Question

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

Hint:

Frequency is inversely proportional to Length AND Diameter! A thicker string vibrates slower, and a longer string vibrates slower. Tension increases frequency, but only by the square root.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We must calculate how the fundamental frequency of a stretched wire changes when three physical parameters (tension, length, and diameter) are altered simultaneously.

Step 2: Detailed Explanation:
The fundamental frequency ($n$) of a stretched vibrating wire is given by the formula:
$n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$
where:
$L$ = length of the wire
$T$ = tension in the wire
$\mu$ = linear mass density (mass per unit length)
Linear mass density ($\mu$) can be expanded in terms of the wire's material density ($\rho$) and its cross-sectional area:
$\mu = \text{Area} \times \rho = (\pi r^2) \rho = \left( \pi \frac{D^2}{4} \right) \rho$
where $D$ is the diameter of the wire.
Substituting this into the frequency formula:
$n = \frac{1}{2L} \sqrt{\frac{T}{\frac{\pi D^2 \rho}{4}}}$
$n = \frac{1}{2L} \frac{2}{D} \sqrt{\frac{T}{\pi \rho}}$
$n = \frac{1}{LD} \sqrt{\frac{T}{\pi \rho}}$
From this expanded formula, we establish the direct proportionalities:
$n \propto \frac{\sqrt{T}}{L \cdot D}$
Now, apply the given changes to find the new frequency $n'$:
- Tension is increased 3 times: $T' = 3T$
- Length is increased 3 times: $L' = 3L$
- Diameter is increased twice: $D' = 2D$
Substitute these scaled values into the proportionality:
$n' \propto \frac{\sqrt{3T}}{(3L) \cdot (2D)}$
$n' \propto \frac{\sqrt{3} \sqrt{T}}{6 \cdot (L \cdot D)}$
$n' = \left( \frac{\sqrt{3}}{6} \right) \times \left( \frac{1}{LD} \sqrt{\frac{T}{\pi \rho}} \right)$
Since the second bracket is the original frequency $n$:
$n' = \frac{\sqrt{3}}{6} n$
Rationalize the fraction to match the options:
$n' = \frac{\sqrt{3}}{2 \times 3} n = \frac{\sqrt{3}}{2 \times \sqrt{3} \times \sqrt{3}} n = \frac{1}{2\sqrt{3}} n$

Step 3: Final Answer:
The new frequency will be $\frac{n}{2\sqrt{3}}$, matching option (c).