Question

Two vibrating metal wires of the same material of length 'L' and '3L' and have radii '2r' and 'r' respectively. They are stretched under same tension. Both the wires vibrate in their fundamental modes. The ratio \(\frac{n_{1}}{n_{2}}\) is [\(n_{1}\) and \(n_{2}\) are the frequencies of first and second wire respectively]

• A. \(1:1\)
• B. \(1.5:1\)
• C. \(2:1\)
• D. \(2.5:1\)

Hint:

Fundamental frequency depends on length and mass per unit length.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Fundamental frequency is \(n=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}\) with \(\mu=\pi r^{2}\rho\).

Step 2: Dependence
For same material and tension, \(n\propto\dfrac{1}{L\,r}\).

Step 3: Calculation
\(\dfrac{n_{1}}{n_{2}}=\dfrac{L_{2}r_{2}}{L_{1}r_{1}}=\dfrac{3L\times r}{L\times2r}=1.5\).

Final Answer:
The frequency ratio is 1.5 to 1. \[ \boxed{\text{(B) }1.5:1} \]