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Hooke’s Law and Simple Harmonic Motion
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Waves
In this course, Professor Steven Balbus (University of Oxford) explores waves. In the first mini-lecture, we introduce Hooke’s Law and solve for a mathematical solution that reveals the concept of Simple Harmonic Motion. In the second mini-lecture, we use a model of many masses on springs to understand properties of waves, including transverse and longitudinal waves, amplitude, wave number, wavelength, waves described in space and time, and wave velocity. The third mini-lecture discusses sound waves and water waves, including examples of sound waves in various mediums with various frequencies and tsunami waves. In the fourth mini-lecture, we give an overview of light waves and electromagnetic radiation. The fifth mini-lecture introduces constructive and destructive wave interference, as well as two-source interference. In the sixth mini-lecture, we turn towards diffraction, detailing Huygens' Construction used to determine the angles at which nulls in a wave appear in order to discuss a few examples with familiar objects. In the seventh mini-lecture, we introduce dispersive waves and go through two examples: surface water waves in deep water and waves such as FM radio waves passing through the Earth’s ionosphere. In the eighth mini-lecture, we briefly explore waves in modern physics, discussing Einstein’s photoelectric effect and the wave-like properties of particles, including photons (light particles).
Hooke’s Law and Simple Harmonic Motion
In this first mini-lecture, we introduce Hooke’s Law, F = -kx. In particular we: (i) equate Hooke’s Law to F = ma and rearrange to arrive at a(t) = -(k/m)x(t); (ii) solve for x(t) with a backwards approach that involves understanding the circular motion of a point mass; and (iii) connect this circular displacement of the point mass to the concept of Simple Harmonic Motion.
Hello.
00:00:06My name is Steven Valve is, um,
00:00:07the civilian professor of astronomy at Oxford University.
00:00:09And I'd like to talk to you today about the physics of waves.
00:00:13Waves, it seems, are everywhere.
00:00:20You're seeing these words because waves of light
00:00:24are carrying them to receptors in your eyes.
00:00:27You're hearing my voice because waves of compression
00:00:31travel in the air and enter your ear,
00:00:35where they move a sensitive membrane back and forth
00:00:38a few 100 times each second.
00:00:43If you are using the WiFi,
00:00:46radio waves are being sent back and
00:00:48forth between the transmitter and your computer,
00:00:51causing a small electromagnetic force
00:00:55to vibrate through empty space several billion times each. Second
00:00:59waves can project a force. They can carry energy, they impart information,
00:01:06and they do all of this simply by causing something to vibrate back and forth.
00:01:14Now it's not always easy to tell what that something is,
00:01:23but there are always vibrations.
00:01:29It makes sense, therefore,
00:01:34to begin our study of waves by studying the physics of simple vibrations.
00:01:36And to do that, let's start by studying something which is known as Hooks Law
00:01:43Book 10 CEO Sick Wiese,
00:01:52Robert Hooke wrote in 16 78 as a Latin expression for what is now known as hooks law.
00:01:56A rough translation would be force is proportional to extension,
00:02:07and this is a natural place to study simple vibrations. In its simplest context,
00:02:14Robert Hooke studied the physics of what amounts to a mass on a spring.
00:02:23This is a system in which the return force
00:02:31caused by the spring is directly proportional to the displacement
00:02:34of the mass.
00:02:41Mathematically the force,
00:02:43which is the mass times The acceleration is equal to some constant K
00:02:45known as the spring constant times the Displacement X with a minus sign.
00:02:53In other words, the acceleration is minus K, divided by the mass M times X,
00:03:00the displacement itself.
00:03:08And the question before us is,
00:03:10if the acceleration is proportional to minus the displacement, what is X of T?
00:03:13What is the displacement as a function of time?
00:03:21Well, in this case, this is a good problem to solve by doing it backwards.
00:03:27In fact,
00:03:33what I'm going to do is begin by writing down X of T at the very start of my discussion,
00:03:34and then I'm going to show that the resulting acceleration from that
00:03:44is in fact just a constant number times minus X empty,
00:03:49which will solve the problem.
00:03:56And it all starts
00:03:58simply by looking
00:04:00at a single point mass going around
00:04:02in a circle.
00:04:06So here you see the diagram of what I have in mind.
00:04:08There's a point mass which is travelling around a
00:04:13circle at Velocity V and the circle has radius.
00:04:15Are we think of our now as a vector with
00:04:21Cartesian coordinates X and Y as shown in the diagram.
00:04:26Now you know, from basic trigonometry
00:04:33with the right angle triangle formed by our X and y,
00:04:37that X is equal to the radius r times the co sign Fada Fada, the angle being shown here
00:04:45and the why coordinate is our times sine theta.
00:04:54Now Fada itself will be the distance travelled by the mass,
00:04:59which is simply its velocity times time in the form of an arc.
00:05:05Around the circle,
00:05:10divided by R theta is V t over our and we call the combination V over R Omega.
00:05:12The Greek letter Omega and Omega is a measure of how many radiance per second
00:05:23the mass travels around the circle. Well, if Fada is omega t then are of t.
00:05:31The vector are now.
00:05:39Each of its coordinates can be expressed in the form of X equals our coast in Omega T.
00:05:42And why is R sine omega T,
00:05:51which then gives us our formula for what the
00:05:55X and Y displacements are as a function of time
00:05:59except T is r cosine omega T y of t is r sine omega T and omega.
00:06:05Remember is simply the velocity divided by the Radius V over R.
00:06:14Now, when the Time T has advanced by one full period, let's call the period p.
00:06:21The mass has gone around the circle completely,
00:06:31and the argument of the co sign, which is theta or omega T,
00:06:35has advanced by Omega T equals omega P, and that is a full two pi radiance.
00:06:41In other words, omega is equal to two pi divided by the period p.
00:06:51This is our angular frequency,
00:06:59the number of radiance per unit time of the motion.
00:07:02Now one over P is the number of full cycles per unit time or, in other words,
00:07:06exactly one cycle per period.
00:07:15P one overpay is the ordinary frequency and is denoted by the letter F
00:07:18Omega then is two pi divided by p or more simply to pi f The units of f R hurts one.
00:07:28Hertz is one cycle per second.
00:07:40Now
00:07:45we know what the acceleration is already because we've studied circular motion.
00:07:47The acceleration A for circular motion
00:07:55is simply the squared over r.
00:07:59The squared over R is simply our itself. Times omega squared
00:08:04in a direction
00:08:12which is pointing towards the centre of the circle.
00:08:14It is directed radial e inward, in other words, in minus the radial direction.
00:08:18That means if I separate the acceleration vector out
00:08:26into its two x and y Cartesian components,
00:08:31the acceleration in the X direction is minus omega squared X,
00:08:37and the acceleration in the Y direction is the same minus omega squared.
00:08:43Why?
00:08:49Well, but this solves the problem.
00:08:51If a in the X direction is equal to minus omega squared X Omega squared is, after all,
00:08:55just a positive constant.
00:09:04The solution is as we've seen X of t equals r cosine omega t
00:09:07the radius are here can be whatever I like.
00:09:15And
00:09:20if we look
00:09:21at y of t
00:09:22r sine omega t, we find that our sign of omega T works exactly the same way.
00:09:24It is just as good a solution.
00:09:32A in the Y direction is minus omega squared y or minus are omega squared sine omega T.
00:09:35What we label X and Y doesn't really matter. In the end,
00:09:49X equals R sine of Omega. T is also a proper mathematical solution to Hook's Law.
00:09:54A equals minus omega Squared X.
00:10:03Well, in fact, the acceleration that goes with a displacement. Let's call it D of T,
00:10:10where D of T is given by a sum of co sign and sign.
00:10:19In particular, let's say D of T is some number a times cosine,
00:10:25omega t plus another number B sine omega t Well,
00:10:31the acceleration that goes with that is simply
00:10:37the acceleration that goes with the co sign
00:10:40added to the acceleration that goes with the sign, which is minus omega squared.
00:10:44In both cases times the quantity a co sign Omega T plus B sine of omega T or,
00:10:51in other words, minus omega squared times the original displacement D of T.
00:11:00And this is the most general solution to our problem.
00:11:09It is called simple harmonic motion
00:11:14signs and co signs will be a central part of our study of waves.
00:11:20So to recap, what have we shown?
00:11:27Well, we have studied hooks law,
00:11:31and by using an analogy of the point mass going around in a circle,
00:11:34we have found that the solution to Hook's Law is a sum of
00:11:41co sign of Omega T and a sign of Omega T,
00:11:49where Omega is known as the angular frequencies.
00:11:54And these are going to be a central part of our study of waves.
00:11:59What we'll do next time is to generalise what
00:12:06we found by looking at what happens when we
00:12:09have many masses on springs and that will lead us to our more general study of waves.
00:12:12
Cite this Lecture
APA style
Balbus, S. (2022, January 12). Waves - Hooke’s Law and Simple Harmonic Motion [Video]. MASSOLIT. https://massolit.io/courses/waves/diffraction
MLA style
Balbus, S. "Waves – Hooke’s Law and Simple Harmonic Motion." MASSOLIT, uploaded by MASSOLIT, 12 Jan 2022, https://massolit.io/courses/waves/diffraction