If x=A 4 t+B 4 t, then d^2 x d t^2 is equal to - Vidyadip

MCQ

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

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

✓

Answer

We have, $x=A \cos 4 t+B \sin 4 t$
Differentiating both sides w.r.t. t, we get
$
\frac{d x}{d t}=A \cdot(-\sin 4 t) \cdot 4+B \cos 4 t \cdot 4
$
Again differentiating both sides of (i) w.r.t. t, we get
$
\begin{array}{l}
\frac{d^2 x}{d t^2}=-4 A(\cos 4 t) \cdot 4+4 B(-\sin 4 t) \cdot 4 \\
=-16 A \cos 4 t-16 B \sin 4 t=-16(A \cos 4 t+B \sin 4 t)=-16 x
\end{array}
$

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