• # question_answer 38) 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)  7C)  8             D)  3

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