# Vertical Spring Forced Pendulum (Lagrange equations)

• Sefrez
In summary, the student is trying to find the equations of motion for a pendulum attached to a spring that can only oscillate in the vertical direction. Using Lagrange equations, he finds that potential energy is (1/2)ky^2+mg(y-l*cos(θ)) and kinetic energy is (1/2)m((vx)^2 + (vy)^2). He also finds that without taking derivatives, the linearized equation is only a function of q1 and q2. He may have done something wrong when trying to solve the problem.
Sefrez

## Homework Statement

The problem states that a pendulum is attached to a spring that can only oscillate in the vertical direction. I am supposed to derive equations of motion for the spring/pendulum system under my chosen generalized coordinates.

## Homework Equations

Lagrange equations of motion

## The Attempt at a Solution

I choose a to describe the system by the height of the pendulum mass and the angle at which it swings. So the two coordinates are y and θ, respectively.

Given a spring constant, k, and the assumption of a massless spring, I formulated the potential energy of the system to be:
U = (1/2)ky^2+mg(y-l*cos(θ))

where l is the length of the pendulum, and I have chosen the potential to be zero at y = 0 & θ = pi/2.

The kinetic energy of the system should be that given only by the mass of the pendulum. I formulated this to be:
T = (1/2)m((vx)^2 + (vy)^2) = (1/2)m((l*θ'*cos(θ))^2 + (y' + l*θ'*sin(θ))^2)

Without attempting to put up all the math on the board, I computed the Lagrange equations on y and θ, and found the two differential equations:
y'' + (k/m)y +l*θ''*sin(θ) + l*(θ')^2*cos(θ)+g = 0
and
θ'' + (1/l)y''*sin(θ) + (1/l)*g*sin(θ) = 0

Thanks.

Sefrez said:
I choose a to describe the system by the height of the pendulum mass and the angle at which it swings. So the two coordinates are y and θ, respectively.

Just to be clear, did you choose y as the height of the pendulum mass or the height of the point where the pendulum is attached to the spring?

Everything looks good to me except I think you might be missing a term involving y'θ' in the following
θ'' + (1/l)y''*sin(θ) + (1/l)*g*sin(θ) = 0

TSny said:
Just to be clear, did you choose y as the height of the pendulum mass or the height of the point where the pendulum is attached to the spring?

Everything looks good to me except I think you might be missing a term involving y'θ' in the following

I chose y as the height of the point where the pendulum is attached to the spring. So the mass potential = 0 at y = 0 & θ = pi/2. Is this good?

As far as the y'θ' term, I was actually surprised to find out at the end that there were two terms of these that exactly canceled out. I could have made a mistake though.

Sefrez said:
I chose y as the height of the point where the pendulum is attached to the spring. So the mass potential = 0 at y = 0 & θ = pi/2. Is this good?
Yes, that's good.
As far as the y'θ' term, I was actually surprised to find out at the end that there were two terms of these that exactly canceled out. I could have made a mistake though.

I seem to still be getting that extra term. Hope I'm not missing something.

When I copy over my work in a typed version I will see if I did anything wrong. It can be easy for me to make the mistake of forgetting to continue the chain rule through when taking a full derivative versus a partial.

Thanks.

Hello,

I have copied over my work, and I still see the term canceling out. I have collected the calculations in a screen shot so that you can see them and see if one of us is doing it incorrectly. See in the attachment, on the last line, the first and last terms.

#### Attachments

• p.png
5.4 KB · Views: 796
OK, Good. It was my mistake. I did overlook a term. Thanks for setting me straight. Hope I didn't cause you too much grief with my oversight.

No problem. Thanks for the help.

I am working on the same problem, I got stuck on the linearization of these equations.

In my form the angle is q2 and the spring length is q1, I have an additional horizontal force working on the mass, my equations look like:

q1''+q2''*l*cos(q2) -q2'^2*l*sin(q2)+k/m(q1-l0)-g=0

l/cos(q2)*q2''+q1''+g*tan(q2)=F/m

After linearization I get:

q1''+l*q2''+k/m(q1-l0)-g=0

l*q2''+q1''+g*q2=F/m

I linearized around q2 = 0, therefore sin(q2)=q2, cos(q2)=1, tan(q2)=q2 and q2'*q2=0

Obviously the linearized equation I get are only a function of q1 and q2 without their derivatives, I probably did something wrong does anyone know what I did wrong?

I am working on this same problem but need to compare the lagrange solution to the Newtonian solution. I am stuck on summing the forces and moments of the problem, anyone have any ideas?

## 1. What is a vertical spring forced pendulum?

A vertical spring forced pendulum is a physical system consisting of a mass attached to a spring that is anchored at one end and free to move vertically. The motion of the mass is also influenced by the force of gravity and can be described using the Lagrange equations.

## 2. How do the Lagrange equations apply to a vertical spring forced pendulum?

The Lagrange equations are a set of equations used in classical mechanics to describe the motion of a system. In the case of a vertical spring forced pendulum, the equations can be used to determine the position and velocity of the mass at any given time, taking into account the forces acting on the system.

## 3. What are the variables involved in the Lagrange equations for a vertical spring forced pendulum?

The variables involved in the Lagrange equations for a vertical spring forced pendulum are the position and velocity of the mass, the length and stiffness of the spring, and the gravitational constant. These variables are used to calculate the kinetic and potential energies of the system, which are then used in the Lagrange equations to determine the motion of the mass.

## 4. How is the motion of a vertical spring forced pendulum affected by changes in the spring parameters?

The motion of a vertical spring forced pendulum is affected by changes in the spring parameters, such as the length and stiffness of the spring. These changes can alter the period and amplitude of the pendulum's oscillation, as well as the energy of the system. A stiffer or longer spring will result in a faster oscillation and a higher energy system.

## 5. What are the practical applications of studying the Lagrange equations for a vertical spring forced pendulum?

Studying the Lagrange equations for a vertical spring forced pendulum has several practical applications, including understanding and predicting the motion of pendulum-based devices such as clocks and metronomes. It can also be applied in engineering and physics to analyze and design systems that involve springs and oscillations, such as shock absorbers and suspension systems.

Replies
7
Views
687
Replies
15
Views
2K
Replies
9
Views
3K
Replies
16
Views
994
Replies
1
Views
3K
Replies
3
Views
2K
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
26
Views
2K