A string has a length of 5 cm between fixed points and has a fundamental frequency of 20 Hz. What is the frequency of the second overtone?

Options

A. 40 Hz
B. 50 Hz
C. 60 Hz
D. 30 Hz

Answer

To solve the problem, we need to find the frequency of the second overtone of a string that has a fundamental frequency of 20 Hz and a length of 5 cm.

### Step-by-Step Solution:

1. **Identify the Fundamental Frequency**:
The fundamental frequency (first harmonic) is given as \( f_1 = 20 \, \text{Hz} \).

2. **Determine the Wavelength for the Fundamental Frequency**:
For a string fixed at both ends, the wavelength \( \lambda \) of the fundamental frequency is given by:
\[ \lambda = 2L \]
where \( L \) is the length of the string. Given \( L = 5 \, \text{cm} = 0.05 \, \text{m} \):
\[ \lambda = 2 \times 0.05 \, \text{m} = 0.1 \, \text{m} = 10 \, \text{cm} \]

3. **Calculate the Wave Speed**:
The wave speed \( v \) can be calculated using the formula:
\[ v = f_1 \cdot \lambda \]
Substituting the known values:
\[ v = 20 \, \text{Hz} \times 0.1 \, \text{m} = 2 \, \text{m/s} \]

4. **Determine the Wavelength for the Second Overtone**:
The second overtone corresponds to the third harmonic (n = 3). For the third harmonic, the relationship between the wavelength \( \lambda' \) and the length of the string is:
\[ L = \frac{3\lambda'}{2} \]
Rearranging gives:
\[ \lambda' = \frac{2L}{3} \]
Substituting \( L = 0.05 \, \text{m} \):
\[ \lambda' = \frac{2 \times 0.05 \, \text{m}}{3} = \frac{0.1 \, \text{m}}{3} = \frac{10 \, \text{cm}}{3} \approx 3.33 \, \text{cm} \]

5. **Calculate the Frequency of the Second Overtone**:
The frequency \( f' \) for the second overtone can be calculated using the wave speed:
\[ v = f' \cdot \lambda' \]
Rearranging gives:
\[ f' = \frac{v}{\lambda'} \]
Substituting the values:
\[ f' = \frac{2 \, \text{m/s}}{\frac{10 \, \text{cm}}{3}} = \frac{2 \, \text{m/s}}{\frac{0.1 \, \text{m}}{3}} = \frac{2 \, \text{m/s} \cdot 3}{0.1 \, \text{m}} = \frac{6}{0.1} = 60 \, \text{Hz} \]

### Final Answer:
The frequency of the second overtone is \( 60 \, \text{Hz} \).

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