Displacement in Simple Harmonic Motion

  • Thread starter chukie
  • Start date
  • #1
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1. Hi! I am new to this forum. I was wondering if someone could help me with this problem. Thanks in advance!

A 0.27 kg mass is suspended on a spring that stretches a distance of 4.9 cm. The mass is then pulled down an additional distance of 12.5 cm and released. What's the displacement from the equilibrium position with the mass attached (in cm) after 0.42 s? Take up to be positive and use g = 9.81 m/s2.

Homework Equations


F=kx
ω(angular velocity)= (k/x)^0.5


The Attempt at a Solution



First I solved for k:
k=mg/x
k=(0.27*9.81)/4.9
k=0.54

Then I solved for the angular velocity:
ω=(0.54/0.27)^0.5
ω=squareroot 2

Then I tried to come up with an equation for displacement using the format:
x(t) = Asin(ωt)

i know that ω=root2
t=0.42

but how do i solve for A?? or what is A? should it be 4.9 or 12.5 or the sum of the two :confused: btw please correct me if im wrong in my calculation. thanks!
 

Answers and Replies

  • #2
458
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4.9 cm is the location of the equilibrium position. They're asking you to find the displacement after t seconds relative to the equilibrium position, so set your equilibrium position to 0. From the equilibrium position, they extended the displacement by 12.5 cm.
 
  • #3
335
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You know anything about S.H.M equations?
 
  • #4
80
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thanks! i got the answer now =)
 

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