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The trajectory of the harmonic oscillator

Simple Harmonic Oscillator SolutionHarmonic Oscillator and QuadratureFind the general solution of the simple harmonic oscillatorSolution to harmonic oscillator with periodic forcingWhat exactly is Green's Function and why can I use it to solve harmonic oscillator problems?solving simple harmonic oscillatorDamped Linear harmonic oscillator , energy questionAbout resonance in the (undamped) harmonic oscillatorEquation of Simple Harmonic OscillatorExpressing Equations in Lagrange Subsidiary Form

1

1

$begingroup$

If we consider the Hamiltonian for the simple harmonic oscillator given by,

$$H(p,x) = fracp^22m+frackx^22$$

where $m$ is the mass, $k$ is the stiffness and $p$ is the momentum, then the equations of motion for the oscillator can be written as,

$$fracdpdt= -fracpartial Hpartial x,,,,,,,,,,fracdxdt=fracpartial Hpartial p$$

The equations of motion in the Lagrange subsidiary form can be written as,

$$fracdp-kx =fracdx(p/m)=fracdt1$$

How can one solve the subsidiary equations to show that the trajectory of the oscillator is,

$$x(t) = A,textsin(omega t+phi)$$

where $omega = sqrt frackm$ and $A$ and $phi$ are constants.

integration ordinary-differential-equations pde physics classical-mechanics

share|cite|improve this question

edited Mar 15 at 17:37

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 16:41

LightningStrikeLightningStrike

274

$endgroup$

add a comment | 

1

1

$begingroup$

If we consider the Hamiltonian for the simple harmonic oscillator given by,

$$H(p,x) = fracp^22m+frackx^22$$

where $m$ is the mass, $k$ is the stiffness and $p$ is the momentum, then the equations of motion for the oscillator can be written as,

$$fracdpdt= -fracpartial Hpartial x,,,,,,,,,,fracdxdt=fracpartial Hpartial p$$

The equations of motion in the Lagrange subsidiary form can be written as,

$$fracdp-kx =fracdx(p/m)=fracdt1$$

How can one solve the subsidiary equations to show that the trajectory of the oscillator is,

$$x(t) = A,textsin(omega t+phi)$$

where $omega = sqrt frackm$ and $A$ and $phi$ are constants.

integration ordinary-differential-equations pde physics classical-mechanics

share|cite|improve this question

edited Mar 15 at 17:37

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 16:41

LightningStrikeLightningStrike

274

$endgroup$

add a comment | 

$begingroup$

If we consider the Hamiltonian for the simple harmonic oscillator given by,

$$H(p,x) = fracp^22m+frackx^22$$

where $m$ is the mass, $k$ is the stiffness and $p$ is the momentum, then the equations of motion for the oscillator can be written as,

$$fracdpdt= -fracpartial Hpartial x,,,,,,,,,,fracdxdt=fracpartial Hpartial p$$

The equations of motion in the Lagrange subsidiary form can be written as,

$$fracdp-kx =fracdx(p/m)=fracdt1$$

How can one solve the subsidiary equations to show that the trajectory of the oscillator is,

$$x(t) = A,textsin(omega t+phi)$$

where $omega = sqrt frackm$ and $A$ and $phi$ are constants.

integration ordinary-differential-equations pde physics classical-mechanics

share|cite|improve this question

edited Mar 15 at 17:37

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 16:41

LightningStrikeLightningStrike

274

$endgroup$

share|cite|improve this question

edited Mar 15 at 17:37

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 16:41

LightningStrikeLightningStrike

274

share|cite|improve this question

edited Mar 15 at 17:37

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 16:41

LightningStrikeLightningStrike

274

edited Mar 15 at 17:37

Rodrigo de Azevedo

13.2k41960

edited Mar 15 at 17:37

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 16:41

LightningStrikeLightningStrike

274

asked Mar 15 at 16:41

LightningStrikeLightningStrike

274

1 Answer
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$begingroup$

A more useful rearrangement of the respective Hamilton's equations is $dotp=-kx,,dotx=fracpm$ (but you can also obtain these from the subsidiary form) so $ddotx=-omega^2 x$.

share|cite|improve this answer

answered Mar 15 at 17:35

J.G.J.G.

31.6k23149

$endgroup$

add a comment | 

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0

$begingroup$

A more useful rearrangement of the respective Hamilton's equations is $dotp=-kx,,dotx=fracpm$ (but you can also obtain these from the subsidiary form) so $ddotx=-omega^2 x$.

share|cite|improve this answer

answered Mar 15 at 17:35

J.G.J.G.

31.6k23149

$endgroup$

add a comment | 

0

$begingroup$

A more useful rearrangement of the respective Hamilton's equations is $dotp=-kx,,dotx=fracpm$ (but you can also obtain these from the subsidiary form) so $ddotx=-omega^2 x$.

share|cite|improve this answer

answered Mar 15 at 17:35

J.G.J.G.

31.6k23149

$endgroup$

add a comment | 

$begingroup$

A more useful rearrangement of the respective Hamilton's equations is $dotp=-kx,,dotx=fracpm$ (but you can also obtain these from the subsidiary form) so $ddotx=-omega^2 x$.

share|cite|improve this answer

answered Mar 15 at 17:35

J.G.J.G.

31.6k23149

$endgroup$

share|cite|improve this answer

answered Mar 15 at 17:35

J.G.J.G.

31.6k23149

answered Mar 15 at 17:35

J.G.J.G.

31.6k23149

answered Mar 15 at 17:35

J.G.J.G.

31.6k23149