What are (a) the lowest frequency (b) the scond lowest frequency and (c) the third 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?

Explanation

To solve the problem of finding the lowest, second lowest, and third lowest frequencies 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 1: Calculate the Linear Density (μ)
The linear density (μ) is defined as the mass per unit length of the wire.

Given:
- Mass (m) = 100 g = 0.1 kg
- Length (L) = 10.0 m

\[ \mu = \frac{m}{L} = \frac{0.1 \, \text{kg}}{10.0 \, \text{m}} = 0.01 \, \text{kg/m} \]

### Step 2: Calculate the Wave Velocity (V)
The velocity of a transverse wave on a string is given by the formula:

\[ V = \sqrt{\frac{T}{\mu}} \]

Where \( T \) is the tension in the wire.

Given:
- Tension (T) = 25 N

Substituting the values:

\[ V = \sqrt{\frac{25 \, \text{N}}{0.01 \, \text{kg/m}}} = \sqrt{2500} = 50 \, \text{m/s} \]

### Step 3: Calculate the Frequencies
The fundamental frequency (first harmonic) for a string fixed at both ends is given by:

\[ f_1 = \frac{V}{2L} \]

Substituting the values:

\[ f_1 = \frac{50 \, \text{m/s}}{2 \times 10.0 \, \text{m}} = \frac{50}{20} = 2.5 \, \text{Hz} \]

The second harmonic (second lowest frequency) is:

\[ f_2 = 2f_1 = 2 \times 2.5 \, \text{Hz} = 5.0 \, \text{Hz} \]

The third harmonic (third lowest frequency) is:

\[ f_3 = 3f_1 = 3 \times 2.5 \, \text{Hz} = 7.5 \, \text{Hz} \]

### Summary of Frequencies
- (a) Lowest frequency \( f_1 = 2.5 \, \text{Hz} \)
- (b) Second lowest frequency \( f_2 = 5.0 \, \text{Hz} \)
- (c) Third lowest frequency \( f_3 = 7.5 \, \text{Hz} \)

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