Period of oscillation of spring system

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SUMMARY

The period of oscillation for a system consisting of two blocks, each of mass M, connected by a spring with force constant k, is determined to be period = 2(pi)sqrt(M/(2k)) when the left block is no longer in contact with the wall. This is derived using the concept of reduced mass, where the effective mass in the oscillation is M/2 due to the spring's configuration. The spring constant is effectively doubled when considering the two halves of the spring acting against a wall in the center of mass frame of reference.

PREREQUISITES
  • Understanding of harmonic motion and oscillation principles
  • Familiarity with the concept of reduced mass in physics
  • Knowledge of spring mechanics and Hooke's Law
  • Basic mathematical skills for manipulating equations involving square roots and constants
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  • Study the concept of reduced mass in multi-body systems
  • Learn about the derivation of the period of oscillation for coupled oscillators
  • Explore animations or simulations of spring-mass systems to visualize motion
  • Investigate the effects of varying spring constants on oscillation periods
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SbCl3
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1. Question
A system consists of two blocks, each of mass M, connected by a spring of force constant k. The system is initially shoved against a wall so that the spring is compressed a distance D from its original uncompressed length. The floor is frictionless. The system is now released with no initial velocity. (See picture)

[part c] Determine the period of oscillation for the system when the left-hand block is no longer in contact with the wall.

Homework Equations



period = 2(pi)sqrt(m/k)

The Attempt at a Solution



The answer given is this: period = 2(pi)sqrt(M/(2k))
The explanation given is "m = reduced mass = M/2".

I don't understand the explanation given. I can't visualize what happens to the right mass M after the left mass M leaves the wall. This is different from all spring problems I have seen, where one end is attached to a wall, so of course I suspect a different answer. Could someone show me the math involved to prove the period is reduced like this?
 

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SbCl3 said:
The answer given is this: period = 2(pi)sqrt(M/(2k))
The explanation given is "m = reduced mass = M/2".

I don't understand the explanation given. I can't visualize what happens to the right mass M after the left mass M leaves the wall. This is different from all spring problems I have seen, where one end is attached to a wall, so of course I suspect a different answer. Could someone show me the math involved to prove the period is reduced like this?

Hi SbCl3! :smile:

Divide the spring into two halves, then you can consider each half to be fixed against a wall (in c.o.m. frame of reference, of course) … the spring constant is doubled (1/K = 1/k + 1/k), and the mass is M :wink:
 
Can anyone please describe the motion qualitatively? I cannot visualize this problem. After the blow, the spring is maximally compressed and the block on the right moves to the right, away from the wall. I know that the left mass leaves the wall the first time that the right mass has its maximum speed to the right and the spring is at its equilibrium length. But I have no idea how the motion is after that.

All help appreciated.
Thanks
 
Please help this question has been giving me nightmares.
 
does anyone have a link to an animation
 
guys?
 

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