PHYS 211 Lecture 5 - Oscillations: simple harmonic motion 5 - 1 Lecture 5 - Oscillations: simple harmonic motion Text: Fowles and Cassiday, Chap. 3 Consider a power series expansion for the potential energy of an object 2 V(x) = Vo + V1x + V2x + ..., where x is the displacement of an object from its equilibrium position, and Vi, i = 0,1,2... are fixed coefficients. The leading order term in this series is unimportant to the dynamics of the object, since F = -dV/dx and the derivative of a constant vanishes. Further, if we require the equilibrium position x = 0 to be a true minimum in the energy, then V1 = 0. This doesn’t mean that potentials with odd powers in x must vanish, but just says that their minimum is not at x = 0. Thus, the simplest function that one can write is quadratic: 2 V = V2x [A slightly more complicated variant is |x| = (x2)1/2, which is not smooth at x = 0]. For small excursions from equilibrium, the quadratic term is often the leading-order piece of more complex functions such as the Morse or Lennard-Jones potentials. Quadratic potentials correspond to the familiar Hooke’s Law of ideal springs V(x) = kx 2/2 => F(x) = -dV/dx = -(2/2)kx = -kx, where k is the spring constant. Objects subject to Hooke’s Law exhibit oscillatory motion, as can be seen by solving the differential equation arising from Newton’s 2nd law: F = ma => -kx = m(d 2x/dt 2) or d 2x /dt 2 + (k/m)x = 0. We solved this equation in PHYS 120: x(t) = A sin( ot + o) Other functional forms such as cosine can be changed into this form using a suitable choice of the phase angle o. The angular frequency of the motion, o is given by 1/2 o = (k/m) Note that subscripts have been placed on and , for reasons that will become apparent later. With o determined by the intrinsic characteristics of the system (k, m), the constants A and o, [the amplitude and phase angles, respectively] can be determined by x,v and some value(s) of t. For examples, if we know x(0) = xi and v(0) = vi, then x(0) = A sin o = xi v(0) = dx /dtút = 0 = + oA cos( ot + o)út = 0 = oA cos o © 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited. PHYS 211 Lecture 5 - Oscillations: simple harmonic motion 5 - 2 Dividing, x /v = sin / [ cos ] = tan / i i o o o o o 2 2 2 2 2 2 But we also have A sin o + A cos o = xi + (vi/ o) 2 2 2 =>A = xi + (vi/ o) We can easily calculate the instantaneous kinetic energy K and potential energy V : 2 2 2 2 2 K = 1/2 mv = 1/2 m [ oA cos( ot + o)] = 1/2 m o A cos ( ot + o) 2 2 2 V = 1/2 kx = 1/2 kA sin ( ot + o) 2 But o = k/m 2 2 => K = 1/2 m(k/m)A cos ( ot + o) 2 2 = 1/2 kA cos ( ot + o) From these, the total energy E = K + V is 2 2 2 2 E = K + V = 1/2 kA cos ( ot + o) + 1/2 kA sin ( ot + o) = 1/2 kA2 In other words: (i) E is constant, independent of t (ii) E is equal to the potential energy at the maximal extension x = A, where the velocity v vanishes. Now, the instantaneous values of K and V change with time K , V ot o What about the average values <K> and <V>? To construct the average of a function over some variable t we must evaluate <f> = òf dt / òdt i.e., we sum over the values of the function, then divide by the range of the independent variable. © 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited. PHYS 211 Lecture 5 - Oscillations: simple harmonic motion 5 - 3 For convenience, we choose the time interval in our evaluation of <K> to be the period T. Further, we drop the o since the average over one complete cycle is independent of where the cycle starts. Thus, 2 2 2 2 2 <K> = (m o A / 2) ò sin ( ot)dt / T = (kA / 2T) ò sin ( ot)dt 2 2 2 2 <V> = (kA / 2) ò cos ( ot)dt / T = (kA / 2T) ò cos ( ot)dt Changing variables to = ot 2 2 2 2 <K> = (kA / 2 oT) ò sin d <V> = (kA / 2 oT) ò cos d Now o = 2p /T by definition, since the motion executes a 2p change in phase angle over the period. Thus <K> = (kA2 / 4p ) ò sin2 d with 0 £ £ 2p <V> = (kA2 / 4p ) ò cos2 d These integrals have the same value, which can be obtained from 2p = ò 1d = ò (sin2 + cos2 ) d = 2 ò sin2 d => ò sin2 d = ò cos2 d = p Then <K> = <V> = (kA2 / 4p )p = kA2 / 4 Not unexpectedly, we find <K> + <V> = 2 • (kA2 / 4) = kA2 / 2 = E because K and V depend on independent quantities. One last quantity that we want to evaluate is the displacement from equilibrium. Because the motion is oscillatory around x = 0, we expect <x> = A <sin( ot)> = 0. The mean square displacement can be obtained by noting <V> = (k/2) <x2> such that <x2> = (2/k) <V> = (2/k) • (kA2 / 4) or <x2> = A2 / 2. This is greater than a random distribution in x (where any location is equally likely); <x 2> = ò x 2 dx / òdx = (1/3)A 3 / A = A 2 / 3 (random) Why? Because the oscillator moves through the regions of small x rather quickly, but moves around the extrema at |x| = A only slowly, thus spending more time at large x. © 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited..