Question

the lagrangian of a particle of mass m, moving in one direction is given by L=1/2mx2-bx, where b is a positive constant the coordinate of particle x(t) at time is given by​

Answer 1

Answer:

the lagrangian of a particle of mass m, moving in one direction is given by L=1/2mx2-bx, where b is a positive constant the coordinate of particle x(t) at time is given by

Explanation:

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Answer 2

Answer:

The coordinate of the particle x(t) is given by $x(t)=-\frac{bt^{2} }{2}+\alpha t+ \beta t$ where α and β are constants.

Explanation:

Euler - Lagrange Equation:

If a Lagrangian is given, then using this equation, the equation of motion described by the given Lagrangian can be found.

Let the Lagrangian be $L(q,\dot{q},t)$ where 'q' represents the position, 'p' the momentum and 't' the time.

The Euler - Lagrange Equation is given by,

                                     $\frac{\partial L}{\partial q} - \frac{d}{dt} \{\frac{\partial L}{\partial \dot{q}} \}=0$

Lagrangian given here is

                                  $L=\frac{m\dot{x^{2}} }{2}-bx$

Step 1:

                                 $\frac{\partial L}{\partial x} = -b$

Step 2:

                                 $\frac{\partial L}{\partial \dot{x}} = m\dot{x}$

Step 3:

Putting these values in the Euler-Lagrange Equations we get,

                                [tex]-b-\frac{d}{dt}\{m\dot{x} \}=0\\ \frac{d}{dt}\{m\dot{x} \}=-b[/tex]

Step 4:

Integrating with respect to 't' we get,

                               $m\dot{x}=-bt+\alpha$

where $\alpha$ is an integrating constant which can be found using initial conditions.

Step 5:

Integrating again with respect to 't' we obtain,

                             $mx(t)=-b\frac{t^{2} }{2}+\alpha t+\beta$

where $\beta$ is an integrating constant which can be found using initial conditions.

The coordinate x(t) at any time t is given by  $\bold{x(t)=-\frac{bt^{2} }{2m} + \alpha t + \beta}$