New Notes Blog: A particle moving in one dimension (`q` ) has the following Lagrangian: `L=1/2 A ((dq)/(dt))^2-Bq^2` Where `A` and `B` are constants. What is the...

Wednesday, 25 June 2014

A particle moving in one dimension ( $\displaystyle{q}$ ) has the following Lagrangian: $\displaystyle{L}=\frac{{1}}{{2}}{A}{{\left(\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}\right)}}^{{2}}-{B}{{q}}^{{2}}$ Where $\displaystyle{A}$ and $\displaystyle{B}$ are constants. What is the...

The equations of motion is to find the Euler-Lagrange equations. These equations have the form:

$\displaystyle\frac{{d}}{{\left.{d}{t}}}\frac{{{d}{L}}}{{{d}\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}}}=\frac{{{d}{L}}}{{{d}{q}}}$

First,

$\displaystyle\frac{{{d}{L}}}{{{d}\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}}}=\frac{{{d}}}{{{d}\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}}}{\left[\frac{{1}}{{2}}{A}{{\left(\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}\right)}}^{{2}}-{B}{{q}}^{{2}}\right]}$

$\displaystyle\frac{{{d}{L}}}{{{d}\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}}}={A}\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}$

Now,

$\displaystyle\frac{{{d}{L}}}{{{d}{q}}}=\frac{{{d}}}{{{d}{q}}}{\left[\frac{{1}}{{2}}{A}{{\left(\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}\right)}}^{{2}}-{B}{{q}}^{{2}}\right]}$

$\displaystyle\frac{{{d}{L}}}{{{d}{q}}}=-{2}{B}{q}$

Now simplify the Euler-Lagrange equation.

$\displaystyle\frac{{d}}{{\left.{d}{t}}}\frac{{{d}{L}}}{{{d}\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}}}=\frac{{{d}{L}}}{{{d}{q}}}$

$\displaystyle{\left(\frac{{d}}{{{\left.{d}{t}\right.}}}\right)}{A}\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}=-{2}{B}{q}$

$\displaystyle{A}\frac{{{{d}}^{{2}}{q}}}{{{{\left.{d}{t}\right.}}^{{2}}}}=-{2}{B}{q}$

$\displaystyle\frac{{{{d}}^{{2}}{q}}}{{{{\left.{d}{t}\right.}}^{{2}}}}=\frac{{-{2}{B}{q}}}{{A}}$

This is the equation of motion. The acceleration is proportional to the position. The exact equation for the position of the particle as a function...

The equations of motion is to find the Euler-Lagrange equations. These equations have the form:

$\displaystyle\frac{{d}}{{\left.{d}{t}}}\frac{{{d}{L}}}{{{d}\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}}}=\frac{{{d}{L}}}{{{d}{q}}}$

First,

$\displaystyle\frac{{{d}{L}}}{{{d}\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}}}={A}\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}$

Now,

$\displaystyle\frac{{{d}{L}}}{{{d}{q}}}=\frac{{{d}}}{{{d}{q}}}{\left[\frac{{1}}{{2}}{A}{{\left(\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}\right)}}^{{2}}-{B}{{q}}^{{2}}\right]}$

$\displaystyle\frac{{{d}{L}}}{{{d}{q}}}=-{2}{B}{q}$

Now simplify the Euler-Lagrange equation.

$\displaystyle\frac{{d}}{{\left.{d}{t}}}\frac{{{d}{L}}}{{{d}\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}}}=\frac{{{d}{L}}}{{{d}{q}}}$

$\displaystyle{\left(\frac{{d}}{{{\left.{d}{t}\right.}}}\right)}{A}\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}=-{2}{B}{q}$

$\displaystyle{A}\frac{{{{d}}^{{2}}{q}}}{{{{\left.{d}{t}\right.}}^{{2}}}}=-{2}{B}{q}$

$\displaystyle\frac{{{{d}}^{{2}}{q}}}{{{{\left.{d}{t}\right.}}^{{2}}}}=\frac{{-{2}{B}{q}}}{{A}}$

This is the equation of motion. The acceleration is proportional to the position. The exact equation for the position of the particle as a function of time can be found from solving this differential equation.

The solution to this equation has the general solution of

$\displaystyle{q}{\left({t}\right)}={q}_{{0}}{\cos{{\left(\sqrt{{\frac{{{2}{B}}}{{A}}}}{t}+\phi\right)}}}$

This is the Where $\displaystyle{q}_{{0}}$ and $\displaystyle\phi$ can be found from initial conditions.

New Notes Blog

Wednesday, 25 June 2014

A particle moving in one dimension ( $\displaystyle{q}$ ) has the following Lagrangian: $\displaystyle{L}=\frac{{1}}{{2}}{A}{{\left(\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}\right)}}^{{2}}-{B}{{q}}^{{2}}$ Where $\displaystyle{A}$ and $\displaystyle{B}$ are constants. What is the...

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Wednesday, 25 June 2014

A particle moving in one dimension ( ) has the following Lagrangian: Where and are constants. What is the...

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Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

A particle moving in one dimension ( $\displaystyle{q}$ ) has the following Lagrangian: $\displaystyle{L}=\frac{{1}}{{2}}{A}{{\left(\frac{{{d}{q}}}{{{\left.{d}{t}\right.}}}\right)}}^{{2}}-{B}{{q}}^{{2}}$ Where $\displaystyle{A}$ and $\displaystyle{B}$ are constants. What is the...

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?