JEE Advanced 2024 — Paper 1 — PHY-II solution

JEE Advanced2024Paper 1PHY-II

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 $v_{0} = \frac{1}{2 L} \frac{T}{μ}$ , which of the following statement(s) is(are) correct?

A
With a node at $O$ , the minimum frequency of vibration of the composite string is $v_{0}$
B
With an antinode at $O$ , the minimum frequency of vibration of the composite string is $2 v_{0}$
C
When the composite string vibrates at the minimum frequency with a node at $O$ , it has 6 nodes, including the end nodes
D
No vibrational mode with an antinode at $O$ is possible for the composite string

Solution

Wave speeds on the two segments: $v_{L} = T / μ$ on the left segment (length $L$ ) and $v_{R} = T / (4 μ) = v_{L} /2$ on the right segment (length $2 L$ ).

Case 1: node at $O$ . Each segment then vibrates with both ends as nodes. Allowed frequencies are $f_{L} = \frac{m v _{L}}{2 L}$ and $f_{R} = \frac{n v _{R}}{2 ( 2 L )} = \frac{n v _{L}}{8 L}$ for integers $m, n \geq 1$ . Matching $f_{L} = f_{R}$ : $\frac{m}{2 L} = \frac{n}{8 L}$ , i.e. $n = 4 m$ . The minimum is $m = 1$ , $n = 4$ , giving $f_{m i n} = \frac{v _{L}}{2 L} = v_{0}$ . Option (A) is correct.

With $m = 1$ the left segment has 2 nodes (its endpoints). With $n = 4$ the right segment has 5 nodes (its 2 endpoints plus 3 interior nodes). The point $O$ is shared, so total distinct nodes $= 2 + 5 - 1 = 6$ . Option (C) is correct.

Case 2: antinode at $O$ . Each segment now has a node at the far fixed end and an antinode at $O$ . Frequencies are $f_{L} = \frac{( 2 m - 1 ) v _{L}}{4 L}$ and $f_{R} = \frac{( 2 n - 1 ) v _{L}}{16 L}$ . Equating: $4 (2 m - 1) = (2 n - 1)$ . The right side is odd but the left side is even, so no integer solution exists. Hence no mode can have an antinode at $O$ . Option (D) is correct; option (B) is incorrect.

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