A particle of mass \(m\) on the Earth’s surface is confined to move on the quadratic curve \(y=ax^2\) where \(y\) is up. Find the Lagrangian of the particle, the equation of motion and the solution for small oscillations.
The kinetic energy is
\( K.E. = \dfrac {1}{2} m(\dot x^2 + \dot y^2) \)
The potential energy is
\( P.E. = mgy\)
The Lagrangian is
\( \mathscr L = \dfrac {1}{2} m(\dot x^2 + \dot y^2) - mgy \)
We can eliminate \(y\) with the change of variables
\( y = ax^2 \)
\( \dot y = 2ax \dot x \rightarrow \dot y^2 = 4a^2x^2 \dot x^2 \)
The Lagrangian can be re-written as
\( \mathscr L = \dfrac {1}{2} m\dot x^2(1 + 4a^2x^2) - mgax^2 \)
To find the equation of motion we find the derivatives
\( \dfrac{\partial \mathscr L}{\partial \dot x} = m\dot x(1 + 4a^2x^2) \)
\( \dfrac {d}{dt} \dfrac{\partial \mathscr L}{\partial \dot x} = m\ddot x (1 + 4a^2x^2) + 8ma^2x\dot x^2 \)
\( \dfrac{\partial \mathscr L}{\partial x} = 4ma^2x\dot x^2 - 2mgax \)
The equation of motion is
\( m\ddot x (1 + 4a^2x^2) + 8ma^2x\dot x^2 = 4ma^2x\dot x^2 - 2mgax \)
We can rearrange the equation as follows
\( \ddot x (1 + 4a^2x^2) + 4a^2x\dot x^2 + 2gax = 0 \)
In the case of small oscillations we ignore second or higher order terms and the equation becomes
\( \ddot x + 2gax = 0 \)
The solution to this simplified equation is the familiar simple harmonic oscillator
\( x = Acos(\omega t + \phi) \)
Where \( \omega = \sqrt{2ga} \)