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PHYS 211 Lecture 5 - Oscillations: simple harmonic 5 - 1

Lecture 5 - Oscillations:

Text: Fowles and Cassiday, Chap. 3

Consider a power series expansion for the potential of an object

2 V(x) = Vo + V1x + V2x + ..., where x is the of an object from its equilibrium , and Vi, i = 0,1,2... are fixed coefficients. The leading order term in this series is unimportant to the of the object, since F = -dV/dx and the 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 ’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 o. The angular 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 and phase , 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,

xi/vi = sin o / [ o cos o] = tan 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 K and 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 v vanishes.

Now, the instantaneous values of K and V change with

K , V

ot o

What about the average values and ? To construct the average of a function over some variable t we must evaluate = 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 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 = (m o A / 2) ò sin ( ot)dt / T = (kA / 2T) ò sin ( ot)dt

2 2 2 2 = (kA / 2) ò cos ( ot)dt / T = (kA / 2T) ò cos ( ot)dt

Changing variables to = ot 2 2 2 2 = (kA / 2 oT) ò sin d = (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 = (kA2 / 4p ) ò sin2 d with 0 £ £ 2p = (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 = = (kA2 / 4p )p = kA2 / 4

Not unexpectedly, we find + = 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

= A = 0.

The mean square displacement can be obtained by noting = (k/2) such that = (2/k) = (2/k) • (kA2 / 4) or = A2 / 2.

This is greater than a random distribution in x (where any location is equally likely);

= 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.