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NEET Sample Paper NEET Sample Test Paper-39

  • 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 the same and equal to \[1\,g{{m}^{-1}}.\]When both the strings vibrate simultaneously the number of beats is

    A)  5              

    B)  7

    C)  8             

    D)  3

    Correct Answer: B

    Solution :

      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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