Answer:
Option B
Explanation:
Given ,$x=A \cos (nt+\alpha)$ ...........(i)
On differentiating x w.r.t .'t' we get
$\frac{dx}{dt}=A\frac{d}{d} \cos (nt+\alpha)$
=$-A \sin (nt+\alpha) \frac{d}{dt}(nt+\alpha)$
=$-A \sin(nt+\alpha)(n)$
$\Rightarrow$ $\frac{dx}{dt}=-An\sin(nt+\alpha)$
$\Rightarrow$ $\frac{dx}{dt}=-\frac{x n \sin (nt+\alpha)}{(\cos (nt+\alpha)}$ [From Eq.(i)]
$\Rightarrow$ $\frac{dx}{dt}=-nx \tan (nt+\alpha)$ ........(ii)
Again differentiating w.r.t 't' , we get
$\frac{d^{2}x}{dt^{2}}=-\left[nx \frac{d}{dt} \tan (nt+\alpha)+\tan (nt+\alpha) \frac{d}{dx} nx\right]$
$=-\left[nx \sec^{2}(nt+\alpha)\frac{d}{dt}(nt+\alpha)+\tan(nt+\alpha)\frac{dx}{dt} \times n\right]$
$=-\left[nx \sec^{2}(nt+\alpha)n+n \tan(nt+\alpha)x-nx \tan(nt+\alpha)\right]$
[From Eq.(ii)]
$=-\left[n^{2}x \sec^{2}(nt+\alpha)-n^{2}x \tan^{2}(nt+\alpha)\right]$
$=-n^{2}x[\sec^{2}(nt+\alpha)-\tan^{2}(nt+\alpha)]$
$= -n^{2}x \times 1$ $[ \because \sec^{2} \theta-\tan^{2} \theta=1]$
$\Rightarrow$ $\frac{d^{2}x}{dt^{2}}=-n^{2}x$
$\Rightarrow$ $\frac{d^{2}x}{dt^{2}}+n^{2}x=0$
