Step 1: Understanding the Concept:
The frequency of a string depends on length, tension, and linear mass density. Linear mass density further depends on diameter and volume density.
Step 2: Key Formula or Approach:
\[ n = \frac{1}{2L} \sqrt{\frac{T}{m}} = \frac{1}{2L} \sqrt{\frac{T}{\rho \times A}} = \frac{1}{2L} \sqrt{\frac{T}{\rho \times \frac{\pi D^{2}}{4}}} \]
\[ n \propto \frac{1}{L \times D} \sqrt{T} \]
: Detailed Explanation:
Let \(n_{1} = n\).
New length \(L_{2} = 3L_{1}\).
New diameter \(D_{2} = 3D_{1}\).
New tension \(T_{2} = 3T_{1}\).
\[ n_{2} = n \times \frac{L_{1}}{L_{2}} \times \frac{D_{1}}{D_{2}} \times \sqrt{\frac{T_{2}}{T_{1}}} \]
\[ n_{2} = n \times \frac{1}{3} \times \frac{1}{3} \times \sqrt{3} \]
\[ n_{2} = n \times \frac{\sqrt{3}}{9} = \frac{n \sqrt{3}}{3 \times 3} = \frac{n}{3\sqrt{3}} \]
Step 3: Final Answer:
The new frequency is \(\frac{n}{3\sqrt{3}}\).