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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2004

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

    A)  20 cm

    B)                         10 cm

    C)  40cm                                   

    D)  30cm

    Correct Answer: A

    Solution :

    Given equation is \[y=0.8\cos \left( \frac{\pi x}{20} \right)\,\sin \,\,200\pi t\] Comparing it with \[y=a\,\cos \frac{2\pi x}{\lambda }\sin \,\omega t,\]we get                 \[\frac{2\pi }{\lambda }=\frac{\pi }{2}\] \[\Rightarrow \]               \[\lambda =40cm\] So, separation between two nodes \[=\frac{\lambda }{2}=20cm\]


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