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JEE Main & Advanced Sample Paper JEE Main Sample Paper-17

  • question_answer
    A string of length 1.5 m with its two ends clamped is vibrating in the fundamental mode. The amplitude at the center of the string is 4  mm. The minimum distance, between two points having amplitude 2 mm, is

    A)  1 m                                       

    B)  75 cm

    C)  60 cm                  

    D)  50 cm

    Correct Answer: A

    Solution :

     \[{{A}_{s}}=2A\sin kx\] \[\Rightarrow \]\[2=4\sin kx\] \[\Rightarrow \]\[kx=\frac{\pi }{6},\frac{5\pi }{6}\] \[\therefore \]  \[{{x}_{2}}-{{x}_{1}}-\left[ \frac{\pi }{6}-\frac{5\pi }{6} \right]\times \frac{1}{k}=\frac{\lambda }{3}\] As the string is vibrating in fundamental mode \[L=\frac{\lambda }{2}\] \[\Rightarrow \]                               \[\lambda =2L=3m\] So, required separation between two points, \[{{x}_{2}}-{{x}_{1}}=1\,m.\]


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