Answer Happy

du d2U Let us define k = dr 1=0, so that = kr. Then from Eq. (1) and Newton's second law we have dU = F = -kr = ma = m- (3) de dt2 The equation –ku = mis called a differential equation, because it has derivatives in it. The function that satisfies this differential equation is z(t) = A cos (wt+0), where w2 = k/m. (4)

(a) Show that Equation (4) is a solution of the differential equation in Equation (3) by substituting r(t) from Equation (4) into the differential equation.