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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    If \[x=A\cos 4t+B\sin 4t\] then \[\frac{{{d}^{2}}x}{d{{t}^{2}}}\] is equal to :

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


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