In the figure given: \[{{l}_{1}}={{l}_{2}}=l,{{\mu }_{1}}=\frac{{{\mu }_{2}}}{9}=\mu .\] The tension in the strings is T. Here, \[\mu \]is the mass per unit length. What is the lowest frequency such that the junction is an antinode?
A) \[\frac{4}{4l}\sqrt{\frac{T}{\mu }}\]
B) \[\frac{3}{4l}\sqrt{\frac{T}{\mu }}\]
C) \[\frac{5}{4l}\sqrt{\frac{T}{\mu }}\]
D) \[\frac{7}{4l}\sqrt{\frac{T}{\mu }}\]
Correct Answer: A
Solution :
\[{{f}_{1}}=\frac{1}{4l}\sqrt{\frac{T}{\mu }},\frac{2}{4l}\sqrt{\frac{T}{\mu }},\frac{5}{4l}\sqrt{\frac{T}{\mu }}\]etc. (Just like closed pipe) \[{{f}_{2}}=\frac{1}{3(4l)}\sqrt{\frac{T}{\mu }},\frac{2}{3(4l)}\sqrt{\frac{T}{\mu }},\frac{5}{3(4l)}\sqrt{\frac{T}{\mu }}\]etc. or\[{{f}_{2}}=\frac{1}{12l}\sqrt{\frac{T}{\mu }},\frac{2}{4l}\sqrt{\frac{T}{\mu }},\frac{5}{12l}\sqrt{\frac{T}{\mu }}\]etc. Therefore, the lowest frequency is\[\frac{1}{4l}\sqrt{\frac{T}{\mu }}\]
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