Two uniform strings of mass per unit length $\mu$ and $4 \mu$, and length $L$ and $2 L$, respectively, are…

Two uniform strings of mass per unit length $\mu$ and $4 \mu$, and length $L$ and $2 L$, respectively, are joined at point $\mathrm{O}$, and tied at two fixed ends $\mathrm{P}$ and $\mathrm{Q}$, as shown in the figure. The strings are under a uniform tension $T$. If we define the frequency $v_0=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$, which of the following statement(s) is(are) correct?

1. With a node at $\mathrm{O}$, the minimum frequency of vibration of the composite string is $v_0$.
2. With an antinode at $\mathrm{O}$, the minimum frequency of vibration of the composite string is $2 v_0$.
3. When the composite string vibrates at the minimum frequency with a node at $\mathrm{O}$, it has 6 nodes, including the end nodes.
4. No vibrational mode with an antinode at $\mathrm{O}$ is possible for the composite string.

Solution

$\mathrm{C}_1=\sqrt{\frac{\mathrm{T}}{\mu}}, \mathrm{C}_2=\sqrt{\frac{\mathrm{T}}{4 \mu}}=\frac{\mathrm{C}_1}{2}$ $\underline{\text { For node at } \mathrm{O} \text { : }}$ $\mathrm{L}=\frac{\mathrm{n} \lambda_1}{2}, 2 \mathrm{~L}=\frac{\mathrm{m} \lambda_2}{2}$ ( $\mathrm{n}, \mathrm{m}$ are integers) $\begin{aligned} & \lambda_1=\frac{2 \mathrm{~L}}{\mathrm{n}}, \lambda_2=\frac{4 \mathrm{~L}}{\mathrm{~m}} \\ & \frac{\mathrm{C}_1}{\lambda_1}=\frac{\mathrm{C}_2}{\lambda_2} \\ & \Rightarrow \frac{\mathrm{C}_1}{\frac{2 \mathrm{~L}}{n}}=\frac{\frac{\mathrm{C}_1}{2}}{\frac{4 \mathrm{~L}}{m}} \\ & \Rightarrow 4 \mathrm{n}=\mathrm{m}\end{aligned}$ For minimum frequency, $\mathrm{n}=1, \mathrm{~m}=4$ $\therefore v_{\min }=\frac{\mathrm{C}_1 \times 1}{2 \mathrm{~L}}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}=v_0$ The string will look like Total no. of nodes $=6$ including the end nodes $\underline{\text { For antinode at } \mathrm{O} \text { : }}$ $\begin{aligned} & \mathrm{L}=(2 \mathrm{n}+1) \frac{\lambda_1}{4} ; 2 \mathrm{~L}=(2 \mathrm{n}+1) \frac{\lambda_2}{4} \quad \text { (n, m are integers) } \\ & \lambda_1=\frac{4 \mathrm{~L}}{(2 \mathrm{n}+1)} ; \lambda_2=\frac{8 \mathrm{~L}}{(2 \mathrm{~m}+1)} \\ & \frac{\mathrm{C}_1}{\lambda_1}=\frac{\mathrm{C}_2}{\lambda_2} \\ & \frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{\lambda_1}{\lambda_2} \\ & 2=\frac{\frac{4 \mathrm{~L}}{(2 \mathrm{n}+1)}}{\frac{8 \mathrm{~L}}{(2 \mathrm{~m}+1)}} \\ & 4=\frac{(2 \mathrm{~m}+1)}{(2 \mathrm{n}+1)} \Rightarrow \text { even }=\frac{\text { odd }}{\text { odd }} \Rightarrow \text { This node is not possible }\end{aligned}$

Asked in: JEE Advanced 2024 (Paper 1)

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