VMMC VMMC Medical Solved Paper-2003

  • question_answer
    Two waves each of amplitude a and frequency \[f\] have a phase difference \[\frac{\pi }{2}\]The amplitude and frequency of resultant wave due to their superposition will be:

    A) \[\frac{a}{\sqrt{2}},\frac{f}{2}\]

    B)        \[\frac{a}{\sqrt{2}},f\]                  

    C) \[2a,\frac{f}{2}\]                             

    D)        \[\sqrt{2}a,f\]

    Correct Answer: D

    Solution :

    The two given waves are \[{{y}_{1}}=a\sin 2\pi ft\] \[{{y}_{2}}=a\sin \left( 2\pi ft+\frac{\pi }{2} \right)\] \[\therefore \] Resultant displacement \[y={{y}_{1}}+{{y}_{2}}\]                 \[=a\sin 2\pi ft+a\,\sin \left( 2\pi \,ft\,+\frac{\pi }{2} \right)\] or            \[y=2a\,\sin \left( 2\pi ft+\frac{\pi }{4} \right)\cos \frac{\pi }{4}\] \[=\frac{2a}{\sqrt{2}}\sin \left( 2\pi ft+\frac{\pi }{4} \right)\] \[=\sqrt{2}a\sin \left( 2\pi ft+\frac{\pi }{4} \right)\]  Thus, the resultant wave has amplitude \[\sqrt{2}a\] and frequency\[f\].


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