Question

If $x=a\cos \omega t+b\sin \omega t$ then prove that $\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$.

Answer 1

Given,

$x=a\cos \omega t+b\sin \omega t$.......(1).

Differentiating both sides with respect to $t$ we get,

$\frac{dx}{dt}=\omega (-a\sin \omega t+b\cos \omega t)$

Again differentiating both sides with respect to $t$ we get,

$\frac{d^{2}x}{d{t}^2}=-{\omega }^2(a\cos \omega t+b\sin \omega t)$

or, $\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$ [ Using (1)]

or, $\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$.