NEET NEET SOLVED PAPER 2015 (C)

  • question_answer
    When two displacements represented by \[{{y}_{1}}=a\sin (\omega t)\]   and  \[{{y}_{2}}=b\cos \,(\omega t)\]  are superimposed, the motion is

    A)  not a simple harmonic

    B)  simple harmonic with amplitude \[\frac{a}{b}\]

    C)  simple harmonic with amplitude \[\sqrt{{{a}^{2}}+{{b}^{2}}}\]

    D)  simple harmonic with amplitude \[\frac{(a+b)}{2}\]

    Correct Answer: C

    Solution :

    Given, \[{{y}_{1}}=a\sin \omega t\] \[{{y}_{2}}=b\,\cos \omega t=b\sin \left( \omega t+\frac{\pi }{2} \right)\] The resultant displacement is given by \[y={{y}_{1}}+{{y}_{2}}=\sqrt{{{a}^{2}}+{{b}^{2}}}\sin (\omega t+\phi )\] Hence, the motion of superimposed wave is simple harmonic with amplitude \[\sqrt{{{a}^{2}}+{{b}^{2}}}.\]


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