RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • question_answer
    If the equation of motion of standing Waves is y = 0.3 sin (314 t - 1.57 x), then the velocity of standing wave will be:

    A)  400 m/s                              

    B)  300 m/s

    C)  200 m/s              

    D)         100 m/s

    Correct Answer: C

    Solution :

    Here: Equation of the standing wave is \[y=0.3\,\sin \,(314t-1.57x)\]                      ...(1) The standard equation of standing wave is \[y=a\,\sin \,2\pi \left[ \frac{t}{T}-\frac{x}{\lambda } \right]\]                                     ...(2) Comparing the given equation (1) and standard equation (2), we get                 \[y=a.3\,\sin \,\left[ \frac{314t}{2\pi }-\frac{1.57}{\lambda } \right]\]         \[=\frac{0.30}{100}\sin 2\pi \left( 50\,t-\frac{x}{4} \right)\] or\[\frac{1}{T}=50\,and\,\lambda =4\] Hence, the velocity of the standing wave is                 \[\upsilon =n\lambda =\frac{1}{T}\times 4=50\times 4=200\,m/s\]


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