Answer :
Answer:
[tex]15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =[/tex] ± [tex]170 cos(5t)[/tex]
[tex]y(0)=0[/tex], [tex]y'(0)=0[/tex]
Step-by-step explanation:
See the attached image
This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: [tex]F_{r} (t)[/tex] that correspond to the force of resistance on the mass by the action of the spring and [tex]F(t)[/tex] that is an external force with unknown direction (that does not specify in the enounce).
For determinate [tex]F_{r} (t)[/tex] we can use Hooke's Law given by the formula [tex]F_{r} (t) = k y(t)[/tex] where [tex]k[/tex] correspond to the elastic constant of the spring and [tex]y(t)[/tex] correspond to the relative displacement of the mass-spring system with respect of his rest state.
We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...
[tex]k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m} = 441.45 \frac{N}{m}[/tex]
Now we apply Newton's 2nd Law and obtaint that...
[tex]F_{r} (t)[/tex] ± [tex]F(t)[/tex] = [tex]ma(t)[/tex]
[tex]F_{r} (t) = ky(t) = 441.45y(t)[/tex]
[tex]F(t) = 170 cos(5t)[/tex]
[tex]m = 15 kg[/tex]
[tex]a(t) = \frac{d^{2}y(t)}{dt^{2} }[/tex]
Finally... [tex]15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =[/tex] ± [tex]170 cos(5t)[/tex]
We know from the problem that there's not initial displacement and initial velocity, so... [tex]y(0)=0[/tex] and [tex]y'(0)=0[/tex]
Finally the Initial Value Problem that models the situation describe by the problem is
[tex]\left \{ 15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.[/tex]
