A) ? 16x
B) 16 x
C) x
D) ? x
Correct Answer: A
Solution :
\[x=A\cos 4t+B\sin 4t\] 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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