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Simple Harmonic Motion and equilibrium

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data

    A 93-kg box hangs from the ceiling of a room—suspended from a spring with a force constant of 540 N/m. The unstressed length of the spring is 0.505 m.
    (a) Find the equilibrium position of the box.

    (b) An identical spring is stretched and attached to the ceiling and the box, and is parallel with the first spring. Find the frequency of the oscillations when the box is released.
    (c) What is the new equilibrium position of the box once it comes to rest?

    2. Relevant equations
    ky=mg (for part a)

    ky + ky = mg (for part c)

    2(pi)f = √(K/m)

    3. The attempt at a solution

    I solved part a and c first. and found the answers that a) 2.19 and c) 1.35

    for part b....When I tried to use the new K (determined the answer of part c) from to find the frequency....the answer is always equivalent to √(K/m) but not (√(k/m))/(2pi).....Where does the 2pi go?
     
  2. jcsd
  3. Nov 19, 2012 #2
    New force constant K = 2 x old force constant = 2k
     
  4. Nov 19, 2012 #3
    What frequency are you supposed to find? In Hertz's or in radians per second?
     
  5. Nov 19, 2012 #4
    YEs, I did know it..but it's still
     
  6. Nov 19, 2012 #5
    radians per second.
     
  7. Nov 19, 2012 #6
    oh nevermind guys

    I just found out that 1Hz = 2pi radians per second....
     
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