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) = r_h (t) + r_p (t)

r(t) = C₁ e^wt + C₂ e^-wt +sin wt

r₀ = C₁+ C₂

r'(0) = V₀

C₁ =-+

C₂ =-+

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

B)

r(0) = 0

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

t

e^wt

e^-wt

Therefor,

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

C)

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

D)

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

[-] (-w) e^-wt +cos wt = 0

Cos wt = [ -] e^-wt

R(t) =sin wt

E)

W = 2 , V₀ = 2.4 , 2.45 , 2.5

0t5

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

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