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 \[={{f}_{2}}-{{f}_{1}}=144-137=7\] beatsYou need to login to perform this action.
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