Question

If x = a cos nt − b sin nt, then $\frac{d^{2}x}{d{t}^2}$is

(a) n² x
(b) −n² x
(c) −nx
(d) nx

Answer 1

(b) −n²x

^{Here,
$$\begin{gathered} x=a\cos nt-b\sin nt \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.t\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}=-an\sin nt-bn\cos nt \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.t\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}\mathit{x}}{\mathit{d}{\mathit{t}}^{\mathit{2}}}=-a{n}^2\cos nt+b{n}^2\sin nt \\ =-n^2\left(a\cos nt-b\sin nt\right) \\ =-n^{2}x \end{gathered}$$}