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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • 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 stilt 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, \[{{f}_{1}}-f=6Hz\]  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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