Two wires $A$ and $B$ of lengths in the ratio $1: 2$ and masses in the ratio $2: 1$ are stretched by same…

Two wires $A$ and $B$ of lengths in the ratio $1: 2$ and masses in the ratio $2: 1$ are stretched by same tension. The ratio of the fundamental frequencies of wires $A$ and $B$ is
  1. $2 \sqrt{2}: 1$
  2. $1: \sqrt{2}$
  3. $1: 1$
  4. $\sqrt{2}: 1$

Solution

Fundamental frequency in a stretched wire is given as, $f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$
Ratio of fundamental frequencies of wire $A$ and $B$, $\begin{aligned} \frac{f_A}{f_B} & =\frac{1}{2 L_A} \sqrt{\frac{T_A}{\mu_A}} \times \frac{2 L_B}{T} \sqrt{\frac{\mu_B}{T_B}} \\ & =\frac{L_B}{L_A} \cdot \sqrt{\frac{T_A}{T_B}} \sqrt{\frac{\mu_B}{\mu_A}}\end{aligned}$ Here $\frac{L_B}{L_A}=\frac{2}{1}$ $\begin{aligned} \frac{\mu_A}{\mu_B}=\frac{m_A / L_A}{m_B / L_B}=\frac{m_A}{m_B} \cdot \frac{L_B}{L_A}=\frac{2}{1} \times \frac{2}{1} & \\ & {\left[\because \frac{m_A}{m_B}=\frac{2}{1} \text { and } \frac{L_B}{L_A}=\frac{2}{1}\right] }\end{aligned}$ or $\quad \frac{\mu_A}{\mu_B}=\frac{2}{1} \times \frac{2}{1}=\frac{4}{1}$ $\Rightarrow \quad \frac{\mu_B}{\mu_A}=\frac{1}{4}$ So, $\quad \frac{f_A}{f_B}=\frac{2}{1} \cdot \sqrt{1} \cdot \sqrt{\frac{1}{4}}=1: 1$

Asked in: AP EAMCET 2022 (08 Jul Shift 2)

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