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S6.5 Energy conservation in SHM

[A] Why SHM entails energy conservation

Analysis

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Analysis

  • Consider the mass-spring system once more.
  • The potential energy is controlled by the displacement x:
    \[ U = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{1}{2} k x^2} \]
  • The kinetic energy is controlled by the velocity $\dot{x}$:
    \[ K = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{1}{2} m \dot{x}^2} \]
  • From our solution to the general SHM equation
    \[ x=\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{x_m \cos(\omega t + \phi)} \hspace{0.2cm}\mbox{\rm and}\hspace{0.2cm} \dot{x}=\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{-x_m \omega \sin(\omega t + \phi)} \]
  • The PE and KE at time t are then

    \[ U(t)=\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{[k x_m^2/2] \cos^2(\omega t + \phi)} \]
    and
    \[ K(t)=\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{[m x_m ^2\omega^2/2] \sin^2(\omega t + \phi)} \]

  • From Key Point 6.4 we have
    \[ \omega ^2 = \frac{k}{m} \]
  • Thus the total energy at instant t is
    \begin{eqnarray*} E(t) &=& U(t) + K(t) \\ &=& [k x_m^2/2] \cos^2 (\omega t + \phi) + [m x_m^2\omega^2/2] \sin^2(\omega t + \phi)\\ &=& [k x_m^2/2] \cos^2 (\omega t + \phi) + [k x_m^2/2] \sin^2(\omega t + \phi)\\ &=& [k x_m^2/2] [\cos^2 (\omega t + \phi) + \sin^2 (\omega t + \phi)]\\ &=& k x_m^2/2 \end{eqnarray*}
  • We may conclude as follows:
 

Key Point 6.7

In a mechanical system exhibiting SHM the sum of the kinetic and potential energy remains constant at its initial value:
\[ K + U = E = \frac{1}{2} m x_m^2 \omega^2 =\frac{1}{2} k x_m^2 \]

Commentary

Commentary

  • Note that E is a constant which does depend on the initial conditions.
  • We can set the system going with any energy we choose by adjusting the initial position, or the initial speed.
  • The total energy can be recognised as the maximum kinetic energy (or maximum potential energy) the system has in its oscillation cycle:
    \[ E= \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{KE_{\mbox{\rm max}}= PE_{\mbox{\rm max}}} \]
 

[B] Variation of K and U during the SHM cycle

  • The exchange of KE & PE as function of t is shown in the figure
  • Each energy has a maximum twice in each cycle.
Graphic - No title - econ1_t
  • The exchange of KE & PE as function of x. is shown in the figure
  • One can make this picture less abstract by thinking of a particle sliding on a frictionless parabolic surface.
Graphic - No title - econ2_x

IWell, well...quantum tunneling at last

[C] From energy conservation to the SHM equation

Analysis

Analysis

  • Suppose we have a system whose potential and kinetic energies can be written in the form

    \[ U= \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{1}{2} S x^2} \hspace{0.2cm}\mbox{\rm and}\hspace{0.2cm} K= \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{1}{2} I v^2} \]

    where S and I are constants, while (for this argument) it will help to use v for $\dot{x}$.

  • The mass spring system is a special case of this general form
  • This time we assume energy conservation so that
    \[ E = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{U + K = \frac{1}{2} S x^2 +\frac{1}{2} I v^2 =\mbox{ constant}} \]
  • Now differentiate this result with respect to time.
  • We need to use the chain rule:

    \[ \frac{d}{dt} x^2 = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{d}{dx} x^2 \times \frac{dx}{dt}} = 2xv \]
    and

    \[ \frac{d}{dt}v ^2 = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{d}{dv} v^2 \times \frac{dv}{dt}} = 2va \]

    where a is the acceleration.

  • Now the statement E is constant is equivalent to
    \[ \frac{dE}{dt} \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{=0} \]
  • Combining these results gives:
    \[ 0 = \frac{dE}{dt} = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{1}{2} S \times 2 x v + \frac{1}{2} I \times 2 va} \]
  • Cancelling common factors and setting $a=\ddot{x}$ gives
    \[ \ddot{x}= -\omega^2 x \hspace{0.4cm}\mbox{\rm where}\hspace{0.4cm} \omega= \sqrt{\frac{S}{I}} \]
    which is the SHM equation.
  • We conclude the following:
 

Key Point 6.8

If the total energy of a system can be written as
\[ E = U + K = \frac{1}{2} S x^2 +\frac{1}{2} I \dot{x}^2 =\mbox{ constant} \]
where S and I are constants, then the coordinate x will exhibit SHM with frequency
\[ \omega= \sqrt{\frac{S}{I}} \]

Commentary

Commentary

  • S and I may be thought of as measures of ’stiffness’ and ’inertia’
  • Expressing the energy of a system this way allows one to deal with SHM without having to think about forces!
 

Mechanics without forces?

Mechanics without forces?

Mechanics without forces seems a contradiction in terms. But classical mechanics can be formulated entirely without forces. These formulations (known as Lagrangian Mechanics and Hamiltionian Mechanics) provide the most appropriate techniques for dealing with harder problems. Some of you will meet these methods in later years. One nice thing to look forward to: they allow you to dispense with vectors!
 

TAverage Energy

Learning Resources

Textbook: HRW 15.4
Course Questions:
Self-Test Questions: