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JEE Advance - Physics (2024 - Paper 1 Online - No. 6)

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?

JEE Advanced 2024 Paper 1 Online Physics - Waves Question 2 English
With a node at $\mathrm{O}$, the minimum frequency of vibration of the composite string is $v_0$.
With an antinode at $\mathrm{O}$, the minimum frequency of vibration of the composite string is $2 v_0$.
When the composite string vibrates at the minimum frequency with a node at $\mathrm{O}$, it has 6 nodes, including the end nodes.
No vibrational mode with an antinode at $\mathrm{O}$ is possible for the composite string.

Explanation

JEE Advanced 2024 Paper 1 Online Physics - Waves Question 2 English Explanation 1

$$C_1=\sqrt{\frac{T}{\mu}}, C_2=\sqrt{\frac{T}{4 \mu}}=\frac{C_1}{2}$$

For node at O :

$$\begin{aligned} & \mathrm{L}=\frac{\mathrm{n} \lambda_1}{2}, 2 \mathrm{~L}=\frac{\mathrm{m} \lambda_2}{2}(\mathrm{n}, \mathrm{m} \text { are integers }) \\ & \lambda_1=\frac{2 \mathrm{~L}}{\mathrm{n}}, \lambda_2=\frac{4 \mathrm{~L}}{\mathrm{~m}} \\ & \frac{\mathrm{C}_1}{\lambda_1}=\frac{\mathrm{C}_2}{\lambda_2} \\ & \Rightarrow \frac{\mathrm{C}_1}{\frac{2 \mathrm{~L}}{\mathrm{n}}}=\frac{\frac{\mathrm{C}_1}{2}}{\frac{2 \mathrm{~L}}{\mathrm{~m}}} \\ & \Rightarrow 4 \mathrm{n}=\mathrm{m} \end{aligned}$$

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

JEE Advanced 2024 Paper 1 Online Physics - Waves Question 2 English Explanation 2

Total no. of nodes $$=6$$ including the end nodes

For antinode at O :

$$\begin{aligned} & \mathrm{L}=(2 \mathrm{n}+1) \frac{\lambda_1}{4} ; 2 \mathrm{~L}=(2 \mathrm{n}+1) \frac{\lambda_2}{4} \quad \text { ( } \mathrm{n}, \mathrm{m} \text { are integers) } \\ & \lambda_1=\frac{4 \mathrm{~L}}{(2 \mathrm{n}+1)} ; \lambda_2=\frac{8 \mathrm{~L}}{(2 \mathrm{~m}+1)} \\ & \frac{C_1}{\lambda_1}=\frac{C_2}{\lambda_2} \\ & \frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{\lambda_1}{\lambda_2} \\ & 2=\frac{\frac{4 L}{(2 n+1)}}{\frac{8 \mathrm{~L}}{(2 m+1)}} \\ \end{aligned}$$

$$4=\frac{(2 m+1)}{(2 n+1)} \Rightarrow \text { even }=\frac{\text { odd }}{\text { odd }} \Rightarrow \text { This node is not possible }$$

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