Step 1: Understanding the Question:
The fundamental frequency of a vibrating string depends on its length, tension, and its mass per unit length (which relates to diameter).
Step 2: Key Formula or Approach:
Frequency $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$
Mass per unit length $\mu = \text{volume per unit length} \times \text{density} = \pi \left(\frac{D}{2}\right)^2 \cdot 1 \cdot \rho = \frac{\pi D^2 \rho}{4}$.
So, $n = \frac{1}{2L} \sqrt{\frac{T}{\pi D^2 \rho / 4}} = \frac{1}{LD} \sqrt{\frac{T}{\pi \rho}}$.
This shows $n \propto \frac{\sqrt{T}}{LD}$.
Step 3: Detailed Explanation:
Initial frequency: $n \propto \frac{\sqrt{T}}{LD}$.
New parameters:
$T' = 3T$, $L' = 3L$, $D' = 2D$.
New frequency $n'$:
\[ n' \propto \frac{\sqrt{3T}}{(3L)(2D)} \]
\[ n' \propto \frac{\sqrt{3}}{6} \cdot \frac{\sqrt{T}}{LD} \]
Ratio:
\[ \frac{n'}{n} = \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{2 \times 3} = \frac{\sqrt{3}}{2 \times \sqrt{3} \times \sqrt{3}} = \frac{1}{2\sqrt{3}} \]
\[ n' = \frac{n}{2\sqrt{3}} \]
Step 4: Final Answer:
The new frequency will be $\frac{n}{2\sqrt{3}}$.