going here: [[Standing Waves]] If you came here from the mechanics' pages, don't fret! You can get back here: [[Mechanics - A Contents Page]]