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Energy conservation in oscillatory motion

  • Thread starter mansfief
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
1,137
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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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