A particle moving under a conservative force

[jamie.j1989]

Homework Statement

From, Classical mechanics 5th edition, Tom W.B. Kibble, Frank H. Berkshire
Chapter 2, problem 30

A particle moving under a conservative force oscillates between x1₁ and x₂. Show that the period of oscillation is

τ = 2[itex]\int[/itex][itex]^{x_{2}}_{x^{1}}[/itex][itex]\sqrt{\frac{m}{2(V(x_{2})-V(x))}}[/itex]dx

Homework Equations

m[itex]\ddot{x}[/itex] + F(x) = 0

F(x) = -[itex]\frac{d}{dx}[/itex]V(x)

The Attempt at a Solution

m[itex]\ddot{x}[/itex] + F(x) = 0

→ m[itex]\ddot{x}[/itex] -[itex]\frac{d}{dx}[/itex]V(x) = 0

→ [itex]\int[/itex][itex]^{x_{2}}_{x_{1}}[/itex]m[itex]\ddot{x}[/itex]dx = V(x₂)-V(x₁)

Im not sure if I've started right and if I have I don't know how to go forward with the [itex]\ddot{x}[/itex]

 

[CAF123]

Rewrite $$m\ddot{x} + \frac{dV}{dx} = 0 \,\,\text{as}\,\,\frac{d}{dt}\left(\frac{1}{2}m\dot{x}^2 + V(x)\right) = 0$$ and carry on from there.

 

[jamie.j1989]

When I try that I just end up with

m[itex]\int[/itex][itex]\dot{x}[/itex][itex]\frac{d\dot{x}}{dx}[/itex]dx + V(x) = 0

by parts on the integral just sends me in a circle?

 

[CAF123]

If $$\frac{d}{dt}\left(\frac{1}{2}m\left(\frac{dx}{dt}\right)^2 + V(x) \right) = 0\,\,\,\text{then}\,\,\,\frac{1}{2}m\left(\frac{dx}{dt}\right)^2 + V(x) = \text{const}$$ Use what you know about the particle at the boundaries of its oscillations (i.e at ##x_1## and ##x_2##) to obtain the constant.

Once you have this, you can separate variables to find T.

 

[paranormal]

term.



I would approach this problem by first understanding the concepts involved. A conservative force is one that can be expressed as the gradient of a potential energy function, meaning that the work done by this force is independent of the path taken by the particle. This is known as the principle of conservation of energy.

In this problem, we are given that the particle oscillates between two points, x1 and x2. This means that the total energy of the particle, which is the sum of its kinetic and potential energy, remains constant. Using the principle of conservation of energy, we can express this as:

E = 1/2 mv^2 + V(x) = constant

Where m is the mass of the particle, v is its velocity, and V(x) is the potential energy function.

Next, we can use the definition of the period of oscillation, τ, which is the time taken for the particle to complete one full cycle, to solve for it. This can be expressed as:

τ = 2π/ω

Where ω is the angular frequency, given by:

ω = √(k/m)

Where k is the spring constant. We can rewrite this equation in terms of the potential energy function, V(x), as:

ω = √(1/m * d^2V/dx^2)

Using this, we can now solve for the period of oscillation, τ, as:

τ = 2π/√(1/m * d^2V/dx^2)

To solve for d^2V/dx^2, we can use the fact that F(x) = -dV/dx, which means that d^2V/dx^2 = -dF/dx. Substituting this into the equation for τ, we get:

τ = 2π/√(1/m * (-dF/dx))

Using the given equation for F(x), we can rewrite this as:

τ = 2π/√(1/m * (-d/dx(V(x2) - V(x1)))

Which simplifies to:

τ = 2π/√(1/m * ∫x1^x2 (V(x2) - V(x1))dx)

Finally, using the fundamental theorem of calculus, we can rewrite this as:

τ = 2π/√(1/m * V(x2)