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3 : 5

4 : 9

1 : 2

1 : 4

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Step-by-Step Solution Step 1: Understand the Problem We have two wires made of the same material (hence same density ρ) but with different diameters. They are subject to the same tension T. We want to find the ratio of the number of antinodes (loops) formed when both wires resonate at a common frequency. The correct ratio given is 1 : 2. Step 2: Express Mass per Unit Length for Each Wire Let the radius of the thinner wire be r, and the mass per unit length of the thinner wire be μ1. For the thicker wire with radius 2r, let the mass per unit length be μ2. The mass per unit length μ is given by: $$ \mu = \rho \times \text{(cross-sectional area)} $$ where ρ is the density of the material. Therefore: $$ \mu_1 = \rho \cdot \pi r^2 $$ $$ \mu_2 = \rho \cdot \pi (2r)^2 = \rho \cdot 4\pi r^2 = 4\mu_1 $$ Step 3: Relate Wave Speed in Each Wire to Its Mass per Unit Length The speed of a wave on a string under tension T is: $$ V = \sqrt{\frac{T}{\mu}}. $$ Since both wires have the same tension T, but different μ, we can write: $$ V_1 = \sqrt{\frac{T}{\mu_1}}, \quad V_2 = \sqrt{\frac{T}{\mu_2}}. $$ With μ2 = 4μ1, we have: $$ V_2 = \sqrt{\frac{T}{4\mu_1}} = \frac{1}{2}\sqrt{\frac{T}{\mu_1}} = \frac{V_1}{2}. $$ Step 4: Determine the Fundamental Frequencies of Each Wire For a wire of length L, fixed at both ends, the fundamental frequency f is given by: $$ f = \frac{V}{2L}. $$ Thus: $$ f_1 = \frac{V_1}{2L}, \quad f_2 = \frac{V_2}{2L}. $$ Using V2 = V1/2, we get: $$ f_2 = \frac{V_1/2}{2L} = \frac{V_1}{4L}. $$ Step 5: Find the Common Resonance Condition and the Resulting Number of Loops When both wires are made to vibrate simultaneously, they each produce their own series of harmonics. The frequency at which both resonate together is effectively where a common frequency can be formed from their harmonic series. Labeling the harmonic of wire 1 by n1 and that of wire 2 by n2: $$ n_1 \cdot f_1 = n_2 \cdot f_2. $$ Since f2 = f1/2, a straightforward scenario is n1 = 1 and n2 = 2. This implies wire 1 is vibrating in its first harmonic (one loop) and wire 2 in its second harmonic (two loops). Hence, the ratio of the number of antinodes (loops) is: $$ 1 : 2. $$ Final Answer The ratio of the number of antinodes (loops) for the two wires is 1 : 2.

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