A string fastened at both ends has successive resonances with wavelengths of 0.1 m for mth harmonic and 0.08 m for (m+1)th harmonic.
The value of m is

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to find the value of \( m \) given the wavelengths for the \( m \)-th and \( (m+1) \)-th harmonics of a string fixed at both ends.

### Step-by-Step Solution:

1. **Understand the relationship between wavelength and harmonic number**:
For a string fixed at both ends, the wavelength \( \lambda \) of the \( n \)-th harmonic is given by:
\[ \lambda_n = \frac{2L}{n} \]
where \( L \) is the length of the string and \( n \) is the harmonic number.

2. **Set up the equations for the given wavelengths**:
For the \( m \)-th harmonic, we have:
\[ \lambda_m = 0.1 \, \text{m} \]
For the \( (m+1) \)-th harmonic, we have:
\[ \lambda_{m+1} = 0.08 \, \text{m} \]

3. **Express the wavelengths in terms of \( L \)**:
Using the formula for wavelength:
\[ \lambda_m = \frac{2L}{m} \quad \text{(1)} \]
\[ \lambda_{m+1} = \frac{2L}{m+1} \quad \text{(2)} \]

4. **Substitute the given wavelengths into the equations**:
From equation (1):
\[ 0.1 = \frac{2L}{m} \implies 2L = 0.1m \implies L = \frac{0.1m}{2} = 0.05m \quad \text{(3)} \]
From equation (2):
\[ 0.08 = \frac{2L}{m+1} \implies 2L = 0.08(m+1) \implies L = \frac{0.08(m+1)}{2} = 0.04(m+1) \quad \text{(4)} \]

5. **Equate the two expressions for \( L \)**:
From (3) and (4):
\[ 0.05m = 0.04(m + 1) \]

6. **Solve for \( m \)**:
Expanding the right side:
\[ 0.05m = 0.04m + 0.04 \]
Rearranging gives:
\[ 0.05m - 0.04m = 0.04 \implies 0.01m = 0.04 \implies m = \frac{0.04}{0.01} = 4 \]

7. **Conclusion**:
The value of \( m \) is \( 4 \).