Answer to Question #212170 in Mechanics | Relativity for Bilqis

Question

Consider the case of a one-dimensional harmonic oscillator on the x-axis. Determine the Hamiltonian of the system using the Lagrangian transformation, then determine the equation of motion!

Answer 1

simple harmonic oscillator

Hamiltonian simple harmonic oscillator -

Hamiltonian of a system is defined as -

$H=\Sigma{p_i}{q_i}$$-L$

in the above equation ${p_i}=Generalised \ momentum$

$q_{i}=$ $Generalised\ velocity$

$L=Lagrangian$

now we know that $L=T-V......1)$ , and motion is along x-axis .

$T=\dfrac{MX'^{2}}{2}$

$V=\dfrac{KX^{2}}{2}$

Now putting $T\$ and $V$ in equation 1 , we get =

$L=$ $\dfrac{MX'^{2}}{2}$ $-\dfrac{KX^{2}}{2}$

Now , similarly Hamiltonian equation will become -

$H=$ $\Sigma{p_i}{q_i}-L$

$H=\Sigma{p_i}{q_i}$ $-\dfrac{Mx'^{2}}{2}+\dfrac{Kx^{2}}{2}$

$P_{i}=$ actually momentum of spring along X-axis .

$Q_{i}=$ actually generalised velocity along X-axis.

From lagrangian $\dfrac {\partial l}{\partial x}=p_{x}$ , $\dfrac {\partial l}{\partial q_{c}'}=P_{i}$

$\dfrac {\partial l}{\partial x'}=$ $\dfrac{\partial( Mx'^{2}+Kx^{2})}{2}$

$\dfrac {\partial l}{\partial x'}=mx'$

$\implies$ $p_{x}=mx'\implies$ $x'=\dfrac{Px}{m}$

now substituting the value of $x'$ in Hamiltonian equation , we get -

$H=P_x\dfrac{P_x}{m}-\dfrac{{(mx')}^{2}}{2m}+\dfrac{Kx^{2}}{2}$

$H=\dfrac{P_x^{2}}{m}-\dfrac{P_x^{2}}{2m}+\dfrac{Kx^{2}}{2}$

$H=$ $\dfrac{P_x^{2}}{2m}$ $+\dfrac{Kx^{2}}{2}$

The above equation is Hamiltonian equation for simple harmonic oscillator .