Derive angular frequency for mass spring system

Click For Summary
SUMMARY

The discussion focuses on deriving the angular frequency for a mass-spring system with two masses, m1 and m2, connected at the ends of a spring. The participants clarify the relationship between the extension of the spring, denoted as x, and the positions of the masses, x1 and x2. The correct expression for the extension is established as x = x2 - x1 - l, where l is the natural length of the spring. The equations of motion for both masses are confirmed to be m1(d²x2/dt²) - d²x1/dt² = sx, leading to a clear understanding of how to combine these equations to derive the angular frequency.

PREREQUISITES
  • Understanding of harmonic motion and angular frequency
  • Familiarity with Newton's second law of motion
  • Knowledge of spring mechanics and Hooke's Law
  • Basic calculus, specifically differentiation and second derivatives
NEXT STEPS
  • Study the derivation of angular frequency in mass-spring systems
  • Learn about coupled oscillators and their equations of motion
  • Explore the implications of reduced mass in multi-body systems
  • Investigate the role of damping in oscillatory systems
USEFUL FOR

Physics students, mechanical engineers, and anyone studying dynamics of mass-spring systems will benefit from this discussion.

so_gr_lo
Messages
69
Reaction score
10
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
 
Physics news on Phys.org
Can you express the extension ##x## in terms of ##x_1##, ##x_2##, and ##l##?
 
  • Like
Likes   Reactions: topsquark
Are the masses connected in series or to either ends of the spring?
 
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
 
VVS2000 said:
Are the masses connected in series or to either ends of the spring?
Yes the masses are at the ends
 
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##.
 
  • Like
Likes   Reactions: topsquark
Is d2x/dt2 = d2x2/dt2 - d2x1/dt2 ?
 
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##?
 
  • Like
Likes   Reactions: topsquark
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.
 

Similar threads

Replies
3
Views
848
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
1
Views
1K
  • · Replies 31 ·
2
Replies
31
Views
2K
Replies
17
Views
3K
Replies
5
Views
3K
Replies
3
Views
2K
Replies
9
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K