
#1
Apr2612, 01:35 AM

P: 1

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[itex]\frac{(d^{2}x)}/{(dt^{2})}[/itex] F=kx 2. Attempt: Part 1: m[itex]\frac{d^{2}x}{dt^{2}}[/itex] = k(x_{2}x_{1})kx_{1} m[itex]\frac{d^{2}x}{dt^{2}}[/itex] = k(x_{1}x_{2})kx_{2} Part 2: x_{1}=A_{1}cosωt x_{2}=A_{2}cosωt and then substitute? Not sure if I even am getting anywhere with this.. 



#2
Apr2612, 03:04 AM

P: 1,200

from looking at it there are two modes, one strech mode (both moving in same direction) and one antistrech (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 in to 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. 



#3
Apr2612, 05:46 AM

P: 84

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 q_{1}=x_{2}x_{1} and q_{2} =x_{1}+x_{2} Solve in terms of these and you can work back to the original variables. 


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