PHYS212 : Lectures

Subsections

• 1.2.1 The Differential Equation and its Solutions
• 1.2.2 Mass and Spring
• 1.2.3 Physical Pendulum

1.2 Simple Harmonic Oscillations

1.2.1 The Differential Equation and its Solutions

• In describing diverse systems, both classical and quantum mechanical, the differential equation
  

  $$\frac{d^2}{dt^2} \psi(t) = \ddot{\psi}(t) = -\omega^2 \psi(t)$$ (1.1)

  
  
  arises. We will consider several examples.

• The general solutions of Eq. 1.1 can be expressed in several forms. We will use three in particular,
  

  $$\psi(t) = \left\{ \begin{array}{l} A \cos (\omega t) + B \s... ...\left\{ C e^{i (\omega t + \phi)} \right\} \end{array}\right.$$ (1.2)

  
  

• We include the third form above, which is actually the same as the second form, because complex exponentials are sometimes easier to work with mathematically. In dealing with such cases, we will do calculations with the complex exponentials and take the real part at the end.

• The time dependent complex phase $\omega t + \phi$ tracks the angular position of $\psi(t)$ in the complex plane, which is sweeping out a circular orbit of radius $\vert\psi\vert$ with angular velocity $\omega$ (counterclockwise positive), starting with initial phase $\phi$ at $t = 0$.

• The constants $A$, $B$, $C$, and $\phi$ of Eq. 1.2 are determined by initial conditions.

• To find relationships between $A$, $B$, $C$, and $\phi$, we can use the identity
  
  $$\cos(x + y) = \cos x \cos y - \sin x \sin y$$
  
  
  which yields
  
  $$C \cos (\omega t + \phi) = C \cos \phi \cos(\omega t) - C \sin \phi \sin(\omega t)$$
  
  
  Comparing that with
  
  $$A \cos (\omega t) + B \sin (\omega t)$$
  
  
  reveals
  

  $$A = C \cos \phi$$

  $$B = -C \sin \phi.$$ (1.3)

  
  
  Eq.'s 1.3 can be solved for $C$ and $\phi$,
  

  $$C = \sqrt{A^2 + B^2}$$

  $$\tan \phi = -\frac{B}{A}$$ (1.4)

  
  

1.2.2 Mass and Spring

• The force exerted by an ideal spring is given by Hooke's law
  

  $$F(x) = - kx.$$ (1.5)

  
  

• Newton's Second Law...
  
  $$m \ddot{x} = -k x$$
  
  
  gives the equation of motion
  

  $$\ddot{x} = -\frac{k}{m} x$$ (1.6)

  
  
  which looks remarkably like Eq. 1.1.

• We can conclude
  

  $$\omega = \sqrt{\frac{k}{m}}.$$ (1.7)

  
  

• Exercise

  A frictionless mass-spring system with $k = 90$ N/m and $m = 10$ kg has an initial displacement of 0.10 m from its equilibrium position and an initial velocity of $v = 0.18$ m/s.

  (a) What is the frequency of oscillation of the system.

  (b) Express its angular position as a function of time in the three forms of Eq.'s 1.2. That is, find $\omega$, $A$, $B$, and $C$, and $\phi$.

1.2.3 Physical Pendulum

Figure 1.1: A physical pendulum.

• The restoring torque
  

  $$\tau = -Lmg \sin \theta$$ (1.8)

  
  
  is not a simple Hooke's law torque.

• However, the Taylor expansion of $\sin \theta$ is
  

  $$\sin \theta = \theta + \frac{\theta^3}{3!} + \frac{\theta^5}{5!} + ...$$ (1.9)

  
  
  It follws that to first order in $\theta$
  

  $$\sin \theta \approx \theta$$ (1.10)

  
  
  so for small angles $\theta « 1$ (in radians), we have
  

  $$\tau = -Lmg \theta.$$ (1.11)

  
  

• Newton's Second Law...
  
  $$I \ddot{\theta} = -Lmg \theta$$
  
  
  gives the equation of motion
  

  $$\ddot{\theta} = -\frac{Lmg}{I} \theta.$$ (1.12)

  
  
  which takes the form of Eq. 1.1.

• Comparing with Eq. 1.1, we conclude
  

  $$\omega = \sqrt{\frac{Lmg}{I}}.$$ (1.13)

  
  

Copyright © 2003-2009, Lewis A. Riley

Updated Wed Jan 18 09:51:28 2006