Question

Two wires \(A\) and \(B\) of lengths in the ratio \(1:2\) and masses in the ratio \(2:1\) are stretched by the same tension. The ratio of the fundamental frequencies of wires \(A\) and \(B\) is

• A. \(2\sqrt{2}:1\)
• B. \(1:\sqrt{2}\)
• C. \(1:1\)
• D. \(\sqrt{2}:1\)

Hint:

For stretched strings, \[ f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}. \] When tensions are equal, frequency varies inversely with length and with the square root of linear mass density.

Answer 1

The Correct Option is C

Step 1: Relate frequency to length and linear density.
f ∝ 1/(L√μ). Given L_A:L_B = 1:2 and M_A:M_B = 2:1 → μ ∝ M/L.

Step 2: Compute linear density ratio.
μ_A/μ_B = (2/1)/(1/2) = 4:1.

Step 3: Find frequency ratio.
f_A/f_B = (L_B/L_A)√(μ_B/μ_A) = 2 × √(1/4) = 2 × 1/2 = 1. Ratio = 1:1.

Step 4: Final Answer:
1:1.