Verified Solution β€’ 2025–26 Session very hard

QJEEA-2024-06: JEE Advanced 2024 JEEA-2024-06 Solution: Q.6 Two uniform strings of mass per unit length πœ‡ and 4πœ‡, and …

Q.6 Two uniform strings of mass per unit length πœ‡ and 4πœ‡, 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   Ξ½β‚€ = (1/2L)√(T/ΞΌ), which of the following statement(s) is(are) correct?


Multiple Choice Options

  • A (A) With a node at O, the minimum frequency of vibration of the composite string is 𝜈0.
  • B (B) With an antinode at O, the minimum frequency of vibration of the composite string is 2𝜈0.
  • C (C) When the composite string vibrates at the minimum frequency with a node at O, it has 6 nodes, including the end nodes.
  • D (D) No vibrational mode with an antinode at O is possible for the composite string.

Step-by-Step Verified Solution

Solution:

  1. String Properties:
  • String PO: Length L, mass/length ΞΌ, wave speed v₁ = √(T/ΞΌ)
  • String OQ: Length 2L, mass/length 4ΞΌ, wave speed vβ‚‚ = √(T/4ΞΌ) = (1/2)√(T/ΞΌ)
  1. Node at O (Option A and C):
  • For a node at O, the frequencies of the two strings must be equal.
  • f₁ = v₁/λ₁ = √(T/ΞΌ)/λ₁
  • fβ‚‚ = vβ‚‚/Ξ»β‚‚ = (1/2)√(T/ΞΌ)/Ξ»β‚‚
  • Equating f₁ and fβ‚‚: Ξ»β‚‚ = λ₁/2
  • For minimum frequency, use the longest possible wavelengths.
  • For PO, λ₁ = 2L/n₁ and for OQ, Ξ»β‚‚ = 4L/nβ‚‚ where n1 and n2 are integers.
  • 4L/nβ‚‚ = 2L/2n₁ => nβ‚‚ = 4n₁
  • The lowest combination is n1 = 1 and n2 = 4.
  • f₁ = √(T/ΞΌ)/(2L) = Ξ½β‚€
  • fβ‚‚ = (1/2)√(T/ΞΌ)/(4L/4) = Ξ½β‚€
  • Therefore, the minimum frequency with a node at O is Ξ½β‚€.
  • PO has 1 loop (n₁ = 1), OQ has 4 loops (nβ‚‚ = 4). Total loops = 5, total nodes = 6.
  • (A) is correct, (C) is correct.
  1. Antinode at O (Option B and D):
  • For an antinode at O, the frequencies must be equal, and the displacements must match.
  • f₁ = √(T/ΞΌ)/λ₁
  • fβ‚‚ = (1/2)√(T/ΞΌ)/Ξ»β‚‚
  • Equating frequencies: 2Ξ»β‚‚ = λ₁
  • For an antinode at O:
  • PO: λ₁ = 4L/n (n is odd)
  • OQ: Ξ»β‚‚ = 8L/m (m is odd)
  • Substituting: 2(8L/m) = 4L/n => 4n = m
  • This implies m must be a multiple of 4, which contradicts m being odd.
  • Therefore, no combination of odd integers n and m satisfies the conditions.
  • No vibrational mode with an antinode at O is possible.
  • (D) is correct, (B) is incorrect.

SO A, C and D are correct

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