Derive angular frequency for mass spring system

In summary, the individual masses are connected in series and the extension of the spring is expressed in terms of x1/2, x2-x1, and l. The relation between x and x_2-x_1 is solved by substituting between x = x2-x1 and x = x.
  • #1
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Homework Statement
Two masses m1 and m2 on the x axis are connected by a spring. The spring has stiffness s, length l and extension x. m2 is at position x2 and m1 at position x1. The equations of the motion are

m1d^2x1/dt^2 = sx and m2d^2x2/dt^2 = -sx

Combine these to show that the angular frequency is w = sqrt(s/M)

Where M = m1m2/ m1 + m2 (the reduced mass)
Relevant Equations
m1d^2x1/dt^2 = sx and m2d^2x2/dt^2 = -sx

w = sqrt(s/M)

M = m1m2/ m1 + m2
tried writing the x position as

x = Acos(wt) (ignoring the phase)

so that d2x / dt2 = -w2x

Substituting that into the individual motion equations would get the required result for the individual masses, but I am not sure how to combine the equations to get the reduced mass
 
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  • #2
Can you express the extension ##x## in terms of ##x_1##, ##x_2##, and ##l##?
 
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  • #3
Are the masses connected in series or to either ends of the spring?
 
  • #4
Expresign extension in term of x1/2 gives

x2-x1 = x

Which could be substituted into each of the motion equatiosn but I'm not sure how that helps
 
  • #5
VVS2000 said:
Are the masses connected in series or to either ends of the spring?
Yes the masses are at the ends
 
  • #6
so_gr_lo said:
Expresign extension in term of x1/2 gives

x2-x1 = x

Which could be substituted into each of the motion equatiosn but I'm not sure how that helps
##x_2 - x_1## is the distance from one end of the spring to the other end of the spring. This distance will include the natural length of the spring ##l## and the extension ##x##.

Once you express ##x## in terms of ##x_1## and ##x_2##, you can see how ##\ddot x## is related to ##\ddot x_2## and ##\ddot x_1##.
 
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  • #7
Is d2x/dt2 = d2x2/dt2 - d2x1/dt2 ?
 
  • #8
so_gr_lo said:
Is d2x/dt2 = d2x2/dt2 - d2x1/dt2 ?
Yes. But I can't tell if you arrived at this correctly. What equation did you use for the relation between ##x## and ##x_2 - x_1##?
 
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  • #9
Here is a figure to help you out with setting the equations. When the spring is unstretched, the masses are at the dotted lines, i.e. ##x_1-x_1=L##.
Twomasseswithspring.png
 
  • #10
TSny said:
Yes. But I can't tell if you arrived at this correctly. What equation did you use for the relation between ##x## and ##x_2 - x_1##?
I used between x = x2-x1
 
  • #11
so_gr_lo said:
I used between x = x2-x1
This is not correct.
##x## represents the amount of stretch of the spring from its unstretched length ##l##. So, if ##x = 0##, then ##x_2 - x_1## = ##l##. Your equation doesn't satisfy this condition.
 
  • #12
So x = x2-x1-l
 
  • #13
so_gr_lo said:
So x = x2-x1-l
Yes, that's the correct relation.
 
  • #14
Okay, but if I substitute that into the equations I get

m1(d2x2/dt2) - d2x1/dt2) = sx

and similar for m2, how does this help with combining them?
 
  • #15
so_gr_lo said:
Okay, but if I substitute that into the equations I get

m1(d2x2/dt2) - d2x1/dt2) = sx

This is not correct. You already had the correct equations of motion for ##m_1## and ##m_2## in the "Relevant Equations" section of your first post.

You also have ##\ddot x = \ddot x_2 - \ddot x_1##. Use the equations of motion to express the right side in terms of ##s##, ##x##, and the masses.
 

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