Energy conservation in oscillatory motion

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SUMMARY

The discussion focuses on calculating the speed of a mass attached to a spring during oscillatory motion, specifically a 0.321-kg mass with a spring constant of 13.3 N/m. The mass is initially displaced 0.256 m from equilibrium and the speed at 0.128 m from equilibrium is determined to be 1.43 m/s. The conservation of mechanical energy principle is applied, equating potential and kinetic energies at two different states of the system to find the unknown speed.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Knowledge of mechanical energy conservation principles
  • Familiarity with oscillatory motion and angular frequency calculations
  • Proficiency in solving quadratic equations
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  • Study the derivation of the formula for mechanical energy in oscillatory systems
  • Learn about the relationship between potential energy and kinetic energy in harmonic motion
  • Explore the concept of angular frequency and its applications in oscillatory systems
  • Investigate real-world applications of energy conservation in spring systems
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for practical examples of energy conservation principles in action.

mansfief
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Homework Statement



A 0.321-kg mass is attached to a spring with a force constant of 13.3 N/m. If the mass is displaced .256 m grom the equilibrium and released, what is its speed when it is .128 m from equilibrium. The answer is 1.43 m/s

Homework Equations



k=(w^2)m
w=2pi/T

The Attempt at a Solution


I used the formula (1/2)mv^2= (1/2)kd^2. I solved for k and got 6.44 rad/s and then i solved the equation for v. I then used .128 for my d. I plugged in my answer and got 8.24. I also used d as .256 and got 1.64
 
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k is given as 13.3

here's how energy conservation works

for one state of system ... write all mechanical (potential + kinetic) energies associated with it
the do same for other state
the 2 energies shall be same so equate them and find unknown

so here find energy when spring is initially stretched and when it has extension .128 and equate them
 

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