<aside> 🪳 Normal mode: of a system is one in which all parts of the system are oscillating at the same frequency. Normal modes also have the property that they are orthogonal in a vector space. There is no coupling or exchange of energy between them, thus if a system is oscillating in one of its normal modes it stays in that mode

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The most general motion of the system can be described as a superposition of motion in its normal modes

A simple system

💼 Case: Consider two identical simple pendula consisting of point masses $m$ suspended by massless strings of length $l$ a distance $L$ apart. The masses are connected together by massless springs of natural length $L$ and spring constant $k$

The equilibrium position is $x_1=0$ and $x_2=0$ where the string is at its natural length.

We will consider small oscillations about this equilibrium position $x_1\sim x_2 \ll l$

• If a system obeys a symmetry the normal modes are eigenfunctions of that symmetry,.

  In this case with mirror symmetry the normal modes are either symmetric or antisymmetric

  • 👎 The antisymmetric normal mode corresponds to both pendula oscillating with the same amplitude and in the same direction.
  • 👍 The symmetric normal mode corresponds to both oscillating in opposite directions

• We can work out the frequencies of the normal mode

  • 👎 In the antisymmetric mode, the spring never extends or compresses and it is as if it was not there $\Rightarrow$ the frequency is that of the individual pendula $\omega^2=g/l$
  • 👍 In the symmetric mode, the centre of the spring does not move and it is as if each pendulum is restored by the sum of gravity and a spring constant 2k $\omega ^2 =g/l+2k/m$

Lagrangian and equations of motion

To obtain the Lagrangian we have to calculate the gravitational potential energy of each pendulum when it is displacement a small horizontal distance $x$ from vertical.

• This corresponds to an angle of displacement $\theta \approx x/l$ at which point the height of the mass $m$ below the pivot is $l\cos\theta \approx l(1-\frac 12 \theta^2)=l-\frac 12 x^2/l$.

• Neglecting the constant term we have a potential energy $\frac 12 mg/(l(x_1^2+x^2_2)$

• When the two masses are displaced by $x_1$ and $x_2$ the spring is stretched to length $x_1-x_2+L$ and so its potential energy is $\frac 12 k(x_2-x_1)^2$

• Thus the Lagrangian is

  $$ L=\frac 12 m(\dot x^2_1+\dot x^2_2)-\frac{mg}{2l}(x^2_1+x^2_2)-\frac 12 k(x_2-x_1)^2 $$

• The equations of motion corresponding to $x_1$ and $x_2$ are then

  $$ \begin{aligned} \frac{\text d}{\text dt} \left ( \frac{\partial L}{\partial \dot x_1} \right )&=m\ddot x_1 =\frac{\partial L}{\partial x_1}=-\frac{mg}{l}x_1+k(x_2-x_1) \\ \frac{\text d}{\text dt} \left ( \frac{\partial L}{\partial \dot x_2} \right )&=m\ddot x_2 =\frac{\partial L}{\partial x_2}=-\frac{mg}{l}x_2+k(x_1-x_2)

  \end{aligned} $$

We will solve these equations in 3 ways of increasing formality and generality

Informal method

• By adding both equations we get

  $$ m(\ddot x_1+\ddot x_2)=-\frac{mg}{l}(x_1+x_2) $$

  or

  $$ x_1+x-2=-\frac gl (x_1+x_2) $$

  💎 Conclusion: the combination $x_1+x_2$ oscillated with $\omega^2=g/l$

• By subtracting we get

  $$ m(\ddot x_1-\ddot x_2)=-\frac{mg}{l}(x_1-x_2)-2k(x_1-x_2) $$

  or

  $$ x_1-x_2=-\left ( \frac{g}{l}+\frac{2k}{m} \right )(x_1-x_2) $$

  💎 Conclusion: the combination $x_1-x_2$ oscillates with $\omega^2=g/l+2k/m$

More formal method

🗒️ Note: A normal mode implies that all components of the system oscillate at the same frequency, but with different amplitudes or phases.

• We assume that in the $n$th normal mode the frequency is $\omega^2_n$ and write the displacement of each component as

  $$ x_j=C_je^{i\omega_n t} \quad i\in\N^* $$

  where $C_j$s are constants

  🗒️ Note: in the present problem $C_j$ can be real but in general it should be complex to allow different phase between different components of a system