# I Solved Problem 1 And I Have Trouble With Problem 2 Part A

I solved problem 1 and I have trouble with problem 2/ part A .. My answers are attached I need help with problem 2 and matlab code would be helpful

m – mw2r = -mg sin(wt)

When m = contestant and w = constant

D2 r – r w2 = -g sin(wt)

(D2 – w2 ) = 0

Therefore, m =

r(t) = A sin wt + B cost wt

r'(t) = A w cos wt – B w sin wt

r” (t) = – A w2 sin wt – B w2 cost wt

mr” (t) – mw2 r = -mg sin wt

A = g/2w2

r(t) = rh (t) + rp (t)

r(t) = C1 ewt + C2 e-wt + sin wt

r0 = C1+ C2

r'(0) = V0

C1 = – +

C2 = – +

R(t) = [ – + ] e-wt + [ – + ] ewt +  sin wt

r(0) = 0

r(t) = [ – ] e-wt + [ – ] ewt + sin wt

t

ewt

e-wt

Therefor,

r(t) = [ – ] e-wt + sin wt

This because to be simple harmonic A = B = 0 and r(0) = 0. We have A = -B this is only possible when v0 = g/2w

For minimum length of the rod r'(t) = 0

[ – ] (-w) e-wt + cos wt = 0

Cos wt = [ ½ – ] e-wt

R(t) = sin wt

W = 2 , V0 = 2.4 , 2.45 , 2.5

0 t 5

r0 = 0

r(t) = [ 9.81/16 – 2.4 /4 ] e-2t + 9.8/8 sin 2t

r(t) = [ 9.81/16 – 2.45 /4 ] e-2t + 9.8/8 sin 2t

r(t) = [ 9.81/16 – 2.50 /4 ] e-2t + 9.8/8 sin 2t

• Attachment 1
• Attachment 2