## Essential University Physics: Volume 1 (3rd Edition)

As we know that $E=\frac{1}{2}mv^2+\frac{1}{2}kx^2$ We differentiate both sides to obtain: $\frac{dE}{dt}=\frac{1}{2}m\frac{d}{dt}v^2+\frac{1}{2}k\frac{d}{dt}x^2=0$ $\implies \frac{1}{2}m2v\frac{dv}{dt}+\frac{1}{2}2x\frac{dx}{dt}=0$ As $\frac{dx}{dt}=v$ and $\frac{dv}{dt}=a$ $\implies mva+kxv=0$ $v(ma+kx)=0$ $ma=-kx$ $\implies m\frac{d^2x}{dt}=-kx$ This is same as equation 13.3, hence proved.