The lagrangian of a system with one degree of freedom is given by L = ...
is not a cyclic coordiante and hence p
x is not conserved.
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The lagrangian of a system with one degree of freedom is given by L = ...
Lagrangian and Canonical Momentum:
The given Lagrangian for the system is L = 2x^2 - 3x^2. To find the canonical momentum px conjugate to x, we can use the relation:
px = ∂L/∂x
Taking the derivative of the Lagrangian with respect to x, we have:
px = 4x - 6x
Simplifying, we get:
px = -2x
Hamiltonian:
The Hamiltonian H of a system is defined as the Legendre transformation of the Lagrangian with respect to the velocities. In this case, since we have a system with one degree of freedom, the Lagrangian does not depend on the velocity.
H = px * x - L
Substituting the expressions for px and L, we have:
H = (-2x) * x - (2x^2 - 3x^2)
Simplifying, we get:
H = -2x^2 - 2x^2 + 3x^2
H = -x^2
Conservation of px:
To determine whether px is conserved, we need to check if dpx/dt = 0. Taking the derivative of px with respect to time, we have:
dpx/dt = d/dt (-2x)
Since x is a function of time, we have:
dpx/dt = -2(dx/dt)
If dx/dt = v is the velocity, then dpx/dt = -2v. Since dx/dt = v, we can see that dpx/dt is not zero unless v = 0, i.e., the velocity is zero. Therefore, px is not conserved.
Correct Answer:
From the above calculations, we can conclude that the correct answer is option 'C': H = 2x^2 - 3x^2 and px is not conserved.