Question

Two wires made up of same material are of equal lengths but their radii are in the ratio \(1: 2\). On stretching each of these two strings by the same tension, the ratio between their fundamental frequency is \(\ldots \ldots \ldots .\) (A) \(1: 2\) (B) \(2: 1\) (C) \(1: 4\) (D) \(4: 1\)

Short answer

The ratio between the fundamental frequencies of the two wires, when stretched with the same tension, is \(2:1\). The correct option is (B).

Step-by-step solution

Write the formula for fundamental frequency

The formula for the fundamental frequency (f) of a stretched wire is given by: \[f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}\] where - L: length of the wire - T: tension in the wire - μ (mu): linear mass density of the wire, which is equal to the mass (m) of the wire divided by its length (L), μ = m/L

Express the linear mass density in terms of radius and volume density

Since the wires are made of the same material, their volume density (ρ, rho) is the same. The volume (V) of a wire is given by: \[V = \pi r^2 L \] where r: radius of the wire The mass (m) of the wire can be expressed as: \[m = \rho V = \rho (\pi r^2 L)\] As we know μ = m/L, we can write μ in terms of r and ρ: \[\mu = \dfrac{m}{L} = \pi r^2 \rho\]

Substitute the expression of mu in the formula of fundamental frequency

Now, we substitute the expression for μ in the formula of f: \[f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\pi r^2 \rho}}\]

Write the formula for both the wires

Let's denote the radii of the wires as r1 and r2. According to the problem, their ratio is given as: \[\dfrac{r_1}{r_2} = \dfrac{1}{2}\] Now, using the formula for fundamental frequency, we have: \[f_1 = \dfrac{1}{2L} \sqrt{\dfrac{T}{\pi r_1^2 \rho}}, \quad f_2 = \dfrac{1}{2L} \sqrt{\dfrac{T}{\pi r_2^2 \rho}} \]

Find the ratio of fundamental frequencies

Now, we will find the ratio of the fundamental frequencies: \[\dfrac{f_1}{f_2} = \dfrac{\dfrac{1}{2L} \sqrt{\dfrac{T}{\pi r_1^2 \rho}}}{\dfrac{1}{2L} \sqrt{\dfrac{T}{\pi r_2^2 \rho}}} = \dfrac{\sqrt{\dfrac{1}{r_1^2}}}{\sqrt{\dfrac{1}{r_2^2}}}\] Remembering that the ratio of r1 to r2 is \(1:2\), so \(r_2 = 2r_1\). we can substitute to get: \[\dfrac{f_1}{f_2} = \dfrac{\sqrt{\dfrac{1}{r_1^2}}}{\sqrt{\dfrac{1}{(2r_1)^2}}} = \dfrac{1}{\sqrt{\dfrac{1}{4}}} = 2\] So, the ratio between their fundamental frequencies is 2:1. The answer is (B) \(2:1\).