Question

A wire of length '\(L\)', diameter '\(d\)', density of material '\(ρ\)' is under tension '\(T\)' has fundamental frequency of vibration '\(n_A\)'. Another wire of length '\(2L\)', diameter '\(3d\)', density of material '\(2ρ\)' is vibrated under tension '\(2T\)', the fundamental frequency of vibration becomes '\(n_B\)'. The ratio \(n_B:n_A\) is

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

Hint:

n = (1/2L) sqrt(T/mu) with mass per unit length mu = rho pi d squared over 4.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The fundamental frequency of a stretched wire is \(n = \frac{1}{2L}\sqrt{\frac T\mu}\), where \(\mu = \rho\cdot\frac{\pi d^2}{4}\) is the mass per unit length.

Step 2: Write the dependence:
\[ n \propto \frac{1}{L}\sqrt{\frac{T}{\rho d^2}} \]

Step 3: Ratio:
Wire A: \(L, T, \rho, d\). Wire B: \(2L, 2T, 2\rho, 3d\).
\[ \frac{n_B}{n_A} = \frac{L}{2L}\sqrt{\frac{2T}{T}\cdot\frac{\rho}{2\rho}\cdot\frac{d^2}{9d^2}} = \frac12\sqrt{2\cdot\frac12\cdot\frac19} = \frac12\cdot\frac13 = \frac16 \]

Step 4: Why the other options are wrong.
1:2 comes from using only the length ratio. 1:4 and 1:8 do not come out of the ratios of tension, density and diameter together.

Final Answer:
The ratio \(n_B : n_A\) is 1:6, option (C). \[ \boxed{1:6} \]