Question:

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 \(v_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\), which of the following statements is(are) correct?
Alternative_Text

Updated On: Mar 7, 2025
  • With a node at \(O\), the minimum frequency of vibration of the composite string is \(v_0\).
  • With an antinode at \(O\), the minimum frequency of vibration of the composite string is \(2v_0\).
  • 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.
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The Correct Option is A, C, D

Solution and Explanation

Wave Motion in Strings

The two strings have different mass densities:

  • \(\mu_1 = \mu\)
  • \(\mu_2 = 4\mu\)
  • Lengths: \(L_1 = L, L_2 = 2L\)

The wave speeds in the strings are:

\( v_1 = \sqrt{\frac{T}{\mu_1}}, \quad v_2 = \sqrt{\frac{T}{\mu_2}} = \sqrt{\frac{T}{4\mu}} = \frac{v_1}{2} \)

The ratio of wavelengths is given by:

\( \frac{\lambda_1}{\lambda_2} = \frac{v_1}{v_2} = \frac{\mu_2}{\mu_1} = \frac{4\mu}{\mu} = 2 \)

1. Minimum Frequency (Node at O)

At the minimum frequency, the lengths of the two strings must fit an integral number of half-wavelengths:

\( L_1 = \frac{p_1 \lambda_1}{2}, \quad L_2 = \frac{p_2 \lambda_2}{2} \)

Substituting \( \lambda_2 = \frac{\lambda_1}{2} \):

\( \frac{p_1}{p_2} = 4 \)

Let \( p_1 = 4 \) and \( p_2 = 1 \), then:

\( f = \frac{v_1}{2L} = v_0 \)

This is the fundamental frequency, with 6 nodes (including P, O, and Q).

2. Antinode at O

For an antinode at O, the lengths of the strings must satisfy the condition for continuity of displacement:

\( L_1 = \frac{p_1 \lambda_1}{4}, \quad L_2 = \frac{p_2 \lambda_2}{4} \)

Substituting \( \lambda_2 = \frac{\lambda_1}{2} \):

\( \frac{p_1}{p_2} = 2 \)

The condition for continuity of vibration cannot be satisfied with integer values of \( p_1 \) and \( p_2 \), so no vibration mode with an antinode at O is possible.

Final Answer:

(A), (C), (D)

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