1. Define System Parameters and Wave Speeds

Let the string from P to O be string 1 and the string from O to Q be string 2.

• String 1 (PO):

  • Length: $L_1 = L$
  • Mass per unit length: $\mu_1 = \mu$

• String 2 (OQ):

  • Length: $L_2 = 2L$
  • Mass per unit length: $\mu_2 = 4\mu$

• Tension: $T$ (uniform throughout)

The speed of a transverse wave on a string is given by $v = \sqrt{T/\mu}$.

• Wave speed on string 1: $v_1 = \sqrt{\frac{T}{\mu_1}} = \sqrt{\frac{T}{\mu}}$
• Wave speed on string 2: $v_2 = \sqrt{\frac{T}{\mu_2}} = \sqrt{\frac{T}{4\mu}} = \frac{1}{2} \sqrt{\frac{T}{\mu}} = \frac{v_1}{2}$

The reference frequency is given as $v_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{v_1}{2L}$.

2. Analysis of Vibration Modes

For standing waves to form on the composite string, the frequency of vibration must be the same for both segments.

Case 1: Node at the junction O

If the junction point O is a node, then both string segments PO and OQ behave as strings fixed at both ends. The ends P and Q are already fixed.

• For string 1 (PO): The allowed frequencies are given by the formula for a string fixed at both ends: $f_1 = n_1 \frac{v_1}{2L_1} = n_1 \frac{v_1}{2L} = n_1 v_0 \quad (n_1 = 1, 2, 3, ...)$

• For string 2 (OQ): The allowed frequencies are: $f_2 = n_2 \frac{v_2}{2L_2} = n_2 \frac{v_1/2}{2(2L)} = n_2 \frac{v_1}{8L} = \frac{n_2}{4} \left(\frac{v_1}{2L}\right) = \frac{n_2}{4} v_0 \quad (n_2 = 1, 2, 3, ...)$

For a stable standing wave on the composite string, the frequencies must be equal: $f = f_1 = f_2$. $n_1 v_0 = \frac{n_2}{4} v_0 \implies 4n_1 = n_2$

To find the minimum frequency, we need the smallest positive integer values for $n_1$ and $n_2$ that satisfy this condition. The smallest value for $n_1$ is 1, which gives $n_2 = 4$.

The minimum frequency is therefore: $f_{min} = 1 \cdot v_0 = v_0$

Evaluation of Option A: The minimum frequency of vibration of the composite string with a node at O is $v_0$. This statement is correct.

Evaluation of Option C: For the minimum frequency ($f = v_0$), we have $n_1=1$ and $n_2=4$.

• String 1 (PO) vibrates in its fundamental mode ($n_1=1$). The number of nodes is $n_1+1=2$ (at P and O).
• String 2 (OQ) vibrates in its 4th harmonic ($n_2=4$). The number of nodes is $n_2+1=5$ (at O, Q, and 3 intermediate points).

The total number of distinct nodes is the sum of nodes on both strings minus 1 (for the common node at O). Total nodes = (Nodes on PO) + (Nodes on OQ) - 1 = $2 + 5 - 1 = 6$. The nodes are at P, O, Q, and three points between O and Q. This statement is correct.

Case 2: Antinode at the junction O

If the junction O is an antinode, then each string segment behaves as a string fixed at one end (P or Q) and free at the other (O).

• For string 1 (PO): The allowed frequencies are given by the formula for a string fixed at one end and free at the other: $f_1 = (2n_1-1) \frac{v_1}{4L_1} = (2n_1-1) \frac{v_1}{4L} = (2n_1-1) \frac{v_0}{2} \quad (n_1 = 1, 2, 3, ...)$

• For string 2 (OQ): The allowed frequencies are: $f_2 = (2n_2-1) \frac{v_2}{4L_2} = (2n_2-1) \frac{v_1/2}{4(2L)} = (2n_2-1) \frac{v_1}{16L} = (2n_2-1) \frac{v_0}{8} \quad (n_2 = 1, 2, 3, ...)$

For a standing wave to exist, the frequencies must be equal: $f = f_1 = f_2$. $(2n_1-1) \frac{v_0}{2} = (2n_2-1) \frac{v_0}{8}$ $4(2n_1-1) = (2n_2-1)$ $8n_1 - 4 = 2n_2 - 1$ $8n_1 - 3 = 2n_2$

In this equation, $n_1$ and $n_2$ must be positive integers. The left-hand side ($8n_1 - 3$) is always an odd integer. The right-hand side ($2n_2$) is always an even integer. An odd integer can never be equal to an even integer. Therefore, there are no integer solutions for $n_1$ and $n_2$ that satisfy this condition. This means that no vibrational mode with an antinode at O is possible for this composite string.

Evaluation of Option D: This statement is correct.

Evaluation of Option B: This statement claims a minimum frequency of $2v_0$ for a mode with an antinode at O. Since no such mode is possible, this statement is incorrect.

Conclusion

The correct statements are A, C, and D.