The tension, length, diameter and density of a string B are double than that of another string A. Which of the following overtones of B is same as the fundamental frequency of A?

Options

A. (a)1st
B. (b)2nd
C. (c)3rd
D. (d)4th

Explanation

To solve the problem, we need to determine which overtone of string B corresponds to the fundamental frequency of string A, given that the tension, length, diameter, and density of string B are all double those of string A.

### Step-by-Step Solution:

1. **Understanding the Frequency Formula**:
The frequency \( f \) of a vibrating string can be expressed as:
\[ f = \frac{V}{\lambda} \]
where \( V \) is the wave speed and \( \lambda \) is the wavelength.

2. **Wave Speed Formula**:
The wave speed \( V \) on a string is given by:
\[ V = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string and \( \mu \) is the linear mass density (mass per unit length).

3. **Linear Mass Density**:
The linear mass density \( \mu \) can be calculated as:
\[ \mu = \frac{m}{L} \]
where \( m \) is the mass of the string and \( L \) is its length. The mass can also be expressed in terms of volume and density:
\[ m = \text{density} \times \text{volume} = \text{density} \times (\text{cross-sectional area} \times L) \]

4. **Calculating Frequencies**:
- For string A:
\[ f_A = \sqrt{\frac{T_A}{\mu_A}} \cdot \frac{1}{\lambda_A} \]
The wavelength for the fundamental frequency (first harmonic, \( n=1 \)) is:
\[ \lambda_A = 2L_A \]
Thus:
\[ f_A = \sqrt{\frac{T_A}{\mu_A}} \cdot \frac{1}{2L_A} \]

- For string B:
Given that tension, length, diameter, and density of string B are double those of A:
\[ T_B = 2T_A, \quad L_B = 2L_A, \quad \mu_B = \frac{2 \cdot \text{density}_A \cdot \text{cross-sectional area}}{2L_A} = \frac{2\mu_A}{2} = \mu_A \]
The wavelength for the fundamental frequency of string B is:
\[ \lambda_B = 2L_B = 4L_A \]
Thus:
\[ f_B = \sqrt{\frac{T_B}{\mu_B}} \cdot \frac{1}{\lambda_B} = \sqrt{\frac{2T_A}{\mu_A}} \cdot \frac{1}{4L_A} \]

5. **Equating Frequencies**:
We need to find \( n \) such that:
\[ f_B = f_A \]
From the expressions derived, we can equate:
\[ \sqrt{\frac{2T_A}{\mu_A}} \cdot \frac{1}{4L_A} = \sqrt{\frac{T_A}{\mu_A}} \cdot \frac{1}{2L_A} \]
Simplifying this gives:
\[ \frac{1}{4} \sqrt{2} = \frac{1}{2} \]
This implies that \( n = 2 \) corresponds to the second harmonic of string B.

6. **Finding the Overtone**:
The overtone is defined as \( n - 1 \). Therefore, if the second harmonic corresponds to \( n = 2 \), the first overtone is:
\[ \text{First overtone} = n - 1 = 2 - 1 = 1 \]

### Final Answer:
The first overtone of string B is the same as the fundamental frequency of string A.