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BHU PMT BHU PMT Solved Paper-2004

  • question_answer
    The equation of stationary wave along a stretched string is given by \[y=5\,\,\sin \frac{\pi x}{3}\cos \,\,40\,\,\pi t\] Where \[x\] and \[y\] are in centimeter and \[t\] in second. The separation between two adjacent nodes is:                                                                                                                                                                                      [BHU PMT-2004]

    A)  6 cm                                     

    B)  4 cm

    C)  3 cm                                     

    D)  1.5 cm

    Correct Answer: C

    Solution :

                     Key Idea: Distance between adjacent nodes in half of wavelength. The standard equation of stationary wave is                                  \[y=2\,a\,\sin \frac{2\pi }{\lambda }x\,\cos \frac{2\pi }{\lambda }\,ct\]   ?(1) Where \[a\]is amplitude, \[\lambda \] is wavelength. Given equation is                                 \[y=5\,\sin \frac{\pi \,x}{3}\cos \,\,40\,\pi \,t\]  ?(2) Comparing Eqs. (1) and (2), we get                                                 \[\frac{2\pi }{\lambda }=\frac{\pi }{3}\] \[\Rightarrow \]                               \[\lambda =6\,\,cm.\] Distance between adjacent nodes \[=\frac{\lambda }{2}=\frac{6}{2}=3\,\,cm\]


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