JEE Advanced 2024 · Paper 1 · Physics Q6

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PhysicsMultiple correct+4 / -2Question 23 of 51

Two uniform strings of mass per unit length $\mu$ and $4\mu$ , and length $L$ and $2L$ , 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 $\nu_0 = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$ , which of the following statement(s) is(are) correct?

A horizontal composite string stretched between two vertical fixed walls: end P on the left wall and end Q on the right wall. The left part, from P to the junction point O (marked with a dot), is a thin string labelled μ with length L. The right part, from O to Q, is a thicker string labelled 4μ with length 2L. Dimension arrows below the string mark L from P to O and 2L from O to Q.

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

\nu_0

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

2\nu_0

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

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

Answer and worked solution

Official answer: A, C, D

Wave speeds: $v_1 = \sqrt{T/\mu}$ and $v_2 = \sqrt{T/4\mu} = \dfrac{v_1}{2}$ . Note $\nu_0 = \dfrac{v_1}{2L}$ .

Node at O. Each part is fixed at both ends:

$f = n_1\dfrac{v_1}{2L} = n_1\nu_0$ ;
$f = n_2\dfrac{v_2}{2(2L)} = \dfrac{n_2\nu_0}{4}$ .

So $n_2 = 4n_1$ . The minimum is $n_1 = 1$ , $n_2 = 4$ , giving $f = \nu_0$ . (A) true.

Nodes: P and O on the first part; the second part has 4 loops, i.e. 5 nodes including O and Q. Total $2 + 5 - 1 = 6$ . (C) true.

Antinode at O. Each part is fixed-free:

$f = (2m-1)\dfrac{v_1}{4L} = \dfrac{(2m-1)\nu_0}{2}$ ;
$f = (2p-1)\dfrac{v_2}{4(2L)} = \dfrac{(2p-1)\nu_0}{8}$ .

Equating gives $4(2m-1) = 2p - 1$ : even = odd, impossible. (D) true, (B) false.

Topics:Standing waves in strings Speed of waves

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