Oscillation of two masses connected to springs and a fixed point

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The discussion focuses on the oscillation of two masses connected by identical springs with spring constant k, where one end of a spring is fixed to a wall. The equations of motion are derived using Newton's second law, resulting in two differential equations: m(d²x₁/dt²) = k(x₂ - x₁) - kx₁ and m(d²x₂/dt²) = k(x₁ - x₂) - kx₂. The solution involves identifying two modes of oscillation: a stretch mode and an anti-stretch mode, leading to the general solution expressed in terms of normal variables q₁ = x₂ - x₁ and q₂ = x₁ + x₂, which simplifies the analysis of the system's motion.

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  • Newton's second law of motion
  • Simple harmonic motion (SHM) concepts
  • Understanding of differential equations
  • Basic knowledge of oscillatory systems
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Q: Two masses m are connected by identical springs of constants k and they lie on a perfectly smooth surface. The extremity of one spring is fixed on the wall, the other one is loose.

Find the equations for the motion of the system.
Find the frequencies of oscillations.

1. Relevant equations:

F=
m\frac{(d<sup>2</sup>x)}/{(dt<sup>2</sup>)}

F=kx

2. Attempt:

Part 1:
m\frac{d<sup>2</sup>x}{dt<sup>2</sup>} = k(x2-x1)-kx1

m\frac{d<sup>2</sup>x}{dt<sup>2</sup>} = k(x1-x2)-kx2

Part 2:

x1=A1cosωt
x2=A2cosωt

and then substitute?Not sure if I even am getting anywhere with this..
 

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from looking at it there are two modes, one strech mode (both moving in same direction) and one anti-strech (moving in opposite directions), from memory of dynamics class a long time ago you might want to try some substitutions, which reflect these anticipated modes, in the general equation.

For example, the stretch mode looks as if x2 will move with twice the amplitude of x1, so you could guess a solution X = x1 + 2x2, and sub this into see if it reduces to a SHM equation in X only. I can't see your equations and I don't have any pen or paper so I can't test this idea. Someone else will hopefully confirm it.
 
your second differential equation in 'part 1' is wrong - the only force acting on the mass is the single spring attached to the end mass.

The general solution is got by adding and subtracting the two differential equations and then expressing them in terms of the 'normal variables' which are q1=x2-x1 and q2 =x1+x2

Solve in terms of these and you can work back to the original variables.
 

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