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BHU PMT BHU PMT (Screening) Solved Paper-2005

  • 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 cm and t in second. The separation between two adjacent nodes is:

    A)  1.5cm                                  

    B)  3 cm

    C)   6 cm                                    

    D)  4 cm

    Correct Answer: B

    Solution :

                      Key Idea: The separation between two adjacent nodes is half of the wavelength. The standard equation of stationary wave is \[y=2a\sin \frac{2\pi x}{\lambda }\cos \frac{2\pi t}{T}\]      ...(1) where a is amplitude,\[\lambda \]is wavelength and T is periodic time. Given, equation is                 \[y=5\sin \frac{\pi x}{3}\cos 40\pi t\]      ?. (2) Comparing Eq. (1) with Eq. (2), we get                 \[\frac{2\pi x}{\lambda }=\frac{\pi x}{3}\] \[\Rightarrow \]               \[\lambda =6\,cm\] Distance between adjacent nodes is                 \[\frac{\lambda }{2}=\frac{6}{2}=3\,cm\]


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