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 a node?
A) \[\frac{1}{2l}\sqrt{\frac{T}{\mu }}\]
B) \[\frac{1}{l}\sqrt{\frac{T}{\mu }}\]
C) \[\frac{4}{l}\sqrt{\frac{T}{\mu }}\]
D) \[\frac{2}{l}\sqrt{\frac{T}{\mu }}\]
Correct Answer: A
Solution :
\[{{f}_{1}}=\frac{1}{2l}\sqrt{\frac{T}{\mu }},\frac{2}{2l}\sqrt{\frac{T}{\mu }},\frac{3}{2l}\sqrt{\frac{T}{\mu }}\]etc. (Just like open pipe) \[{{f}_{2}}=\frac{1}{2l}\sqrt{\frac{T}{9\mu }},\frac{2}{2l}\sqrt{\frac{T}{9\mu }},\frac{1}{3l}\sqrt{\frac{T}{9\mu }}\]etc. or\[{{f}_{2}}=\frac{1}{6l}\sqrt{\frac{T}{\mu }},\frac{2}{3l}\sqrt{\frac{T}{\mu }},\frac{1}{2l}\sqrt{\frac{T}{\mu }}\]etc. We see that \[\frac{1}{2l}\sqrt{\frac{T}{\mu }}\]is the lowest frequency at which \[{{f}_{1}}\] and \[{{f}_{2}}\] both are equal. This corresponds to fundamental frequency (or first harmonic) of wire-1 and second overtone (or third harmonic of wire-2).
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