Solution.
F=kx⟹k=xF=12=2;
W=mg⟹3.2=32m⟹m=0.1slugs.
β=0.4;
mx′′+βx′+kx=0;
0.1x′′+0.4x′+2x=0;
x′′+4x′+20x=0;
2λ=4;λ=2,ω2=20
x(t)=e−λt(C1cos((ω2−λ2t)+C2sin(ω2−λ2t);
x(t)=e2t(C1cos(4t)+C2sin(4t);
x′(0)=0;x(0)=−1;C1=−1;
x′(0)=−2C1+4C2=2+4C2⟹C2=−21;
x(t)=e−2t(−cos(4t)−21sin(4t));
x(2π)=e−π(−cos(2π)−21sin(2π))=e−π.
Answer: a)x(t)=e−2t(−cos(4t)−21sin(4t));
b)e−π.
Comments
Leave a comment