Two pendula connected by a spring - normal modes

In summary, the equations of motion for a two-mass system with springs are given by m(d^2x_1/dt^2) = -(mg/l)x_1 + k(x_2-x_1) and m(d^2x_2/dt^2) = -(mg/l)x_2 + k(x_1-x_2). The k matrix is equal to [mg/l + k -k; -k mg/l + k] and substituting x_1 and x_2 with A_1e^(i\omega t) and A_2e^(i\omega t), respectively, results in an angular frequency of \omega = sqrt(mg/l). The correct units for \
  • #1
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Homework Statement
Two identical pendula each of length l and with bobs of mass m are free to oscillate in the same plane. The bobs are joined by a massless spring with a small spring constant k, such that the tension in the spring is k times its extension. Find the equations governing the two bobs, and write them in the form ##m\frac{d^2x}{dt^2} = -Kx## and write down the K matrix. Subsitute a normal mode solution ##x=af((t)# and show that this satisifes the equation of motion if a is an eigenvector. Solve the corresponding equation for f(t). How many eigenvectors does K have? Write them and find a general solution. At t=0, both pendula are at rest with x1=x2=a. Describe the subsequent motion of the two pendula.
Relevant Equations
##Ae^iwt##
##ma=m\frac{d^2x}{dt^2}##
I found the equations of motion as

##m\frac{\mathrm{d}^2x_1 }{\mathrm{d} t^2} = -\frac{mg}{l}x_1 + k(x_2-x_1)##

and
##m\frac{\mathrm{d}^2x_2 }{\mathrm{d} t^2} = -\frac{mg}{l}x_2 + k(x_1-x_2)##

I think the k matrix might be
##\begin{bmatrix}
mg/l + k & -k \\
-k & mg/l + k
\end{bmatrix}##
Then when I substitute
##x_1 = A_1e^(i\omega t)##
and
##x_2 = A_2e^(i\omega t)##
and found ##\omega = sqrt{\frac{mg}{l}}##
But how do I do the rest?
 
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  • #2
Your result for omega has the wrong units.

What are the eigenvalues and eigenvectors of K?
 

FAQ: Two pendula connected by a spring - normal modes

What is the concept of "normal modes" in a system of two pendula connected by a spring?

The concept of normal modes refers to the natural oscillations that occur in a system when it is disturbed from its equilibrium position. In the case of two pendula connected by a spring, the normal modes represent the different patterns of motion that the system can exhibit.

What factors affect the normal modes of a system of two pendula connected by a spring?

The normal modes of a system of two pendula connected by a spring are affected by the mass, length, and stiffness of the pendula, as well as the strength of the spring that connects them. These factors determine the frequency and amplitude of the normal modes.

How do the normal modes of a system of two pendula connected by a spring relate to each other?

The normal modes of a system of two pendula connected by a spring are coupled, meaning that they influence each other's motion. This is because the pendula are connected by the spring, so the motion of one affects the motion of the other.

What are the implications of the normal modes in a system of two pendula connected by a spring?

The normal modes of a system of two pendula connected by a spring have important implications in fields such as physics, engineering, and mathematics. They can help us understand and analyze the behavior of complex systems, and they have practical applications in areas such as vibration control and energy harvesting.

How can the normal modes of a system of two pendula connected by a spring be calculated or observed?

The normal modes of a system of two pendula connected by a spring can be calculated using mathematical equations and principles such as Newton's laws of motion and Hooke's law. They can also be observed experimentally by measuring the frequencies and amplitudes of the system's natural oscillations.

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