Vibration Frequency and Nodes of Two Joined Composite Strings

To determine the correct options, we analyze the wave propagation and boundary conditions for the composite string.

1. Properties of the String Segments

• Left segment (PO):

  • Length $L_1 = L$
  • Linear mass density $\mu_1 = \mu$
  • Tension $T$
  • Wave speed $v_1 = \sqrt{\frac{T}{\mu_1}} = \sqrt{\frac{T}{\mu}}$

• Right segment (OQ):

  • Length $L_2 = 2L$
  • Linear mass density $\mu_2 = 4\mu$
  • Tension $T$
  • Wave speed $v_2 = \sqrt{\frac{T}{\mu_2}} = \sqrt{\frac{T}{4\mu}} = \frac{1}{2}\sqrt{\frac{T}{\mu}} = \frac{v_1}{2}$

The characteristic frequency $\nu_0$ is given as: $\nu_0 = \frac{1}{2L}\sqrt{\frac{T}{\mu}} = \frac{v_1}{2L}$

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2. Analysis of Node at Point O (Options A and C)

If point O is a node, then segment PO and segment OQ vibrate independently as strings fixed at both ends (P & O for PO, and O & Q for OQ).

• For segment PO with fixed ends at $x = 0$ and $x = L$, the possible frequencies are: $\nu = n_1 \frac{v_1}{2L_1} = n_1 \left(\frac{v_1}{2L}\right) = n_1 \nu_0, \quad \text{where } n_1 \in \{1, 2, 3, \dots\}$

• For segment OQ with fixed ends at $x' = 0$ and $x' = 2L$, the possible frequencies are: $\nu = n_2 \frac{v_2}{2L_2} = n_2 \frac{v_1 / 2}{2(2L)} = \frac{n_2}{4} \left(\frac{v_1}{2L}\right) = \frac{n_2}{4} \nu_0, \quad \text{where } n_2 \in \{1, 2, 3, \dots\}$

For the composite string to oscillate in a standing wave mode, both segments must vibrate at the same frequency $\nu$: $n_1 \nu_0 = \frac{n_2}{4} \nu_0 \implies n_2 = 4 n_1$

• Minimum Frequency: Taking the smallest positive integer $n_1 = 1$, we get $n_2 = 4$. $\nu_{\text{min}} = 1 \cdot \nu_0 = \nu_0$ Therefore, Option A is correct.

• Number of Nodes at Minimum Frequency:

  • For $n_1 = 1$ in PO, there are $n_1 + 1 = 2$ nodes (at P and O).
  • For $n_2 = 4$ in OQ, there are $n_2 + 1 = 5$ nodes (at O, 3 interior nodes, and Q).
  • Since point O is shared by both segments, the total number of distinct nodes is: $N = (n_1 + 1) + (n_2 + 1) - 1 = n_1 + n_2 + 1 = 1 + 4 + 1 = 6 \text{ nodes}$ Therefore, Option C is correct.

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3. Analysis of Antinode at Point O (Options B and D)

An antinode occurs at a position where the derivative of the standing wave amplitude with respect to position is zero ($\left.\frac{\partial y}{\partial x}\right|_{\text{at O}} = 0$).

Set up the standing wave equation for both segments:

• Segment PO ($x \in [0, L]$ measured from P): $y_1(x, t) = A_1 \sin(k_1 x) \cos(\omega t), \quad \text{where } k_1 = \frac{\omega}{v_1}$

• Segment OQ ($x' \in [0, 2L]$ measured from Q towards O): $y_2(x', t) = A_2 \sin(k_2 x') \cos(\omega t), \quad \text{where } k_2 = \frac{\omega}{v_2} = \frac{\omega}{v_1 / 2} = 2k_1$

For point O ($x = L$ and $x' = 2L$) to be an antinode: $\left.\frac{\partial y_1}{\partial x}\right|_{x=L} = 0 \implies A_1 k_1 \cos(k_1 L) = 0 \implies k_1 L = (2m+1)\frac{\pi}{2}, \quad m \in \mathbb{Z}_{\ge 0}$ $\left.\frac{\partial y_2}{\partial x'}\right|_{x'=2L} = 0 \implies A_2 k_2 \cos(2 k_2 L) = 0 \implies 2 k_2 L = (2n+1)\frac{\pi}{2}, \quad n \in \mathbb{Z}_{\ge 0}$

Since $k_2 = 2 k_1$, we substitute $2 k_2 L = 4 k_1 L$: $4 k_1 L = 4 (2m+1)\frac{\pi}{2} = 2(2m+1)\pi = (4m+2)\pi$

However, $\cos((4m+2)\pi) = 1 \neq 0$ for any integer $m$. Thus, $\cos(k_1 L) = 0$ and $\cos(4 k_1 L) = 0$ can never be satisfied simultaneously.

Hence, no vibrational mode with an antinode at O is physically possible. Therefore, Option D is correct, and Option B is incorrect.

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Conclusion

The correct options are A, C, and D.