JEE Advanced 2024 Paper 1 Q06 Physics Waves & Optics Sound Waves Hard

JEE Advanced 2024 Paper 1 · Q06 · Sound Waves

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?

[Figure: string PQ with segment of mass density $\mu$ over length $L$ from P to O, and segment of mass density $4\mu$ over length $2L$ from O to Q.]

  1. A. With a node at O, the minimum frequency of vibration of the composite string is $\nu_0$.
  2. B. With an antinode at O, the minimum frequency of vibration of the composite string is $2\nu_0$.
  3. C. When the composite string vibrates at the minimum frequency with a node at O, it has 6 nodes, including the end nodes.
  4. D. No vibrational mode with an antinode at O is possible for the composite string.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, C, D

Solution

Wave speeds: $v_1 = \sqrt{T/\mu}$ in segment PO, $v_2 = \sqrt{T/(4\mu)} = v_1/2$ in OQ. Common frequency $f$, wavelengths $\lambda_1 = v_1/f$ and $\lambda_2 = v_2/f = \lambda_1/2$.

(A,C) Node at O: PO holds an integer number of half-wavelengths: $L = m\,\lambda_1/2$, and OQ holds $2L = n\,\lambda_2/2 = n\,\lambda_1/4$, i.e. $n = 8L/\lambda_1 = 4m$. Smallest pair: $m=1, n=4$, giving $\lambda_1 = 2L$ and $f = v_1/(2L) = \nu_0$. ✓(A) The number of nodes (including ends) is $1 + m + n = 1 + 1 + 4 = 6$. ✓(C)

(B,D) Antinode at O requires $L = (2m-1)\lambda_1/4$ and $2L = (2n-1)\lambda_2/4 = (2n-1)\lambda_1/8$, so $(2n-1) = 4(2m-1)$. Right side is odd, $4(2m-1)$ is divisible by 4 — contradiction. So no antinode-at-O mode exists. ✓(D), ✗(B).

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