Answer:
a) [tex]y(t) =cos (\omega t)[/tex]
b) As t →∞ y(t) becomes undefined as value of cos is undefined at ∞
Step-by-step explanation:
Considering the complete question attached in fig below:
Non-homogeneous differential equation for given undamped system is:
[tex]\frac{d^{2}y}{dt^2} + \omega ^2y = \frac{F_{o}}{m} cos \omega t[/tex] ---(1)
To find general solution, consider the homogeneous part:
[tex]\frac{d^{2}y}{dt^2} + \omega ^{2}y =0\\\implies D^2+\omega =0\\\implies D=\pm \omega i[/tex]
General solution is
[tex]y_{h} = A cos \omega t + B sin \omega t[/tex]
Using method of undetermined co-efficient,find the particular solution:
[tex]y_{p} = \frac{F_{o}}{m} (c_{1} cos \omega t +c_{2} sin \omega t)\\y'_{p} = \frac{F_{o}}{m} \omega (-c_{1} sin \omega t +c_{2} cos \omega t)\\y''_{p} = \frac{F_{o}}{m} \omega^{2} (-c_{1} cos \omega t-c_{2} sin \omega t)[/tex]
Substituting all these values in (1)
[tex]-\frac{F_{o}}{m} \omega^{2} (c_{1} cos \omega t+c_{2} sin \omega t) + \omega^{2}(c_{1} cos \omega t+c_{2} sin \omega t) =-\frac{F_{o}}{m} cos\omega t[/tex]
Equating the terms on both sides
[tex]\omega^2 cos \omega t (-\frac{F_{o}}{m}+1) c_{1} = \frac{F_{o}}{m} cos \omega t\\\omega^2sin \omega t (-\frac{F_{o}}{m} +1) c_{2}=0\\\implies c_{2} =0[/tex][tex]c_{1} = \frac{F_{o}}{\omega^2 (-F_{o} +m})[/tex]
solution of given differential eq. :
[tex]y(t) =y_{h} + y_{p}\\y(t) = A cos \omega t + B sin \omega t +c_1 cos \omega t -----(2)\\y'(t) = \omega (-A sin \omega t + B cos \omega t - c_1 sin \omega t[/tex]
Using initial values :
[tex]y(0) = A cos \omega (0) + B sin \omega (0_ +c_1 cos \omega (0)\\1 = A + c_1\\A = 1 -c_1\\y'(0) = \omega (-A sin \omega (0) + B cos \omega (0) - c_1 sin \omega (0)\\B= 0[/tex]
then (2) becomes
[tex]y(t) = A cos ( \omega t) + c_1 cos (\omega t)\\y(t) = (1- c_1)cos ( \omega t) + c_1 cos (\omega t)[/tex]
[tex]y(t) =cos (\omega t)[/tex]
As t →∞
y(t) becomes undefined as value of cos is undefined at ∞
