JEE Advanced · Physics · 16. Waves & Sound
A massless rod \(B D\) is suspended by two identical massless strings \(A B\) and \(C D\) of equal lengths. A block of mass ' \(m\) ' is suspended point \(P\) such that \(B P\) is equal to ' \(x\) ', if the fundamental frequency of the left wire is twice the fundamental frequency of right wire, then the value of \(x\) is

- A \(1 / 5\)
- B \(1 / 4\)
- C \(41 / 5\)
- D \(31 / 4\)
Answer & Solution
Correct Answer
(A) \(1 / 5\)
Step-by-step Solution
Detailed explanation
\(f \propto v \propto \sqrt{T}\)
Further \(\Sigma \tau_p=0\)
or
or
\(
f_{A B}=2 f_{C D} \Rightarrow T_{A B}=4 T_{C D}
\)
\(
\begin{aligned}
T_{A B}(x) & =T_{C D}(l-x) \\
4 x & =l-x\left(\text { as } T_{A B}-4 T_{C D}\right) \\
x & =\frac{l}{5}
\end{aligned}
\)
Further \(\Sigma \tau_p=0\)
or
or
\(
f_{A B}=2 f_{C D} \Rightarrow T_{A B}=4 T_{C D}
\)
\(
\begin{aligned}
T_{A B}(x) & =T_{C D}(l-x) \\
4 x & =l-x\left(\text { as } T_{A B}-4 T_{C D}\right) \\
x & =\frac{l}{5}
\end{aligned}
\)
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