Effect of Breaking Mass on SHM Time Period & Amplitude

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SUMMARY

The discussion centers on the effects of breaking a mass attached to a ruler on the time period and amplitude of oscillations in Simple Harmonic Motion (SHM). It is established that the amplitude remains unchanged when the mass is broken, as potential energy depends solely on the position of the mass. Conversely, the time period decreases as it is directly proportional to mass, which aligns with the derived formulas: F = mr (ω)^2 and T² = 4π²m/k. The conclusion is that breaking the mass does not affect amplitude but reduces the time period of oscillations.

PREREQUISITES
  • Understanding of Simple Harmonic Motion (SHM)
  • Familiarity with the formulas for oscillation: F = mr (ω)^2 and T² = 4π²m/k
  • Basic knowledge of potential energy in mechanical systems
  • Concept of mass distribution and its effects on oscillatory motion
NEXT STEPS
  • Study the principles of Simple Harmonic Motion in depth
  • Explore the relationship between mass and oscillation frequency
  • Investigate energy conservation in oscillatory systems
  • Learn about the effects of varying mass on the dynamics of SHM
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in the dynamics of oscillatory systems will benefit from this discussion.

chocofingers
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Could anybody temme what wuld happen to the time period and the amplitude of oscillations if a mass attaching to a ruler, lying horizontally on the table, is broken ...then what effect would this have on the amplitude and time period of oscillations?

I say that amplitude would increase because mass is inversely proportional to amplitude and time period would decrease since it is directly proportional to mass...
the two formulas i derived my answers are...

F = mr (omega)^2

and T^2 = 4 x pi^2 x m/k

Am I right?
 
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If I have understood your question (my english is not so good).
Using energy arguments if the mass brakes in two halfs when the potential energy is max, the potential energy is the same for the half attached to the spring since it just depends on the position of the mass, so the amplitude would be the same.
 
Last edited:
u didnt get my question :(

np ... i solved it! :)
 

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