Two uniform strings of mass per unit length $μ$ and $4 μ$ , and length $L$ and $2 L$ , respectively, are joined at point $O$ , and tied at two fixed ends $P$ and $Q$ , as shown in the figure. The strings are under a uniform tension $T$ . If we define the frequency $ν_{0} = \frac{1}{2 L} \sqrt{\frac{T}{μ}},$ which of the following statement(s) is(are) correct [2024]

Question

Smart Achievers · Accepted Answer

(1, 3, 4)

The velocity of a transverse wave in a stretched string, $4 μ$

$C_{1} = \sqrt{\frac{T}{μ}}, C_{2} = \sqrt{\frac{T}{4 μ}} = \frac{C_{1}}{2}$

For node at $O$ ,

$L = \frac{n λ_{1}}{2} and 2 L = \frac{m λ_{2}}{2}$ $(Here, n, m are integers)$

or, $λ_{1} = \frac{2 L}{n} and λ_{2} = \frac{4 L}{m}$

$\frac{C_{1}}{λ_{1}} = \frac{C_{2}}{λ_{2}}$ $\Rightarrow \frac{C_{1}}{\frac{2 L}{n}} = \frac{\frac{C_{1}}{2}}{\frac{4 L}{m}}$

$∴$ $4 n = m$

For minimum frequency, $n = 1, m = 4$

$∴$ $ν_{\min} = \frac{C_{1} \times 1}{2 L} = \frac{1}{2 L} \sqrt{\frac{T}{μ}} = ν_{0}$

So, option (1) is correct.

The string will look like

i.e. 6 nodes including the end nodes so option (3) is correct.

For antinode at $O$ ,

$L = \frac{(2 n + 1) λ_{1}}{4} and 2 L = \frac{(2 m + 1) λ_{2}}{4}$ $(n, m are integers)$

or, $λ_{1} = \frac{4 L}{(2 n + 1)} and λ_{2} = \frac{8 L}{(2 m + 1)}$

$m \cdot \frac{1}{4 L} \sqrt{\frac{T}{μ}} = n \cdot \frac{1}{4 (2 L)} \sqrt{\frac{T}{4 μ}}$

$\Rightarrow \frac{m}{n} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

$∴$ $m = 1, f_{\min} = 1 \cdot \frac{1}{4 L} \sqrt{\frac{T}{μ}} = \frac{ν_{0}}{2}$

So, option (2) is incorrect.

Solution

1 With a node at $O$ , the minimum frequency of vibration of the composite string is $ν_{0}$ .

2 With an antinode at $O$ , the minimum frequency of vibration of the composite string is $2 ν_{0}$ .

3 When the composite string vibrates at the minimum frequency with a node at $O$ , it has 6 nodes, including the end nodes.

4 No vibrational mode with an antinode at $O$ is possible for the composite string.

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Solution

1 With a node at O, the minimum frequency of vibration of the composite string is ν0.

2 With an antinode at O, the minimum frequency of vibration of the composite string is 2ν0.

3 When the composite string vibrates at the minimum frequency with a node at O, it has 6 nodes, including the end nodes.

4 No vibrational mode with an antinode at O is possible for the composite string.

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1 With a node at $O$ , the minimum frequency of vibration of the composite string is $ν_{0}$ .

2 With an antinode at $O$ , the minimum frequency of vibration of the composite string is $2 ν_{0}$ .

3 When the composite string vibrates at the minimum frequency with a node at $O$ , it has 6 nodes, including the end nodes.

4 No vibrational mode with an antinode at $O$ is possible for the composite string.