Oxbridge Notes | Oscillating Systems

Notes for Oscillating Systems (Part 1)

Undamped simple harmonic motion:

This is an ideal case of simple harmonic motion in which the only force acting on an object is a restoring force towards equilibrium:

Letting k/m=ω₀² such that the equation of motion is not specific to the mass/spring system:

$$\boxed{\mathbf{\therefore}\ddot{\mathbf{x}}\mathbf{+}\mathbf{\omega}_{\mathbf{0}}^{\mathbf{2}}\mathbf{x = 0}}$$

Which is the equation of motion for undamped SHM in general. The solutions to this differential equation can be expressed in two forms:

Complex exponential form:	Real (physical) sinusoidal form:
Let $x = Ae^{\alpha t}\therefore\ddot{x} = A\alpha^{2}e^{\alpha t}$: Ae^αt(α²+ω₀²)=0,Ae^αt0 α=iω₀ $$\boxed{\mathbf{\therefore x = A}\mathbf{e}^{\mathbf{i}\mathbf{\omega}_{\mathbf{0}}\mathbf{t}}\mathbf{+ B}\mathbf{e}^{\mathbf{- i}\mathbf{\omega}_{\mathbf{0}}\mathbf{t}}}$$ Where A and B are in general complex. ω₀ defines the period of the motion and is thus called the ‘angular frequency’: $$T = \frac{2\pi}{\omega_{0}} \rightarrow \omega_{0} = \frac{2\pi}{T}$$ ω₀=2πυ Where υ is the oscillatory frequency of the motion and T is the oscillatory period.	Due to the equivalence of exponentials and sinusoids: x=(A+B)cosω₀t+(AB)isinω₀t: We are only interested in the real part of this, so we are looking for the real parts of A+B and (AB)i: If a₀= amplitude and ϕ= phase, then: When t=0: x=A+B Re{A+B}=a₀cosϕ When t=π/2ω₀: x=(AB)i Re{(A+B)i}=a₀sinϕ Re{x}=a₀cosϕcosω₀t+a₀sinϕsinω₀t $$\boxed{\mathbf{\therefore}\mathbf{Re}\left\{ \mathbf{x} \right\}\mathbf{=}\mathbf{a}_{\mathbf{0}}\mathbf{\cos}{\mathbf{(}\mathbf{\omega}_{\mathbf{0}}\mathbf{t + \phi)}}}$$ Or Re{x}=Ccosω₀t+Dsinω₀t $$a_{0} = \sqrt{C^{2} + D^{2}},\phi = \tan^{- 1}{- \frac{D}{C}}$$

Let $x = Ae^{\alpha t}\therefore\ddot{x} = A\alpha^{2}e^{\alpha t}$:

Ae^αt(α²+ω₀²)=0,Ae^αt0

α=iω₀

$$\boxed{\mathbf{\therefore x = A}\mathbf{e}^{\mathbf{i}\mathbf{\omega}_{\mathbf{0}}\mathbf{t}}\mathbf{+ B}\mathbf{e}^{\mathbf{- i}\mathbf{\omega}_{\mathbf{0}}\mathbf{t}}}$$

Where A and B are in general complex.

ω₀ defines the period of the motion and is thus called the ‘angular frequency’:

$$T = \frac{2\pi}{\omega_{0}} \rightarrow \omega_{0} = \frac{2\pi}{T}$$

ω₀=2πυ

Where υ is the oscillatory frequency of the motion and T is the oscillatory period.

Due to the equivalence of exponentials and sinusoids: x=(A+B)cosω₀t+(AB)isinω₀t:

We are only interested in the real part of this, so we are looking for the real parts of A+B and (AB)i:

If a₀= amplitude and ϕ= phase, then:

When t=0:

x=A+B

Re{A+B}=a₀cosϕ

When t=π/2ω₀:

x=(AB)i

Re{(A+B)i}=a₀sinϕ

Re{x}=a₀cosϕcosω₀t+a₀sinϕsinω₀t

$$\boxed{\mathbf{\therefore}\mathbf{Re}\left\{ \mathbf{x} \right\}\mathbf{=}\mathbf{a}_{\mathbf{0}}\mathbf{\cos}{\mathbf{(}\mathbf{\omega}_{\mathbf{0}}\mathbf{t + \phi)}}}$$

Re{x}=Ccosω₀t+Dsinω₀t

$$a_{0} = \sqrt{C^{2} + D^{2}},\phi = \tan^{- 1}{- \frac{D}{C}}$$

Mechanical systems displaying undamped SHM:

All mechanical systems displaying SHM have the defining equation $\ddot{x} + \omega_{0}^{2}x = 0$, but ω₀² is defined differently for each:

Simple pendulum:

$$- F_{T} = m\ddot{s} = ml\ddot{\theta}$$

$$\therefore - mg\sin\theta = ml\ddot{\theta}$$

As long as θ is small, sinθθ:

$$\therefore - g\theta = l\ddot{\theta}$$

$$\mathbf{\therefore}\ddot{\mathbf{\theta}}\mathbf{+}\frac{\mathbf{g}}{\mathbf{l}}\mathbf{x = 0}$$

$$\therefore\omega_{0}^{2} = \frac{g}{l}$$

Torsional pendulum:

$$G = I\ddot{\theta} = - \tau\theta$$

$$\mathbf{\therefore}\ddot{\mathbf{\theta}}\mathbf{+}\frac{\mathbf{\tau}}{\mathbf{I}}\mathbf{x = 0}$$

$$\therefore\omega_{0}^{2} = \frac{\tau}{I}$$

Velocity, acceleration and phasor diagrams:

Phasor diagrams represent the phase relationships between displacement, velocity and acceleration for an oscillating system:

Displacement:	Velocity:	Acceleration:
x=a₀cos(ω₀t+ϕ)	$$\dot{x} = \frac{d}{dt}(x) = - a_{0}\omega_{0}\sin{(\omega_{0}t + \phi)}$$	$$\ddot{x} = \frac{d^{2}}{dt^{2}}(x) = - a_{0}\omega_{0}^{2}\cos{(\omega_{0}t + \phi)}$$
Phase relationship:	Phase relationship:	Phase relationship:
x=a₀cos(ω₀t+ϕ)	$$\dot{x} = a_{0}\cos\left( \omega_{0}t + \phi + \frac{\pi}{2} \right)$$	$$\ddot{x} = a_{0}\cos\left( \omega_{0}t + \phi + \pi \right)$$

The successive phase difference between displacement, velocity and acceleration is therefore π/2:

Argand plane:	Phasor diagram:

Velocity:

Argand plane:	Phasor diagram:

Acceleration:

Argand plane:	Phasor diagram:

Superposition of simple harmonic motion:

The equation of motion for SHM is linear, hence solutions of this equation can be superimposed to give another solution:

This can be proven for the general case of superimposing two simple harmonic oscillations of the same frequency; SHM₁ and SHM₂:

$${SHM}_{1}:\ddot{x_{1}} + \omega_{0}x_{1} = 0,\ \ {SHM}_{2}:\ddot{x_{2}} + \omega_{0}x_{2} = 0$$

Where $x_{1} + x_{2} = x\therefore\ddot{x_{1}} + \ddot{x_{2}} = \ddot{x}$:

$$\lambda_{1}{SHM}_{1} + \lambda_{2}{SHM}_{2} = \lambda_{1}\left( \ddot{x_{1}} + \omega_{0}x_{1} \right) + \lambda_{2}\left( \ddot{x_{2}} + \omega_{0}x_{2} \right)$$

$$= \left( \lambda_{1} + \lambda_{2} \right)\ddot{x} + \left( \lambda_{1} + \lambda_{2} \right)\omega_{0}x$$

$$\mathbf{=}\mathbf{\lambda}_{\mathbf{3}}\ddot{\mathbf{x}}\mathbf{+}\mathbf{\lambda}_{\mathbf{3}}\mathbf{\omega}_{\mathbf{0}}\mathbf{x}$$

The main use of phasor diagrams is to superimpose SHMs of either common or different frequency.

Superposition at a common frequency: General case:
x₁=a₁cos(ω₀t+ϕ₁)

x₂=a₂cos(ω₀t+ϕ₂)

x₁+x₂=acos(ω₀t+ϕ)

Such an analysis could be used when considering interference between waves of equal frequency, such as rays of monochromatic light:

Constructive interference would occur for ϕ₁=ϕ₂, as the magnitude a would have its largest possible value at cos(ϕ₂ϕ₁)=1
Destructive interference would occur for ϕ₁=ϕ₂, as the phase difference between the vectors z₁ and z₂ would be π, such that a would have its smallest possible value at cos(ϕ₂ϕ₁)=0 (i.e. z₁ and z₂ would partially or totally negate each other)

Superposition at different frequencies: Simplified case in which a₁=a₂=a₀ and ϕ₁=ϕ₂=0:
x₁=a₀cos(ω₁t)

x₂=a₀cos(ω₂t)

x₁+x₂=acosωt

Differing amplitudes and phases would simply be dealt with separately in the way illustrated above for superposition at a common frequency – using a phasor diagram:

x₁+x₂=a₀cosω₁t+a₀cosω₂t=a₀(cosω₁t+cosω₂t)

cos(A+B)+cos(AB)=2cosAcosB

$$\therefore\cos C + \cos D = 2\cos\left( \frac{C + D}{2} \right)\cos\left( \frac{C - D}{2} \right)$$

$$\therefore a_{0}\left( \cos{\omega_{1}t} + \cos{\omega_{2}t} \right)\mathbf{= 2}\mathbf{a}_{\mathbf{0}}\mathbf{\cos}\left( \frac{\left( \mathbf{\omega}_{\mathbf{1}}\mathbf{+}\mathbf{\omega}_{\mathbf{2}} \right)\mathbf{t}}{\mathbf{2}} \right)\mathbf{\cos}\left( \frac{\left( \mathbf{\omega}_{\mathbf{1}}\mathbf{-}\mathbf{\omega}_{\mathbf{2}} \right)\mathbf{t}}{\mathbf{2}} \right)$$

The resultant simple harmonic oscillation is an oscillation with a frequency of (ω₁+ω₂)/2 and an amplitude between 0 and 2a₀, modulated by the slower frequency (ω₁ω₂)/2:

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Oscillating Systems Notes

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