Two uniform strings of mass per unit length 𝜇 and 4𝜇, 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 $v_0=\frac{1}{2L}\sqrt{\frac{T}{\mathrm{\mu <!--\; \mu \; -->}}}$ , which of the following statement(s) is(are) correct?

Let us analyze the wave propagation and standing wave conditions on the composite string shown in the figure.

1. Wave Velocities in the Two Strings:
The tension $T$ is uniform throughout both strings.
For the left string (segment PO of length $L$ and mass per unit length $\mu$), the wave speed $v_1$ is:
$v_1=\sqrt{\frac{T}{\mu }}$
For the right string (segment OQ of length $2L$ and mass per unit length $4\mu$), the wave speed $v_2$ is:
$v_2=\sqrt{\frac{T}{4\mu }}=\frac{1}{2}\sqrt{\frac{T}{\mu }}=\frac{v_1}{2}$

2. Condition for a Node at O:
Both boundary ends P and Q are fixed, meaning they must always be nodes.
If the junction point O is also a node, then both segments PO and OQ act independently as strings fixed at both ends.
For segment PO (length $L_1=L$), the allowed frequencies of vibration are:
$f_1=\frac{p{v}_1}{2L}=p\left(\frac{1}{2L}\sqrt{\frac{T}{\mu }}\right)=p{\nu }_0$
where $p=1,2,3,...$ is an integer.
For segment OQ (length $L_2=2L$), the allowed frequencies of vibration are:
$f_2=\frac{q{v}_2}{2\left(2L\right)}=\frac{q(v_1/2)}{4L}=\frac{q}{8}\frac{v_1}{L}=\frac{q}{4}\left(\frac{v_1}{2L}\right)=\frac{q}{4}{\nu }_0$
where $q=1,2,3,...$ is an integer.
For a standing wave to form on the entire composite string, the frequency must be identical in both sections ($f_1=f_2=f$):
$p{\nu }_0=\frac{q}{4}{\nu }_0\Rightarrow q=4p$
Since $p$ and $q$ must be integers, the minimum non-zero frequency occurs for $p=1$, which gives:
$q=4$
Thus, the minimum frequency of vibration with a node at O is:
$f_{\text{min}}=1·{\nu }_0={\nu }_0$

3. Number of Nodes at Minimum Frequency:
- String 1 (PO) vibrates in its fundamental mode ($p=1$), having nodes at its ends P and O (2 nodes total).
- String 2 (OQ) vibrates in its 4th harmonic ($q=4$), having nodes at its ends O, Q and 3 internal nodes (5 nodes total).
- Since the junction O is a common node, the total number of nodes is:
$N_{\text{total}}=2+5-1=6$
Therefore, there are 6 nodes including the end nodes.

4. Checking for an Antinode at O:
If there is an antinode at O, since P and Q are fixed nodes: - For PO: The distance from the node P to the antinode O must be an odd multiple of a quarter-wavelength:
$L=(2n_1-1)\frac{{\lambda }_1}{4}=(2n_1-1)\frac{v_1}{4f}$
- For OQ: The distance from the antinode O to the node Q must be an odd multiple of a quarter-wavelength:
$2L=(2n_2-1)\frac{{\lambda }_2}{4}=(2n_2-1)\frac{v_2}{4f}$
Since $v_2=v_1/2$, substituting this in:
$2L=(2n_2-1)\frac{v_1}{8f}\Rightarrow L=(2n_2-1)\frac{v_1}{16f}$
Equating the two expressions for $L$:
$(2n_1-1)\frac{v_1}{4f}=(2n_2-1)\frac{v_1}{16f}$
Simplifying:
$4(2n_1-1)=2n_2-1$
$8n_1-4=2n_2-1\Rightarrow 2(n_2-4n_1)=-3$
Since the left-hand side is an even integer and the right-hand side is an odd integer ($-3$), there is no integer solution for $n_1$ and $n_2$. Thus, no vibrational mode with an antinode at O is possible.