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

MCQ

If $x = A\cos 4t + B\sin 4t$, then ${{{d^2}x} \over {d{t^2}}} = $

✓
$-16x$
B
$16 x$
C
$x$
D
$-x$

✓

Answer

Correct option: A.

$-16x$

a
(a) $x = A\cos 4t + B\sin 4t$

Differentiate w.r.t. $t,$

$\frac{{dx}}{{dt}} = - 4A\sin 4t + 4B\cos 4t$

Again, differentiate w.r.t. $t,$

$\frac{{{d^2}x}}{{d{t^2}}} = - 16A\cos 4t - 16B\sin 4t$

$\frac{{{d^2}x}}{{d{t^2}}} = - 16[A\cos 4t + B\sin 4t]$.

Hence, $\frac{{{d^2}x}}{{d{t^2}}} = - 16x$.

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