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JEE Main & Advanced Physics Wave Mechanics Question Bank Self Evaluation Test - Waves

  • question_answer
    A stretched wire 60 cm long is vibrating with its fundamental frequency of 256 Hz. If the length of the wire is decreased to 15 cm and the tension remains the same. Then the fundamental freuqency of the vibration of the wire will be

    A)  1024   

    B)  572  

    C) 256      

    D) 64

    Correct Answer: A

    Solution :

    [a] \[{{L}_{0}}=60cm\]            \[{{v}_{0}}=256Hz.\] \[v=\frac{1}{2L}\sqrt{\frac{T}{m}}\]                 \[\therefore \,\,v\propto \frac{1}{L}\]  \[\frac{{{v}_{1}}}{{{v}_{0}}}=\frac{{{L}_{0}}}{{{L}_{1}}}\] \[\Rightarrow \,{{v}_{1}}={{v}_{0}}\frac{{{L}_{0}}}{{{L}_{1}}}=256\times \frac{60}{15}=1024Hz.\]


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