VIEWS: 12 PAGES: 12 POSTED ON: 4/13/2011 Public Domain
Week 1 Lecture 2 Problems 2, 5 What if something oscillates with no obvious spring? What is ? (problem set problem) Start with +ve Try and get to SHM form Ex. Full beer can in lake, oscillating get (where x = distance from equilibrium point and A=cross k sectional area) This is now in SHM form rearrange: so, effective “k” 1 Note that „g‟ in this SHM equation comes from the Archimedes contribution Rotational Oscillations (review) These are systems which oscillate rotationally about an axis, rather than linearly Eg. Torsional Pendulum This stuff is all Trick here: note that linear review from first variables simply change to angular year – look at your ones: old notes - all Fs become ts - all ms become Is - all xs become qs and so on…. Linear: F = -kx F = ma Rotational: restoring angular torque angular torque displacement rotational acceleration torsional spring inertia constant Combining as before: or Defining angular equation for frequency, angular SHM 2 Like the linear form 2 except that x s are now qs Going back to linear SHM again… How do we figure out where the block is at any given time, i.e. what is x(t)? (defining equation for linear SHM) is a 2nd order differential equation. To find x as a function of time (solving the differential equations) rewrite as: where The general solution* to this diff eq‟n is given by (1) To evaluate constants c 1 and c2, we need Boundary Conditions (BCs). So, look again at our oscillating block: Pull it out to A and let go at t = 0 +x So at t = 0, x = A (BC #1) A t= 0 A = max displacement Also at t = 0, v = dx/dt = 0 m (BC #2) equilibrium 3 * This is something that you learn in a differential equations course – if you haven‟t seen it yet you should soon Sub BC #1 into (1) gives or Differentiate equation (1), then sub in BC #2 At t = 0, dx/dt = 0 So, subbing c1 and c2 back into (1) gives This gives us x as a or function of t for this case Note that this equation only holds for the boundary conditions we assumed: at t = 0, x = A and v = 0. Really important to keep in mind! 4 The Phase Constant Usually we write x = Acos(t + ) (rather than just x=Acos t) The value of depends on where we choose t = 0 If we choose t = 0 when x = A, then: x = Acost x (these were the boundary conditions we considered A t on the previous page) However, if we choose t = 0 when x = 0, then our plot of x vs. t looks like: The curve is shifted x right by /2 (but of course A t in the general case this shift can be any value) shifted right by /2 = -/2 Mathematically, this can be written as ( = - /2) Or Therefore, depending on where you define t = 0, SHM can be represented by either a sin or cos function 5 Therefore, there are three equivalent ways to express the solution of 1) here we use BCs to set c 1 and c2 and have no phase constant. Use BCs to get C 2) and (note: since C is the amplitude it will be the same for both 3) of these but will be 900 different) Note: We will use the solution a lot, so we have Hey! Is there a relationship to pick which one of the above three we will use. Between c 1 , c2 and C? We choose this one – mainly because it is the You bet! form used in the textbook French. Other books (like Pain) use the sin form more frequently. You can see the proof of this on page 5 of the courseware or prove it with phasors We will mainly use: (cos and sin are 90o out of phase) C c2 c1 Although we usually use the above cos form - you should be 6 comfortable with finding each of the 3 forms of solution for any oscillating system!!! - see the example on the next page Ex. A spring/mass system is pulled 0.5mm from equilibrium and then given a push to give it an initial velocity of mm/s as shown. Its resulting angular frequency is = 2rad/s. Develop an expression for x(t). 0.5mm equilibrium ___________________________________________________ BCs @ t= 0 x = +0.5 mm (BC#1) and v = + 2√2 mm/s (BC#2) Also =2 rad/s Assume SHM applies Examine the three equivalent methods: Method 1 uses BC#1 gives 0.5 = C1sin 0 + C2cos 0 C2 = 0.5mm BC#2 need to differentiate, giving v = C1cos t – C2sin t Subbing in BC#2 and w=2 rad/s gives 2√2 = C1 - 0 C1 = √2 7 Result is x = √2 sin 2t + 0.5 cos 2t Method 2 uses BC 1 gives BC 2 need to diff, gives Sub in BC 2 and gives get Solving the two eq ns simultaneously gives so Out by /2 radians Method 3 uses (1.57) as BC 1 gives expected BC 2 – diff gives or Solving the two eq ns simultaneously gives so 8 Maple Plots for 3 Equivalent Methods These are all identical! 9 Plotting Displacement, Velocity and Acceleration [back to using x(t)=Acos(t+)] Plots of the displacement x, the velocity v, and the acceleration a, as functions of time ( = 0). Displacement (x) x A t Velocity (v) Note that v v has advanced (left) by /2 compared to x(t) A t vmax at x=0 Acceleration (a) Note that a a a=max has advanced at v=0 by at xmax compared to x(t) 2A t a=0 at vmax at x=0 10 Hey – check out IP demo#1 for this and follow along with the bouncing spring… Representing SHM as a Complex Exponential It is useful to be able to define SHM in terms of a complex exponential. Why? Because when we try and solve more complicated versions of our differential equation (say, ones which include damping terms or forcing terms) the math becomes much easier if we can turn our equation into a complex exponential. The complex exponential can also be represented as a vector (phasor). This is handy when we start to add waves together – adding vectors is much easier than adding a bunch of equations! To go from the SHM equations to a complex exponential, we need to go through a couple of steps, shown below. The next few pages will illustrate these steps. SHM (x = C cost) rotating vector complex number complex exponential and complex vector makes math easier to add 11 easier waves as vectors Simple Harmonic Motion and Circular Motion: The Rotating Vector Representation Imagine particle P (or vector OP) moving with constant speed vo counter-clockwise around the circle as shown (radius A). Point Q is the projection onto the x axis, and as the vector OP rotates, the point Q oscillates in SHM from –xo to +xo. The y position also exhibits SHM At any position: x = Acost y = Asint y P A y = Asint t x O Q xo x = Acost Check out the Virtual Physic Laboratory website on my webpage links. Look under Mechanics, then Simple Harmonic motion. This will help you visualize the motion. Note that the only difference between 12 this demo and the figure above is that this demo has the particle oscillating along the y axis rather than the x axis.