Simple Harmonic Motion and equilibrium

[daivinhtran]

Homework Statement

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?

Homework Equations

ky=mg (for part a)

ky + ky = mg (for part c)

2(pi)f = √(K/m)

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?

 

[grzz]

New force constant K = 2 x old force constant = 2k

 

[voko]

What frequency are you supposed to find? In Hertz's or in radians per second?

 

[daivinhtran]

New force constant K = 2 x old force constant = 2k

YEs, I did know it..but it's still

 

[daivinhtran]

What frequency are you supposed to find? In Hertz's or in radians per second?

radians per second.

 

[daivinhtran]

oh nevermind guys

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