What is the second lowest frequency for standing waves on a wire that is 10.0 m long has a mass of 100 g and is stretched under a tension of 25 N which is fixed at both ends ?

Options

A. 2.5 Hz
B. 5.0 Hz
C. 7.5 Hz
D. 10.0 HZ

Explanation

To find the second lowest frequency for standing waves on a wire that is 10.0 m long, has a mass of 100 g, and is stretched under a tension of 25 N, we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the linear mass density (μ)**:
\[ \text{Mass} = 100 \text{ g} = 0.1 \text{ kg} \]
\[ \text{Length} = 10.0 \text{ m} \]
\[ \mu = \frac{\text{Mass}}{\text{Length}} = \frac{0.1 \text{ kg}}{10.0 \text{ m}} = 0.01 \text{ kg/m} \]

2. **Determine the wave speed (v)**:
The wave speed on the string can be calculated using the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T = 25 \text{ N} \).
\[ v = \sqrt{\frac{25 \text{ N}}{0.01 \text{ kg/m}}} = \sqrt{2500} = 50 \text{ m/s} \]

3. **Find the fundamental frequency (f₁)**:
The fundamental frequency (first harmonic) is given by:
\[ f_1 = \frac{v}{2L} \]
where \( L = 10.0 \text{ m} \).
\[ f_1 = \frac{50 \text{ m/s}}{2 \times 10.0 \text{ m}} = \frac{50}{20} = 2.5 \text{ Hz} \]

4. **Calculate the second harmonic frequency (f₂)**:
The second harmonic (first overtone) frequency is:
\[ f_2 = 2f_1 \]
\[ f_2 = 2 \times 2.5 \text{ Hz} = 5.0 \text{ Hz} \]

### Final Answer:
The second lowest frequency for standing waves on the wire is **5.0 Hz**.

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