Two uniform strings of mass per unit length u and 4u, 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 v0 = 1/ 2L *[√T/u], which of the following statement(s) is(are) correct?

🔍 Question Breakdown

Scenario: Two strings are joined at point O and tied at fixed ends P and Q.

• String 1: Mass/length = μ, Length = L.

• String 2: Mass/length = 4μ, Length = 2L.

• Tension: T in both strings.

• Given: Frequency $f_0=\frac{1}{2\pi }\sqrt{\frac{T}{\mu }}$.

Task: Determine which statements about vibrational modes (nodes/antinodes at O) are correct.

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🎯 Key Concepts

1. Standing Waves: Nodes (zero displacement) and antinodes (max displacement) form based on boundary conditions.

2. Frequency Formula: For a string of length $l$, mass/length $\mu$, and tension $T$:

   $$f=\frac{n}{2l}\sqrt{\frac{T}{\mu }},n=\text{harmonic number}\mathrm{.}$$

3. Composite Strings: Joined strings must vibrate at the same frequency with matching boundary conditions at O.

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📝 Detailed Analysis of Each Option

Option A: With a node at O, the minimum frequency is $f_0$.

Step 1: For a node at O, both strings vibrate in their fundamental modes (1st harmonic).

• String 1 (length $L$):

  $$f_1=\frac{1}{2L}\sqrt{\frac{T}{\mu }}=f_0(\text{since }{f}_0\text{ is defined as }\frac{1}{2\pi }\sqrt{\frac{T}{\mu }})\mathrm{.}$$

• String 2 (length $2L$):

  $$f_2=\frac{1}{2(2L)}\sqrt{\frac{T}{4\mu }}=\frac{1}{4L}\cdot \frac{1}{2}\sqrt{\frac{T}{\mu }}=\frac{f_0}{4}\mathrm{.}$$

Step 2: To match frequencies, String 2 must vibrate in its 4th harmonic ($n=4$):

$$f_2^{(4)}=4\times \frac{f_0}{4}=f_0\mathrm{.}$$

Conclusion: Minimum frequency = $f_0$. ✅ Option A is correct!

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Option B: With an antinode at O, the minimum frequency is $2f_0$.

Step 1: For an antinode at O, both strings have free ends at O, vibrating in odd harmonics.

• String 1 (length $L$): Fundamental frequency with antinode at O:

  $$f_1=\frac{1}{4L}\sqrt{\frac{T}{\mu }}=\frac{f_0}{2}\mathrm{.}$$

• String 2 (length $2L$): Fundamental frequency with antinode at O:

  $$f_2=\frac{1}{4(2L)}\sqrt{\frac{T}{4\mu }}=\frac{1}{8L}\cdot \frac{1}{2}\sqrt{\frac{T}{\mu }}=\frac{f_0}{8}\mathrm{.}$$

Step 2: To match frequencies, the LCM of $f_0\mathrm{/}2$ and $f_0\mathrm{/}8$ is $f_0\mathrm{/}2$.
Conclusion: Minimum frequency = $f_0\mathrm{/}2$, not $2f_0$. ❌ Option B is incorrect!

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Option C: With a node at O, the composite string has 6 nodes (including ends).

Step 1: For node at O:

• String 1 has nodes at P and O (2 nodes).

• String 2 has nodes at O and Q (2 nodes).
  Step 2: Total nodes = P, O, Q (3 nodes).
  Conclusion: Option C claims 6 nodes—way off! ❌ Option C is incorrect!

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Option D: No vibrational mode with an antinode at O is possible.

Step 1: From Option B, the minimum frequency with antinode at O is $f_0\mathrm{/}2$, but String 2 cannot vibrate at this frequency without fractional harmonics (physically impossible).
Step 2: Frequencies never align for antinodes.
Conclusion: No antinode mode exists. ✅ Option D is correct!

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🎉 Final Answer

Correct Options: $\boxed{{\displaystyle A}}$ and $\boxed{{\displaystyle D}}$

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🌟 Key Takeaways

• Node at O: Minimum frequency = $f_0$, with 3 nodes total.

• Antinode at O: Impossible due to mismatched harmonics.

• Always verify boundary conditions and harmonic compatibility!

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