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If $x=A \cos 4 t+B \sin 4 t$, then $\frac{d^{2} x}{d t^{2}}$ is equal to
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1307 Upvotes
Verified Answer
The correct answer is:
$-16 x$
We have,
$$
\begin{aligned}
x &=A \cos 4 t+B \sin 4 t \\
\Rightarrow \frac{d x}{d t} &=A(-\sin 4 t \cdot 4)+B(\cos 4 t \cdot 4) \\
&=-4 A \sin 4 t+4 B \cos 4 t
\end{aligned}
$$
Again,
$$
\begin{aligned}
\frac{d^{2} x}{d t^{2}} &=-4 A(\cos 4 t \cdot 4)+4 B(-\sin 4 t \cdot 4) \\
&=-16 A \cos 4 t-16 B \sin 4 t \\
&=-16[A \cos 4 t+B \sin 4 t]=-16 x
\end{aligned}
$$
$$
\begin{aligned}
x &=A \cos 4 t+B \sin 4 t \\
\Rightarrow \frac{d x}{d t} &=A(-\sin 4 t \cdot 4)+B(\cos 4 t \cdot 4) \\
&=-4 A \sin 4 t+4 B \cos 4 t
\end{aligned}
$$
Again,
$$
\begin{aligned}
\frac{d^{2} x}{d t^{2}} &=-4 A(\cos 4 t \cdot 4)+4 B(-\sin 4 t \cdot 4) \\
&=-16 A \cos 4 t-16 B \sin 4 t \\
&=-16[A \cos 4 t+B \sin 4 t]=-16 x
\end{aligned}
$$
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