Spring oscillation with friction

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SUMMARY

The discussion focuses on the dynamics of a spring-mass system with friction, specifically analyzing the initial compression of the spring, the decrease in maximal stretch or compression after oscillations, and the total number of oscillations before the mass comes to rest. The equilibrium position is determined by the balance of forces acting on the mass, leading to the equation h = 2mg/k for initial compression. The role of kinetic friction, quantified as 1/8, is crucial in calculating energy loss during oscillations.

PREREQUISITES
  • Understanding of Hooke's Law and spring constant (K)
  • Basic principles of energy conservation in mechanical systems
  • Knowledge of friction coefficients and their impact on motion
  • Familiarity with oscillatory motion and equilibrium conditions
NEXT STEPS
  • Study the derivation of Hooke's Law and its applications in oscillatory systems
  • Learn about energy conservation in systems with friction
  • Explore the effects of varying friction coefficients on oscillation damping
  • Investigate the mathematical modeling of oscillatory motion in mechanical systems
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Students and professionals in physics, mechanical engineering, and applied mathematics who are interested in understanding oscillatory systems, energy conservation, and the effects of friction on motion.

mmht579
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I think I'm getting really confused by sign changes... can someone help me with this?

Spring with spring constat K stands vertically, and mass m is placed on top. Mass is gradually lowered to its euqilirium position. With the spring held at this compression length, the system is rotated to horizontal position. Left end of the spring is attached to a wall, and the mass is placed on a table with coefficient of kinetic friction=1/8. The mass is released.

(a) What is the initial compression of the spring?
(b) How much does the maximal stretch or compression of the spring decrease after each half oscillation?
(c) How many times does the mass osciallate back and forth before coming to rest?

(a)mgh= (1/2)kh^2
h= 2mg/k

(b) this is where I am not sure with signs...

conservation of energy:
spring potential energy initial= work friction +spring potential energy
.5k(2mg/k)^2= mg( (2mg/k) + x)/8 + .5kx^2 (?)
i get x= (sq rt.97 -1)mg/(8k)

that seems wrong...
 
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Welcome to PF!
(a) You are asked for the EQUILIBRIUM position, not the position in which all gravitational energy has been converted into spring potential energy!
That is, the equilibrium position of the system is where the sum of all (vertical) forces acting on the mass is 0.
See how this helps you along..
 
oops.. i forgot that i posted...thanks for the help though!
 

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