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S6.3 The SHM equation: a general tour

Preamble

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Preamble

  • We consider the general properties of the SHM equation (Key Point 6.2)
  • We prepare for the mathematics by considering the SHM equation qualitatively.
  • The key features of this equation of motion are that
    • $\ddot{x}$ depends on x  –we have previously considered only situations where the force driving the motion is constant, at least in magnitude.
    • $\ddot{x}$ is always opposite to x, back towards the equilibrium position
    • $\ddot{x}$ is linearly proportional to x, vanishing at x=0
    • the motion must conserve energy (section S3.7).

      More?

      More?

      We must expect that this is so, since the underlying force F(x) is conservative, ie associated with a potential energy function U(x). But we will actually show it is so.
       

  • Energy-conservation tells us that
    • The coordinate x cannot grow arbitrarily big – this would entail gaining some energy
    • x cannot simply come to rest at x=0  –this would entail losing energy
  • We are left only with oscillatory motion about x=0 – which is what we would expect from our experience of near stable equilibrium systems
 

[A] The solution: algebra

Key Point 6.3

The general solution to the SHM equation $\ddot{x}=-\omega^2 x$ is
x(t)=xmcos(ωt+φ)
where xm and φ are constants determined by the initial conditions.

Cosine or sine?

Cosine or sine?

There is no deep reason for choosing a cosine function here rather than a sine function. For any angle θ we can write
cos(θ)=sin(θ+π/2)
So we could choose to write the displacement as a sine function, provided we added π/2 to our assignment of the initial phase φ.
 

Commentary

Commentary

  • To show that
    x=xmcos(ωt+φ)
    is a solution of the SHM equation we differentiate it with respect to time giving the speed:
    \[ v=\dot{x} = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{-\omega x_m \sin(\omega t + \phi)} \]
  • We differentiate again to give the acceleration
    \[ a=\ddot{x} =\dot{v} = -\omega^2 x_m \cos(\omega t + \phi) = - \omega^2 x \]
    where the last step uses the proposed solution again and recovers the SHM equation.
  • Since the proposed solution works (satisfies the SHM equation) for any choices of xm and φ it is the general solution to this equation.
  • These two constants are fixed by the initial conditions …whatever is happening at the time we call t=0: we will come back to this shortly
  • In this system, knowledge of the initial conditions allows us to predict the future

    Should we believe that promise?

    Should we believe that promise?

    The short answer is that we should (at least) be cautious; for a longer answer see S6.7
     

 

[B] Visualisation: the role of amplitude and phase

  • xm is called the amplitude of the motion
  • Since the cosine function varies between ±1 the value of x varies between ±xm
  • The figure shows x(t) for two values of xm, with one twice the other.
  • The ’natural’ unit of time is the period
    T=2π/ω
Graphic - No title - roleofamp
  • φ is called the initial phase of the motion, or the phase constant.
  • We can write
    ωt+φ=ω(t+φ/ω)
  • Thus a solution with a given φ is identical to a zero-φ solution shifted bodily along the time axis.
  • The figure shows x(t) for two values of φ differing by φ=2π/6.
Graphic - No title - roleofphase
  • There is another useful way of visualising the solution to the SHM equation and the role of xm and φ.
  • Think of a particle moving with constant angular velocity ω anticlockwise in a circle of radius xm
  • Focus on the radial vector from the centre to the particle.
Graphic - No title - circshm
ISHM and circular motion
ASimilar motions
APhases of the moon
MHarmonic Oscillations

[C] Visualisation: displacement, velocity and acceleration

  • The accompanying figure shows the behaviour of

    • the displacement:
      x(t)=xmcos(ωt+φ)
    • the velocity:
      \[ v(t)=\dot{x}=-\omega x_m \sin(\omega t + \phi) \]
    • the acceleration:
      \[ a(t)= \ddot{x}=-\omega^2 x_m \cos(\omega t + \phi) = - \omega^2 x (t) \]

    on the same time-axis.

Graphic - No title - xva_t

Terminology

Terminology

As noted earlier x may represent something other than a ’displacement’; then calling $\dot{x}$ a velocity and $\ddot{x}$ an acceleration is a little inappropriate. We do it nevertheless.
 

[D] How initial conditions fix amplitude and phase

Proof

Proof

  • We have for the displacement and the velocity
    \[ x(t)=x_m\cos(\omega t + \phi) \hspace{0.2cm}\mbox{\rm and}\hspace{0.2cm} v(t) = -\omega x_m \sin(\omega t + \phi) \]
  • We set t=0
    \[ x(0)=x_m\cos\phi \hspace{0.4cm}\mbox{\rm and}\hspace{0.4cm} \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{v(0) = -\omega x_m \sin \phi} \]
  • Dividing the second equation by the first gives
    \[ -\frac{v(0)}{\omega x(0)} = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{\sin\phi}{\cos\phi} = \tan{\phi}} \]
  • Rewriting the two equations in the form
    \[ \frac{x(0)}{x_m}=\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\cos\phi} \]
    and
    \[ \frac{v(0)}{\omega x_m} = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{-\sin\phi} \]
  • Squaring and adding
    \[ \left[\frac{x(0)}{x_m}\right]^2 + \left[\frac{v(0)}{\omega x_m}\right]^2 = \cos^2\phi +\sin^2\phi =1 \]
  • Rearranging gives
    \[ \left[x(0)\right]^2 + \left[\frac{v(0)}{\omega }\right]^2 = x_m^2 \]
 
Check It!

[E] All about the ’frequency’

Commentary

Commentary

  • It is helpful to see how the relationship between T and ω arises.
  • Recall the equation for the displacement
    x(t)=xmcos(ωt+φ)
  • Then deduce that a time T later the displacement is
    x(t+T)=xmcos(ω(t+T)+φ)=xmcos(ωt+ωT+φ)
  • Setting T=2π/ω then gives
    x(t+T)=xmcos(ωt+2π+φ)=xmcos(ωt+φ)=x(t)
  • Thus waiting for T=2π/ω takes us back to the same displacement
  • The same argument applies to $\dot{x}$ and $\ddot{x}$  (velocity and acceleration) which are also sine or cosine functions of ωt+φ
  • Waiting for T thus restores the status quo completely.
 
TStarting a Pendulum With Force

Learning Resources

Textbook: HRW 15.3
Course Questions:
Self-Test Questions: