Question

A wire of length L , diameter ' d ' density of material ' e ' is under tension ' T ', having fundamental frequency of vibration $\text{n}_\text{A}$. Another wire of length 2 L , tension 2 T , density 2 e and diameter 3 d has fundamental frequency of vibration $\text{n}_\text{B}$. The ratio $\text{n}_\text{B} : \text{n}_\text{A}$ is

• A. 1 : 2
• B. 1 : 4
• C. 1 : 6
• D. 1 : 8

Hint:

Frequency is inversely proportional to length and diameter, and proportional to the square root of tension. $n \propto \frac{1}{Ld}\sqrt{\frac{T}{\rho}}$.

Answer 1

The Correct Option is C

Step 1: Concept
The fundamental frequency of a vibrating string is $n = \frac{1}{2L} \sqrt{\frac{T}{m}}$, where $m = \text{area} \times \text{density} = \frac{\pi d^2}{4} \rho$.

Step 2: Meaning
Thus, $n = \frac{1}{2L} \sqrt{\frac{T}{\frac{\pi d^2}{4} \rho}} \propto \frac{1}{L \cdot d} \sqrt{\frac{T}{\rho}}$.

Step 3: Analysis
$\frac{n_B}{n_A} = \frac{L_A}{L_B} \cdot \frac{d_A}{d_B} \sqrt{\frac{T_B}{T_A} \cdot \frac{\rho_A}{\rho_B}}$. $\frac{n_B}{n_A} = \frac{L}{2L} \cdot \frac{d}{3d} \sqrt{\frac{2T}{T} \cdot \frac{e}{2e}}$. $\frac{n_B}{n_A} = \frac{1}{2} \cdot \frac{1}{3} \sqrt{1} = \frac{1}{6}$.

Step 4: Conclusion
The ratio $\text{n}_\text{B} : \text{n}_\text{A}$ is $1 : 6$. Final Answer: (C)