RAJASTHAN PMT Rajasthan - PMT Solved Paper-1996

  • question_answer
    Two waves represented by \[{{y}_{1}}=a\,\sin \,\left( \omega t+\frac{\pi }{6} \right),{{y}_{2}}=a\,\cos \,\omega t,\] the resultant amplitude will be:

    A) \[a\]                                     

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

    C)        \[a\sqrt{3}\]                                      

    D)        \[2a\]

    Correct Answer: C

    Solution :

    Equations of waves are \[{{y}_{1}}=a\,\sin \,(\omega t+\pi /6)\] \[{{y}_{2}}=a\,\cos \omega t\] \[{{y}_{2}}=a\,sin(\omega t-\pi /2)\] Comparing these equations with \[y=a\,\sin (\omega t\pm \phi )\] \[{{a}_{1}}=a,\,{{a}_{2}}=a,\,{{\phi }_{1}}=\frac{\pi }{6},{{\phi }_{2}}=\frac{\pi }{2}\] Resultant amplitude                 \[=\sqrt{{{a}_{1}}^{2}+{{a}_{2}}^{2}+2{{a}_{1}}{{a}_{2}}\,\cos ({{\phi }_{2}}-{{\phi }_{2}})}\]                 \[=\sqrt{{{a}^{2}}+{{a}^{2}}+2{{a}^{2}}\,\cos \left( \frac{\pi }{2}-\frac{\pi }{6} \right)}\]                 \[=a\sqrt{3}\]


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