RAJASTHAN PMT Rajasthan - PMT Solved Paper-1996

  • question_answer The displacement equation for two waves undergoing superposition are \[{{y}_{1}}=4\,\sin \,\omega t,{{y}_{2}}=3\sin \,(\omega t+\pi /2),\]then resultant amplitude will be:

    A)  \[5\,cm\]                                           

    B)         \[7\,cm\]                           

    C)         \[1\,cm\]                                           

    D)         \[0\,cm\]

    Correct Answer: A

    Solution :

    Equations of the waves is \[{{y}_{1}}=4\,\sin \,\omega t\],   \[{{y}_{2}}=3\sin \left( \omega t+\frac{\pi }{2} \right)\] Comparing these equations with \[y=a\,\sin \,(\omega t+\phi )\]                \[{{a}_{1}}=4,\,{{a}_{2}}=3,\,\phi =\frac{\pi }{2}\] Resultant amplitude \[R=\sqrt{{{a}_{1}}^{2}+{{a}_{2}}^{2}+2{{a}_{1}}{{a}_{2}}\,\cos \phi }\] \[=\sqrt{{{4}^{2}}+{{3}^{2}}+2\times 4\times 3\cos \frac{\pi }{2}}\] \[=\sqrt{16+9+24\times 0}\] \[=5cm\]


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