MCQ
જો $x = A\cos 4t + B\sin 4t$, તો ${{{d^2}x} \over {d{t^2}}} = $
- ✓$-16x$
- B$16 x$
- C$x$
- D$-x$
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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