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

`d/dt (dL)/(d (dq)/(dt))=(dL)/(dq)`

First,

`(dL)/(d (dq)/(dt))=(d)/(d (dq)/(dt)) [ 1/2 A ((dq)/(dt))^2-Bq^2]`

`(dL)/(d (dq)/(dt))= A (dq)/(dt)`

Now,

`(dL)/(dq)=(d)/(dq) [1/2 A ((dq)/(dt))^2-Bq^2]`

`(dL)/(dq)= -2Bq`

Now simplify the Euler-Lagrange equation.

`d/dt (dL)/(d (dq)/(dt))=(dL)/(dq)`

`(d/(dt))A (dq)/(dt)=-2Bq`

`A (d^2q)/(dt^2)=-2Bq`

`(d^2q)/(dt^2)=(-2Bq)/A`

...

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

`d/dt (dL)/(d (dq)/(dt))=(dL)/(dq)`

First,

`(dL)/(d (dq)/(dt))=(d)/(d (dq)/(dt)) [ 1/2 A ((dq)/(dt))^2-Bq^2]`

`(dL)/(d (dq)/(dt))= A (dq)/(dt)`

Now,

`(dL)/(dq)=(d)/(dq) [1/2 A ((dq)/(dt))^2-Bq^2]`

`(dL)/(dq)= -2Bq`

Now simplify the Euler-Lagrange equation.

`d/dt (dL)/(d (dq)/(dt))=(dL)/(dq)`

`(d/(dt))A (dq)/(dt)=-2Bq`

`A (d^2q)/(dt^2)=-2Bq`

`(d^2q)/(dt^2)=(-2Bq)/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

`q(t)=q_0Cos(sqrt((2B)/A) t+phi)`

This is the Where `q_0` and `phi` can be found from initial conditions.

**Further Reading**