Oscillations of suspended block

[buzzmath]

Homework Statement

A block of mass 2kg is suspended from a fixed support by a spring of strength 2000N/m. The block is subject to the vertical driving force 36cos(pt)N. Given that the spring will yield if its extension exceeds 4 cm, find the range of frequencies that can safely be applied.


2. Relevant equation
x''+2kx'+w^2x=F(t)
p is angular frequency
2mk is the damping constant
mw^2 is the spring constant
mF(t) is the driving force

The Attempt at a Solution

I think this system is undamped so then k=0. Now I have x''+xw^2=F(t)
So w^2 = 2000 and F(t)=36cos(pt)
with this we have a formula for the amplitude which is
a = 36/([(2000-p^2)^2]^1/2)
So I want the amplitude to be -4<=a<=4
rearranging I get p<=(1991)^1/2 and p>=(1991)^1/2
but my book says the answer is p<20rads/s and p>40rads/s what am I doing wrong?
should I have gravity in there somewhere?

 

[I PICKLE THEY]

The quantity a = 36/([(2000-p^2)^2]^1/2) isn't the amplitude of the oscillator, it's the "complex amplitude". You need to take a = 36/([(2000-p^2)^2]^1/2) and plug it into the equation:

Xparticular + Xhomogeneous = X(t)

As far as the damping is concerned, it appears that there isn't any damping, and that mg is incorporated into 36cos(pt) already.