SHM-1 Simple Harmonic Motion A pendulum, a mass on a spring, and many other kinds of oscillators exhibit a special kind of oscillatory motion called Simple Harmonic Motion (SHM). SHM occurs whenever : i. there is a restoring force proportional to the displacement from equilibrium: F ∝ −x ii. the potential energy is proportional to the square of the displacement: PE ∝ x2 iii. the period T or frequency f = 1 / T is independent of the amplitude of the motion. iv. the position x, the velocity v, and the acceleration a are all sinusoidal in time. x vt t(Sinusoidal means sine, cosine, or anything in between.) As we will see, any one of these four properties guarantees the other three. If one of these 4 things is true, then the oscillator is a simple harmonic oscillator and all 4 things must be true. Not every kind of oscillation is SHM. For instance, a perfectly elastic ball bouncing up and down on a floor: the ball's position (height) is oscillating up and down, but none of the 4 conditions above is satisfied, so this is not an example of SHM. A mass on a spring is the simplest kind of Simple Harmonic Oscillator. k Hooke's Law: Fspring = – k x (–) sign because direction of Fspring is opposite to the direction of displacement vector x (bold font indicates vector) k = spring constant = stiffness, units [k] = N / m Big k = stiff spring Definition: amplitude A = |xmax| = |xmin|. Mass oscillates between extreme positions x = +A and x = –A x relaxed: x = 0 mFrestorex Frestorex mx –A+A04/10/2009 © University of Colorado at BoulderSHM-2 Notice that Hooke's Law (F = − kx) is condition i : restoring force proportional to the displacement from equilibrium. We showed previously (Work and Energy Chapter) that for a spring obeying Hooke's Law, the potential energy is U = (1/2)kx2 , which is condition ii. Also, in the chapter on Conservation of Energy, we showed that F = −dU/dx, from which it follows that condition ii implies condition i. Thus, Hooke's Law and quadratic PE (U ∝ x2) are equivalent. We now show that Hooke's Law guarantees conditions iii (period independent of amplitude) and iv (sinusoidal motion). We begin by deriving the differential equation for SHM. A differential equation is simply an equation containing a derivative. Since the motion is 1D, we can drop the vector arrows and use sign to indicate direction. net net2222FmaandF kx makdx kadv/dtdx/dt xdt m==−⇒== ⇒ =−x=− The constants k and m and both positive, so the k/m is always positive, always. For notational convenience, we write 2k/m=ω . (The square on the ω reminds us that ω2 is always positive.) The differential equation becomes 222dxxdt=−ω (equation of SHM) This is the differential equation for SHM. We seek a solution x = x(t) to this equation, a function x = x(t) whose second time derivative is the function x(t) multiplied by a negative constant (−ω2 = −k/m). The way you solve differential equations is the same way you solve integrals: you guess the solution and then check that the solution works. ()x(t) Acos t=ω+ϕ, Based on observation, we guess a sinusoidal solution: kmω=. where A, ϕ are any constants and (as we'll show) A = amplitude: x oscillates between +A and –A ϕ = phase constant (more on this later) Danger: ωt and ϕ have units of radians (not degrees). So set your calculators to radians when using this formula. Just as with circular motion, the angular frequency ω for SHM is related to the period by 22fTπω= π = , T = period. 4/10/2009 © University of Colorado at BoulderSHM-3 (What does SHM have to do with circular motion? We'll see later.) Let's check that is a solution of the SHM equation. (x(t) Acos t=ω+)ϕ Taking the first derivative dx/dt , we get ()dxv(t) A sin tdt==− ω ω +ϕ. Here, we've used the Chain Rule: ()ddcos()dcos t , ( t )dt d d tsin sin( t )θθω+ϕ = θ=ω +ϕθ=− θ⋅ω= −ω ω +ϕ Taking a second derivative, we get ()()[]222222222dx dv da(t) A sin t A cos( t )dt dt dtdxAcos( t )dtdxxdt===−ωω+ϕ=−ωω+=−ω ω +ϕ=−ωϕ This is the SHM equation, with 2kk,mmω= ω= We have shown that our assumed solution is indeed a solution of the SHM equation. (I leave to the mathematicians to show that this solution is unique. Physicists seldom worry about that kind of thing, since we know that nature usually provides only one solution for physical systems, such as masses on springs.) We have also shown condition iv: x, v, and a are all sinusoidal functions of time: ()2x(t) Acos tv(t) A sin( t )a(t) A cos( t )=ω+ϕ=− ω ω+ϕ=− ω ω+ϕ The period T is given by k2 mT2mT kπω= = ⇒ = π. We see that T does not depend on the amplitude A (condition iii). Let's first try to make sense of k/mω=: big ω means small T which means rapid oscillations. According to the formula, we get a big ω when k is big and m is small. This makes sense: a big k (stiff spring) and a small mass m will indeed produce very rapid oscillations and a big ω. 4/10/2009 © University of Colorado at BoulderSHM-4 A closer look at x(t) = A cos(ωt+ϕ) Let's review the sine and cosine functions and their relation to the unit circle. We often define the sine and cosine functions this way: adjcoshypθ= hypotenuseopposite θ oppsinhypθ= adjacent This way of defining sine and cosine is correct but incomplete. It is hard to see from this definition how to get the sine or cosine of an angle greater than 90o. A more complete way of defining sine and cosine, a way that gives the value of the sine and cosine for any angle, is this: Draw a unit circle (a circle of radius r = 1) centered on the origin of the x-y axes as shown: point (x, y) Define sine and cosine as adj xcos xhyp 1θ= = = opp ysin yhyp 1θ= = = This way of defining sin and cos allows us to compute the sin or cos of any angle at all. For instance, suppose the angle is θ = 210o. Then the diagram looks like this: The point on the unit circle is in the third quadrant, where both x and y are negative. So both cosθ = x and sinθ = y are negative For any angle θ, even angles bigger than 360o (more than once around the circle), we can always compute sin and cos. When we plot sin and cos vs angle θ, we get functions that oscillate between +1 and –1 like so: xyθr = 1θxy1point (x, y)4/10/2009 © University of Colorado at BoulderSHM-5 We will almost always measure angle θ in