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S6.4 The SHM equation: applications

[A] The task and the strategy

[B] Application: mass on spring

A mass m rests on a horizontal frictionless surface, and is attached to a wall by a spring of force constant k. Establish the motion if the mass is pulled a small distance (stretching the spring from its natural length), and then released. [Yes: you have seen this example before; it is useful to help establish the sequence of steps in the strategy; you should work through it yourself first; then check what you have done against the solution that follows .]
Graphic - No title - shmqspring

Solution

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Solution

  • Identify stable equilibrium: the spring is unstretched

    Why?

    Why?

    Because the spring PE is lowest there!
     

  • Think about the motion near the point of stable equilibrium: the mass will oscillate back and forth
  • Choose the coordinate x: a displacement of the mass to right, measuring spring extension
Graphic - No title - shmaspring
  • Choose a system to which to apply Newton’s 2nd Law: the mass
  • Identify the force(s) acting on this system:
    F=-kx
  • Write down the equation of motion:
    \[ \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{m\ddot{x} = -k x} \]
  • Rearrange this equation in the standard form and identify ω:
    \[ \ddot{x} = -\omega^2 x \hspace{0.4cm}\mbox{\rm with}\hspace{0.4cm} \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\omega= \sqrt{\frac{k}{m}}} \]
 

Results

Key Point 6.4

A mass m subject to a linear spring-force of spring-constant k will exhibit SHM of angular frequency
\[ \omega= \sqrt{\frac{k}{m}} \]
Check It!

Commentary

Commentary

  • Real springs give rise to a linear restoring force only for small enough distortions.
  • So associated motion is SHM only for small amplitudes
  • For larger amplitudes a variety of things may happen...
 
MTime Evolution of a Four Spring, Three mass system

[C] Application: hydrogen molecule

A molecule of hydrogen can be modelled as two particles of mass mH=1.7×10-27 kg linked by a bond viewed as a spring of spring constant k=5.2×102 Nm-1 Establish the motion if the bond is compressed and then released.
Graphic - No title - shmqhydrogen

Solution

Solution

  • Identify stable equilibrium: the spring (’bond’ is unstretched)
  • Think about the behaviour near the equilibrium point:
    • In the absence of any external forces the CoM will remain at rest
    • The masses will oscillate back and forth in antiphase

      Auntie who?

      Auntie who?

      Saying they oscillate in ’antiphase’ means that when one is moving to the right the other is moving to the left. The two displacements could be written as
      \[ x_1= x_m \cos(\omega t) \hspace{0.2cm}\mbox{\rm and}\hspace{0.2cm} x_2= x_m \cos(\omega t +\pi) = -x_m \cos(\omega t) = -x_1 \]
      Their ’phase’ differs by π.
       
Graphic - No title - shmahydrogen
  • Choose the coordinate x: a displacement of (say) mass 2 to to the right.
  • Choose a system to which to apply Newton’s 2nd Law: the mass 2
  • Identify the force(s) acting on this system:
    • The spring length is changed by 2x

      since both masses move a distance x in opposite directions

    • The force on mass 2 is then
      F=-2kx
  • Write down the equation of motion:

    \[ \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{m\ddot{x} = -2k x} \]

  • Rearrange this equation in the standard form and identify ω:

    \[ \ddot{x} = -\omega^2 x \hspace{0.4cm}\mbox{\rm with}\hspace{0.4cm} \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\omega= \sqrt{\frac{2k}{m}}} \]

  • Insert numbers (…but not before now: recall Guideline 0.6)
    \begin{eqnarray*} \omega& =&\sqrt{\frac{2k}{m}}\\ & = & \sqrt{\frac{2 \times 5.2 \times 10^2 }{1.7 \times 10^{-27}}}\\ & = &7.8 \times 10 ^{14} rad \cdot s^{-1} \end{eqnarray*}
  • So the frequency is

    \[ f= \frac{\omega}{2\pi} = 1.2 \times 10 ^{14} Hz \]

 

Results

The molecular bond (spring) exhibits SHM with angular frequency

\[ \omega= \sqrt{\frac{2k}{m}} =7.8 \times 10 ^{14} rad \cdot s^{-1} \]

Commentary

Commentary

  • This frequency lies in the near-infrared part of the electromagnetic spectrum
  • The bond-stretching-frequencies of other molecules are generally lower because ’other’ atoms are heavier than hydrogen!
  • The frequency provides a distinctive signature of the molecule –apparent in the spectrum of light emitted by or absorbed by gases containing such molecules.
 
IWhat three can do that two cannot

[D] Application: simple pendulum

A simple pendulum comprises a point mass m suspended from a fixed point by a string of length L. Establish the motion if the bob is pulled to one side and then released.
Graphic - No title - shmqpend

Solution

Solution

  • Identify stable equilibrium: the bob hangs vertically

    Why?

    Why?

    Because the gravitational PE is lowest there!
     
  • Think about the motion near the point of stable equilibrium behaviour: the bob oscillates back and forth
  • For small angular displacement θ:
  • [–] horizontal displacement:
    x=LsinθLθ
  • [–] vertical displacement:
    y=L-Lcosθ
  • [–] Remembering Taylor expansion for cosines...
    yL(θ2)/2
Graphic - No title - shmapend

Help? Don’t miss this one!

Help? Don’t miss this one!

The approximations for sinθ and cosθ for small θ are amongst the most important examples of Taylor series.
 
  • So the vertical displacement will be negligible compared with the horizontal displacement,  

    Is that clear?

    Is that clear?

    For small θ the vertical component is proportional to θ2 and the horizontal component is proportional to θ. Thus
    \[ \frac{\mbox{vertical component of displacement}} {\mbox{horizontal component of displacement}} \propto \frac{\theta^2}{\theta} \propto \theta \]
    which is thus small.
     
  • Choose the coordinates: x (horizontal) and y (vertical).
  • Choose a system to which to apply Newton’s 2nd Law: the bob
  • Identify the forces acting on this system
    • The tension T along the string
    • The gravitational force mg downwards
  • Write down the equation of motion:
    • In the x-direction:
      \[ m\ddot{x} = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{-T\sin \theta \simeq -T\theta} \]
    • In the y-direction:
      \[ m\ddot{y} = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{T\cos \theta -mg \simeq T-mg} \]
    • If y is ’small’, then so is $\ddot{y}$ so the y-equation implies
      Tmg
    • Substitute into the x equation:

      \[ m\ddot{x} = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{-mg\theta \simeq -mg \frac{x}{L}} \]

      where the last step uses the fact that

      \[ x= L \tan {\theta} = L\frac{\sin \theta }{\cos \theta} \simeq L \theta \]
      for small θ using the Taylor expansions for sin and cos again.

  • Rearrange the x equation in the standard form and identify ω:
    \[ \ddot{x} = -\omega^2 x \hspace{0.2cm}\mbox{\rm with}\hspace{0.2cm} \omega= \sqrt{\frac{g}{L}} \]
 

Results

Key Point 6.5

A pendulum comprising a point mass suspended from a fixed point by a light string of length L exhibits SHM of angular frequency
\[ \omega= \sqrt{\frac{g}{L}} \]
provided the angular displacement from the vertical is small.
Check It!

Commentary

Commentary

  • A ’light’ string has mass small compared to that of the bob
  • For larger θ the behaviour is not SHM (Q7.11)
  • It can be far-from-simple (S6.7)
 
MAnimated Pendulum

[E] Application: physical pendulum

A physical pendulum is made from a sheet of steel, of mass m, pivoted about an axis through a point a distance d from its centre of mass. The moment of inertia about this axis is I. Establish the motion if the sheet is pushed to one side and then released.
Graphic - No title - shmqcomppend

Solution

Solution

  • Identify stable equilibrium:

    the plate hangs with its CoM immediately under the pivot

  • Think about the motion near the point of stable eqquilibrium behaviour: the plate swings back and forth  through the position of equilibrium
  • Choose the coordinate to play the role of ’x’: the angular displacement θ
Graphic - No title - shmacomppend

Comment

Comment

Remember that in the SHM equation the oscillating quantity doesn’t have to be a displacement; this is the first example we have met where it isn’t. While we could choose to call the angle ’x’ if we really wanted, it is better to use a symbol that reminds us what kind of quantity we are dealing with.
 
  • Choose a system to which to apply Newton’s 2nd Law:the plate
  • Identify the forces acting on this system:
    • The weight of the plate acting downwards, through the CoM
    • The reaction force exerted by the pivot acting through the pivot
  • Write down the equation of motion:
    • We need the angular form of Newton’s 2nd Law (because we have chosen to consider an angular coordinate! (Key Point 5.19)
      \[ \tau = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{I \alpha = I \frac{d^2 \theta}{dt^2}} \]
      where I is the moment of inertia about the pivot axis, and τ is the torque about this axis.
    • The torque due to the weight has magnitude
      τ=mgdsinθ
    • The equation of motion is (after some thought about signs

      \[ I \ddot{\theta} = - \mbox{ (minus!) }mgd\sin \theta \]

      Do you want to understand the minus sign?

      Do you want to understand the minus sign?

      You do? Good!

      First note that if we missed this minus sign we would rapidly find that we had done something wrong. Instead of the standard SHM equation we would find that θ had the equation of motion

      \[ \ddot{\theta} = + \mbox{ (!) } \omega ^2 \theta \]
      This equation gives an altogether different behaviour: θ would go on growing at an ever greater rate!

      This physical nonsense comes about if we fail to capture the fact that the torque has a sense (direction): it is a restoring torque that always acts back towards equilibrium. The direction of a torque is expressed through its vector character, which is not apparent in the equation τ=Iα. In this equation the LHS gives the component of the torque vector along the chosen axis; the RHS gives the rate of change of the angular momentum along the same axis.

      The component of the torque along the axis into the plane is:

      τ=mgdsinθ
      ie it is into the plane (positive) if θ is positive, and out of the plane (negative) if θ is negative.

      The angular momentum along the axis into the plane is:

      \[ L = - I \dot{\theta} \]
      ie it is out of the plane (negative) if $\dot{\theta}$ is positive, and into the plane (postive) if $\dot{\theta}$ is negative.

      Then the equation

      \[ \tau = \frac{dL}{dt} \]
      does indeed imply
      \[ I \ddot{\theta} = - \mbox{ (!) } mgd\sin \theta \]
      Phew!

       

    • Now suppose that θ is small so we can use the Taylor Series expansion for sinθ:
      \[ I \ddot{\theta} = - mgd\theta \]
  • Rearrange the θ equation in the standard form and identify ω:
    \[ \ddot{\theta} = -\omega^2 \theta \hspace{0.2cm}\mbox{\rm with}\hspace{0.2cm} \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\omega= \sqrt{\frac{mgd}{I}}} \]
 

Results

The physical pendulum exhibits SHM of angular frequency

(6.8)
\begin{displaymath} \omega= \sqrt{\frac{mgd}{I}} \end{displaymath}

provided the angular displacement from the vertical is small.

Commentary

Commentary

The simple pendulum result is here as a special case with:
  • d identified as L
  • I identified as mL2
Check It!
  • Are the units OK?
  • Does it make sense?
 

Review

Review

Summary of applications:

Mass-on-spring:

\[ \omega= \sqrt{\frac{k}{m}} \]

Graphic - No title - shmqspring

Hydrogen molecule:

\[ \omega= \sqrt{\frac{2k}{m}} \]

Graphic - No title - shmqhydrogen

Simple pendulum:

\[ \omega= \sqrt{\frac{g}{L}} \]

Graphic - No title - shmqpend

Physical pendulum:

\[ \omega= \sqrt{\frac{mgd}{I}} \]

Graphic - No title - shmqcomppend
 

[F] The core physics of SHM: stiffness versus inertia

Example

Example

Star oscillations

Graphic - No title - staroscillations
  • A typical star exhibits oscillations perpendicular to the galactic plane in addition to its motion across the plane.
  • These oscillations are roughly simple harmonic
  • The parameters controlling the frequency might be
    • the galactic density ρ ?
    • the gravitational constant G ?
    • the star mass M ?
  • A guess at the stiffness:
    stiffnessGρM ?
  • A guess at the inertia:
    inertiaM?
  • The guesstimate for the frequency:
    \[ \omega = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\sqrt{\frac{G\rho M}{M}} =\sqrt{G\rho}} \]
  • The units check:
    \[ \sqrt{G\rho} \rightarrow \sqrt{ kg \cdot m \cdot s^{-2} \cdot m^2 \cdot kg ^{-2} \cdot kg \cdot m^{-3}} \rightarrow s^{-1} \]
  • A full analysis gives:
    \[ \omega = \sqrt{2\pi G\rho} \]
  • Setting G6×10-11 and ρ10-18 (in SI units) leads to a period of order 108 years.
 
TPendulum Period vs Amplitude

Learning Resources

Textbook: HRW 15.5-15.6
Course Questions:
Self-Test Questions: