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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2010

  • question_answer
    The equation of a stationary wave is\[y=0.8cos\] \[\left( \frac{\pi x}{20} \right)\]sin 200 \[\pi \]t and t is in sec. The separation between consecutive nodes will be

    A)  20 cm          

    B)  10 cm

    C)  40 cm           

    D)  30 cm

    Correct Answer: D

    Solution :

     On comparing the given equation with standard equation \[y=2a\sin \frac{2\pi x}{\lambda }.\cos \frac{2\pi vt}{\lambda }\] we get, \[\frac{2\pi }{\lambda }=\frac{\pi }{20}\] \[\Rightarrow \] \[\lambda =40\] Separation between two consecutive nodes \[=\frac{\lambda }{2}=\frac{40}{2}=20\,cm\]


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