Two masses connected by spring, find period of oscillation

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SUMMARY

The discussion focuses on calculating the period of oscillation for two masses connected by a spring on a frictionless horizontal track. The equations of motion derived are m1(x1)'' = -k(x1 - x2 + L) and m2(x2)'' = -k(x2 - x1 - L), where m1 and m2 are the masses, k is the spring constant, and L is the equilibrium length of the spring. The solution involves recognizing the system as two oscillators with the same period, utilizing the concept of reduced mass to simplify the two-body problem into a one-body problem. The discussion emphasizes that differential equations may not be necessary for this scenario.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with harmonic motion and oscillation principles
  • Knowledge of spring constants and Hooke's law
  • Concept of reduced mass in mechanics
NEXT STEPS
  • Study the concept of reduced mass in two-body systems
  • Learn how to derive the period of oscillation for coupled oscillators
  • Explore the mathematical treatment of differential equations in mechanical systems
  • Investigate the effects of varying spring constants on oscillation periods
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Students studying classical mechanics, physics educators, and anyone interested in understanding oscillatory motion in coupled systems.

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Homework Statement



Two masses are connected by spring and slide freely without friction along horizontal track. What is period of oscillation?

Homework Equations


The Attempt at a Solution



My solution:
let x1 be position of mass 1 (m1) and x2 be position of mass 2 (m2) and L be length of spring in equilibrium.
Then, the total stretch of the spring is x2-x1-L. Also, F1 = -F2. Thus:

m1(x1)'' = -k(x1 - x2 + L)
m2(x2)'' = -k(x2 - x1 - L)

Solving for x2 from first eqn and substituting back into second eqn yield:

\frac{d^2}{dt^2}[\frac{m1 m2}{k}(x1)''+(m1+m2)x1] = 0

I am unsure how to proceed from here, any hints? I would like to just multiply both sides by (dt^2)/d^2 but I am unsure if this is mathematically correct? It does simplify the problem though and gives me right answer...
 
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You should be well aware that ##\frac{d^2}{dt^2}## isn't a fraction.
 
I don't think you really need differential equations for this one, since there is no external force there is a point on the spring that neither stretches nor compresses. on either side of this point is 2 springs with different spring constants, so effectively you have two different oscillators, with the same period.
 
Do you know about reduced mass and how to convert a two-body problem into a one-body problem?
 

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