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Manipal Medical Manipal Medical Solved Paper-2003

  • question_answer
    Two identical straight wires are stretched so as to produce 6 beats per second when vibrating simultaneously. On changing the tension slightly in one of them, the beat frequency still remains unchanged. Denoting by\[{{T}_{1}}\]and\[{{T}_{2}},\]the higher and the lower initial tensions in the strings, it could be said that while making the above changes in tension:

    A)  \[{{T}_{1}}\]was decreased

    B)  \[{{T}_{1}}\] was increased

    C)  \[{{T}_{2}}\] was increased

    D)  \[{{T}_{2}}\] was decreased

    Correct Answer: A

    Solution :

     The relation for frequency and tension is given by\[f\propto \sqrt{T}\] As    \[{{T}_{1}}>{{T}_{2}}\] i.e, \[{{f}_{1}}>{{f}_{2}},\] So,   \[{{f}_{1}}-{{f}_{2}}=6\,Hz\] when we increase lower tension\[{{T}_{2}},\]then\[{{f}_{2}}\] will be increased and\[{{f}_{1}}\]will, decrease Hence, \[{{f}_{1}}-{{f}_{2}}=6Hz\]


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