Q92 / 160 · TS EAMCET 2017 · Mathematics
The differential equation of the simple harmonic motion given by $x=A \cos (n t+\alpha)$ is
- A$\frac{d^2 x}{d t^2}-n^2 x=0$
- Correct answerB$\frac{d^2 x}{d t^2}+n^2 x=0$
- C$\frac{d x}{d t}-\frac{d^2 x}{d t^2}=0$
- D$\frac{d^2 x}{d t^2}-\frac{d x}{d t}+n x=0$
Solution
Given, $x=A \cos (n t+\alpha)$
On differentiating $x$ w.r.t. ' $t$ ', we get
$$
\begin{aligned}
& \frac{d x}{d t}=A \frac{d}{d t} \cos (n t+\alpha) \\
& =-A \sin (n t+\alpha) \frac{d}{d t}(n t+\alpha) \\
& =-A \sin (n t+\alpha)(n) \\
\Rightarrow \frac{d x}{d t} & =-A n \sin (n t+\alpha) \\
\Rightarrow \frac{d x}{d t} & =-\frac{x n \sin (n t+\alpha)}{\cos (n t+\alpha)} \\
\Rightarrow \frac{d x}{d t} & =-n x \tan (n t+\alpha)
\end{aligned}
$$
[from Eq. (i)]
Again, differentiating w.r.t ' $t$ ', we get
$$
\begin{aligned}
\frac{d^2 x}{d t^2} & =-\left[n x \frac{d}{d t} \tan (n t+\alpha)+\tan (n t+\alpha) \frac{d}{d t} n x\right] \\
& =-\left[n x \sec ^2(n t+\alpha) \frac{d}{d t}(n t+\alpha)+\tan \right. \\
& \left.(n t+\alpha) \frac{d x}{d t} \times n\right]
\end{aligned}
$$
$$
\begin{array}{r}
=-\left[n \operatorname{xsec}^2(n t+\alpha) n+n \tan (n t+\alpha)\right. \\
\times-n x \tan (n t+\alpha)] \\
\text { [from Eq. (ii)] }
\end{array}
$$
$$
\begin{aligned}
& =-\left[n^2 x \sec ^2(n t+\alpha)-n^2 x \tan ^2(n t+\alpha)\right. \\
& =-n^2 x\left[\sec ^2(n t+\alpha)-\tan ^2(n t+\alpha)\right] \\
& =-n^2 x \times 1 \quad\left[\because \sec ^2 \theta-\tan ^2 \theta=1\right] \\
\Rightarrow \quad & \quad \frac{d^2 x}{d t^2}=-n^2 x \\
\Rightarrow \frac{d^2 x}{d t^2}+n^2 x & =0
\end{aligned}
$$
Hence, the correct answer is option: 'B'.