A) \[-16\text{ }x\]
B) \[16x\]
C) \[x\]
D) \[-x\]
Correct Answer: A
Solution :
Here, \[x=A\cos 4t+B\sin 4t\] \[\therefore \] On differentiating w.r. to t, we get \[\frac{dx}{dt}=-4A\sin \,4t+4B\cos 4t\] Again differentiating w.r. to t, we get \[\frac{{{d}^{2}}x}{d{{t}^{2}}}=-16A\cos \,4t-16B\sin 4t\] \[=-16\,(A\cos \,4t+B\sin 4t)\] \[\frac{{{d}^{2}}x}{d{{t}^{2}}}=-16x\]You need to login to perform this action.
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