JEE Main & Advanced JEE Main Paper (Held On 10 April 2015)

  • question_answer
    A simple harmonic oscillator of angular frequency \[2rad\,{{s}^{-1}}\]is acted upon by an external force F = sint N. If the oscillator is at rest in its equilibrium position at t = 0, its position at later times is proportional to: JEE Main Online Paper (Held On 10 April 2015)

    A) \[\sin t+\frac{1}{2}\cos 2t\]        

    B) \[\sin t-\frac{1}{2}\sin 2t\]

    C)  \[\cot t-\frac{1}{2}\sin 2t\]        

    D) \[\sin t+\frac{1}{2}\sin 2t\]

    Correct Answer: B

    Solution :

                     \[x\propto \sin t-\frac{1}{2}\sin 2t\] at t = 0        x & velocity both has to be zero


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