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

The correct option is A $-16x$

Here, $x=A\cos 4t+B\sin 4t$
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
$\frac{d^{2}x}{d{t}^2}=-16A\cos 4t-16B\sin 4t$
$=-16(A\cos 4t+B\sin 4t)$
$=-16x$