"\\lambda=m\/L"
Equation of motion
"\\lambda y(t)g-\\lambda(L-y(t))g-\\lambda g=L\\lambda\\frac{d^2y(t)}{dt^2}\\to"
"y(t)=\\frac{L+1}{2}+Ae^{\\sqrt{2g\/L}\\cdot t}"
"L=8\\ m"
"y(0)=6\\ m" . So, we have "A=1.5"
"y(t)=\\frac{L+1}{2}+1.5e^{\\sqrt{2g\/L}\\cdot t}=4.5+1.5e^{1.581t}"
For "y(t)=8\\ m\\to t\\approx0.54\\ (s)" . Answer
Comments
Leave a comment