Question

The fundamental frequency of sonometer wire is 'n'. If the tension and length are increased 3 times and diameter is increased twice, the new frequency will be

• A. $\sqrt{\frac{3}{2}}n$
• B. $\frac{\sqrt{3}}{2}n$
• C. $\frac{n}{2\sqrt{3}}$
• D. $2\sqrt{3}n$

Hint:

Frequency is inversely proportional to length and diameter, but directly proportional to the square root of tension.

Answer 1

The Correct Option is C

Step 1: Sonometer Frequency Formula
$n = \frac{1}{2l} \sqrt{\frac{T}{m}} = \frac{1}{2l} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{1}{l \cdot D} \sqrt{\frac{T}{\pi \rho}}$ ($D$ is diameter).
Step 2: Proportionality
$n \propto \frac{\sqrt{T}}{l \cdot D}$.
Step 3: Apply Changes
$T' = 3T, l' = 3l, D' = 2D$.
$n' = \frac{\sqrt{3T}}{(3l) \cdot (2D)} = \frac{\sqrt{3}}{6} \cdot \frac{\sqrt{T}}{l \cdot D} = \frac{\sqrt{3}}{6} n$.
Step 4: Simplifying
$\frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{2 \cdot 3} = \frac{\sqrt{3}}{2 \cdot \sqrt{3} \cdot \sqrt{3}} = \frac{1}{2\sqrt{3}}$.
$n' = \frac{n}{2\sqrt{3}}$.
Final Answer:(C)