Answer:
[tex]13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\[/tex]
#Where [tex]u[/tex] is in meters and [tex]t[/tex] in seconds.
Explanation:
Given that :[tex]m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s[/tex]
From [tex]\omega=kL[/tex] we have:
[tex]k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2[/tex]
From [tex]F_d(t)=-\gamma u\prime(t)[/tex] we have that:
[tex]\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m[/tex]
Now,given that the initial value problem is given by:
[tex]13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\[/tex]
Hence,the position of u at time t is given by
[tex]13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\[/tex], [tex]u[/tex] in meters,[tex]t[/tex] in seconds.
