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Physics · Transverse waves

JEE Advanced 2024 — Paper 1 — Question 23

Two uniform strings of mass per unit length μ\mu and 4μ4 \mu, and length LL and 2L2 L, respectively, are joined at point OO, and tied at two fixed ends PP and QQ, as shown in the figure. The strings are under a uniform tension TT. If we define the frequency v0=12LTμv_{0}=\frac{1}{2 L} \sqrt{\frac{T}{\mu}},

which of the following statement(s) is(are) correct?

Question figure
  1. Option A:

    With a node at O , the minimum frequency of vibration of the composite string is v0v_{0}.

    Correct
  2. Option B:

    With an antinode at OO, the minimum frequency of vibration of the composite string is 2v02 v_{0}.

  3. Option C:

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

    Correct
  4. Option D:

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

    Correct

Answer: A, C, D

Step-by-step solution

L=mλ12L=m \frac{\lambda_{1}}{2} and 2L=nλ222 L=n \frac{\lambda_{2}}{2}

So, 12=mnv1v2=2mn\frac{1}{2}=\frac{m}{n} \frac{v_{1}}{v_{2}}=2 \frac{m}{n}

So, mn=14⇒λ1=2 L\frac{\mathrm{m}}{\mathrm{n}}=\frac{1}{4} \Rightarrow \lambda_{1}=2 \mathrm{~L}

v0=12LTμv_{0}=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}

So, there are total 6 nodes So, L=(2m+1)λ14L=(2 m+1) \frac{\lambda_{1}}{4} 2L=(2n+1)λ242 L=(2 n+1) \frac{\lambda_{2}}{4}

So, 2m+12n+1=14\frac{2 m+1}{2 n+1}=\frac{1}{4}

So, it is not possible

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 1
Subject
Physics
Chapter
Transverse waves
Topic
Standing Waves on a String and Modes of Vibration
Two uniform strings of mass per unit length μ and 4 μ , and length L… | JEE Advanced 2024 PYQ with Solution · DhiX AI