Find the amplitude and frequency of the vibrations

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SUMMARY

The discussion focuses on calculating the amplitude and frequency of a vibrating system consisting of two balls attached to a spring. Initially, a 1.50-kg ball and a 2.00-kg ball are glued together, with the system vibrating at an amplitude of 15.0 cm. Upon the lower ball detaching, the amplitude increases to 23.9 cm due to the reduction in mass. The spring constant is given as 165 N/m, and the frequency can be determined using the formula w=sqrt(k/m).

PREREQUISITES
  • Understanding of harmonic motion and oscillations
  • Familiarity with Hooke's Law (F=-kx)
  • Knowledge of spring constants and mass-spring systems
  • Ability to apply the formula for angular frequency (w=sqrt(k/m))
NEXT STEPS
  • Calculate the frequency of the system after the lower ball detaches using w=sqrt(k/m)
  • Explore the effects of mass changes on oscillation amplitude in spring systems
  • Investigate energy conservation in oscillating systems
  • Learn about the dynamics of multi-body systems in harmonic motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillations, as well as educators looking for examples of mass-spring systems and their behaviors.

JustinLiang
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Homework Statement


A 1.50-kg ball and a 2.00-kg ball are glued together with the lighter below the heavier one. The upper ball is attached to a vertical ideal spring of force constant 165 N/m, and the system is vibrating vertically with an amplitude of 15.0 cm. the glue connecting them breaks at the lowest point. Find the amplitude and frequency of the vibrations after the lower ball has come loose.

Homework Equations


w=sqr(k/m)
F=-kx

The Attempt at a Solution


The answer is 23.9cm but I don't see how the amplitude could be larger if mass is removed... Unless of course I misread the question...

I initially tried F=kx --> mg=kx --> x=0.12m
 
Physics news on Phys.org
Just because there is no net force doesn't mean the object stops, if the system was horizontal then in the middle of the oscillation there would be no net force but the weights would certainly move. You want the work done by the spring pulling up to be canceled by the work done by gravity pulling down and by the spring pushing down.
 

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