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  • question_answer
    Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to \[1\,g{{m}^{-1}}\]. When both the strings vibrate simultaneously the number of beats is [AIPMT (S) 2009]

    A)  5                    

    B)       7                    

    C)  8        

    D)       3

    Correct Answer: B

    Solution :

    Key Idea The number of beats will be the difference of frequencies of the two strings.
    Frequency of first string \[{{f}_{1}}=\frac{1}{2{{l}_{1}}}\sqrt{\frac{T}{m}}\]
    \[=\frac{1}{2\times 51.6\times {{10}^{-2}}}\sqrt{\frac{20}{{{10}^{-3}}}}\]
    \[=137.03\,Hz\]
    Similarly, frequency of second string
    \[=\frac{1}{2\times 49.1\times {{10}^{-2}}}\sqrt{\frac{20}{{{10}^{-3}}}}\]
    \[=144.01\]
    Number of beats\[={{f}_{2}}-{{f}_{1}}=144-137\]
    = 7 beats


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