Question

If $x=A\cos 4t+B\sin 4t$, then $\frac{d^{2}x}{d{t}^2}$ is equal to

• A. $-16x$
• B. $16x$
• C. $x$
• D. $-x$

Answer 1

Answer: (A)

$-16x$

Given, $x=A\cos 4t+B\sin 4t$
On differentiating w.r.t., $t$, we get
$\frac{dx}{dt}=-4A\sin 4t+4B\cos 4t$
Again, differentiating w.r.t. $t$, we get
$\frac{d^{2}x}{d{t}^2}=-6A\cos 4t-16B\sin 4t$
$\Rightarrow \frac{d^{2}x}{d{t}^2}=-16\left(x\right)$.