Question

A light elastic cord of length 2l and stiffness k is held with the ends fixed a distance 21 apart in a horizontal position. A block of mass m is then suspended from the midpoint of the cord. Show that the potential energy of the system is given by the expression: U(y) = 2k1/2-21Vy2t12-may, where y is the vertical sag of the center of the cord. From this show that the equilibrium position is given by a root of equation 114-2au3 +a2t12-2au +a2-0, where u = y/l and a-mg/4k1.

Answer 1

Given Thas Potcnt lal energy , v K CC-20-mgt den ivative bctf gidel ety for the equilbicp dj 12492

9quar ing both sides +y IK 29 을 2 divide both side by614k2- 2 given lot take צ:U a ma HKl