Two uniform wires of a the same material are vibrating under the same tension. If the first overtone of the first wire is equal to the second overtone of the second wire and radius of the first wire is twice the radius of the second wire, then the ratio of the lengths of the first wire to second wire is

Options

A. `(1)/(3)`
B. `(1)/(4)`
C. `(1)/(5)`
D. `(1)/(6)`

Answer

To solve the problem, we need to find the ratio of the lengths of two uniform wires given certain conditions. Let's break it down step by step:

### Step 1: Understand the Problem
We have two uniform wires made of the same material, vibrating under the same tension. The first overtone of the first wire is equal to the second overtone of the second wire. The radius of the first wire is twice that of the second wire.

### Step 2: Define the Variables
Let:
- \( r_1 \) = radius of the first wire
- \( r_2 \) = radius of the second wire
- \( L_1 \) = length of the first wire
- \( L_2 \) = length of the second wire
- \( A_1 \) = cross-sectional area of the first wire
- \( A_2 \) = cross-sectional area of the second wire
- \( T \) = tension in the wires
- \( \rho \) = density of the material
- \( n_1 \) = harmonic number for the first wire
- \( n_2 \) = harmonic number for the second wire

### Step 3: Calculate the Areas
Since the radius of the first wire is twice that of the second wire:
\[ r_1 = 2r_2 \]
The cross-sectional area \( A \) of a wire is given by:
\[ A = \pi r^2 \]
Thus, the areas of the wires are:
\[ A_1 = \pi (r_1)^2 = \pi (2r_2)^2 = 4\pi (r_2)^2 = 4A_2 \]

### Step 4: Identify the Harmonics
The first overtone corresponds to the second harmonic, so:
\[ n_1 = 2 \quad (\text{for the first wire}) \]
The second overtone corresponds to the third harmonic, so:
\[ n_2 = 3 \quad (\text{for the second wire}) \]

### Step 5: Use the Frequency Formula
The frequency \( f \) of a vibrating wire fixed at both ends is given by:
\[ f = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \]
where \( \mu \) (linear mass density) is given by:
\[ \mu = \rho A \]
Thus, we can write:
\[ f_1 = \frac{n_1}{2L_1} \sqrt{\frac{T}{\mu_1}} \quad \text{and} \quad f_2 = \frac{n_2}{2L_2} \sqrt{\frac{T}{\mu_2}} \]

### Step 6: Set the Frequencies Equal
Since the first overtone of the first wire is equal to the second overtone of the second wire:
\[ f_1 = f_2 \]
Substituting the expressions for \( f_1 \) and \( f_2 \):
\[ \frac{2}{2L_1} \sqrt{\frac{T}{\mu_1}} = \frac{3}{2L_2} \sqrt{\frac{T}{\mu_2}} \]

### Step 7: Simplify the Equation
Cancel \( T \) and rearrange:
\[ \frac{2}{L_1} \sqrt{\frac{1}{\mu_1}} = \frac{3}{L_2} \sqrt{\frac{1}{\mu_2}} \]
This leads to:
\[ \frac{L_2}{L_1} = \frac{3\sqrt{\mu_1}}{2\sqrt{\mu_2}} \]

### Step 8: Substitute for Linear Mass Densities
Since \( \mu = \rho A \):
\[ \mu_1 = \rho A_1 = \rho (4A_2) = 4\rho A_2 \quad \text{and} \quad \mu_2 = \rho A_2 \]
Thus:
\[ \sqrt{\mu_1} = \sqrt{4\rho A_2} = 2\sqrt{\rho A_2} \]
Substituting back:
\[ \frac{L_2}{L_1} = \frac{3(2\sqrt{\rho A_2})}{2\sqrt{\rho A_2}} = 3 \]

### Step 9: Find the Ratio of Lengths
Taking the reciprocal gives:
\[ \frac{L_1}{L_2} = \frac{1}{3} \]
Thus, the ratio of the lengths of the first wire to the second wire is:
\[ L_1 : L_2 = 1 : 3 \]

### Final Answer
The ratio of the lengths of the first wire to the second wire is \( 1 : 3 \).

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