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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    When the two waves \[{{y}_{1}}=a\sin \omega t\]and V \[{{y}_{2}}=a\cos \omega t\] are superimposed, then resultant amplitude is:

    A)  a                                            

    B)  \[2a\]

    C)  \[\sqrt{2}a\]                     

    D)  \[\frac{a}{\sqrt{2}}\]

    Correct Answer: C

    Solution :

                     Given,     \[{{y}_{1}}=a\sin \omega t\] and        \[{{y}_{2}}=a\cos \omega t\] \[\Rightarrow \]               \[{{y}_{2}}=a\sin \left( \frac{\pi }{2}-\omega t \right)\]                  ?(ii) From Eqs. (i) and (ii), we get \[{{a}_{1}}=a,\,{{a}_{2}}=a,o|=\frac{\pi }{2}\] Hence, resultant amplitude \[R=\sqrt{a_{1}^{2}+a_{2}^{2}+2{{a}_{1}}{{a}_{2}}\cos \text{o }\!\!|\!\!\text{ }}\]                 \[=\sqrt{{{a}^{2}}+{{a}^{2}}+2{{a}^{2}}\cos \frac{\pi }{2}}\]                 \[=\sqrt{{{a}^{2}}+{{a}^{2}}}=\sqrt{2}\,a\]


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