Solving Spring Under Gravity Homework

[imjess]

Homework Statement

Consider particle of mass m on a spring of length k, suspended vertically( subject to gravity). Choose an appropriate coordinate for this 1 degree of freedom system and find the langragian. Derive the equation of motion and find the equilibrium point. Using a new coordinate u with u=0 at the equilibrium point, find the general solution to the differential equation. What is the period of motion?

Homework Equations

The Attempt at a Solution

Sorry I just feel as if I am doing this wrong... But can you please tell me if I am on the straight lines... F=mg-kx and V=1/2kx^2. And T=1/2mx'^2 And the period is T=2∏/ω

I don't really know what I can do...please help.

 

[anathema]

Hello, thank you for your post. It seems like you are on the right track, but let me guide you through the solution step by step.

Firstly, for this 1 degree of freedom system, we can choose the coordinate x as the displacement of the particle from its equilibrium position. This means that when x=0, the particle is at its equilibrium point.

The lagrangian for this system can be written as L=T-V, where T is the kinetic energy and V is the potential energy. In this case, the kinetic energy is given by T=1/2mx'^2, where x' is the derivative of x with respect to time. The potential energy is given by V=mgh+1/2kx^2, where h is the height of the particle from its equilibrium position.

Using the lagrangian, we can derive the equation of motion by applying the Euler-Lagrange equation, which is given by d/dt(dL/dx')-dL/dx=0. This will give us the following differential equation:

m*x''+kx=mg

To find the equilibrium point, we can set x''=0, which gives us x=mg/k. This is the point where the gravitational force is balanced by the spring force, and the particle is at rest.

Next, we can introduce a new coordinate u=x-xeq, where xeq is the equilibrium point. This means that when u=0, the particle is at its equilibrium point. Substituting this into the differential equation, we get:

m*u''+ku=0

This is a simple harmonic oscillator equation, which has the general solution of u=A*cos(ωt)+B*sin(ωt), where A and B are constants and ω is the angular frequency, given by ω=sqrt(k/m).

Finally, the period of motion can be found using the relation T=2π/ω, which gives us T=2π*sqrt(m/k).

I hope this helps to clarify the solution for you. Let me know if you have any further questions.