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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2001

  • question_answer
    Two waves represented by: \[{{y}_{1}}=a\,\sin \left( \omega t+\frac{\pi }{3} \right)\] and \[{{y}_{2}}=a\,\sin \,\omega t\]superimpose on each other, then resultant amplitude will be:

    A) a                                             

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

    C)  \[2\,a\]                               

    D)         \[\sqrt{3}\,a\]

    Correct Answer: D

    Solution :

    \[{{y}_{1}}=a\,\sin \,(\omega t+\pi /3)\]  \[{{y}_{2}}=a\,\sin \,\omega t\] Comparing these equations with \[\begin{align}   & {{y}_{2}}=a\,\sin \,(\omega t\,\text{)} \\  & {{a}_{1}}-{{a}_{2}}=a,\text{=}\pi \text{/3} \\ \end{align}\] Resultant amplitude \[R=\sqrt{{{a}_{1}}^{2}+a_{2}^{2}+2{{a}_{1}}a{{ & }_{2}}\cos \,}\] \[=\sqrt{{{a}^{2}}+{{a}^{2}}+2{{a}^{2}}\cos \,\frac{\pi }{3}}\] \[=\sqrt{3}a\]


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