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Haryana PMT Haryana PMT Solved Paper-2004

  • 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,          \[{{f}_{1}}-{{f}_{2}}=6Hz\] when we increase lower tension \[{{T}_{2}}\], then \[{{f}_{2}}\] will be increased and \[{{f}_{1}}\] will decrease Hence,  \[{{f}_{2}}-{{f}_{1}}=6Hz\]


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