What are (a) the lowest frequency. (b) the second 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 275 N?

Answer

To solve the problem of finding the lowest, second lowest, and third lowest frequencies for standing waves on a wire, we can follow these steps:

### Step 1: Calculate the mass per unit length (μ)
The mass per unit length (μ) is given by the formula:
\[ \mu = \frac{m}{L} \]
where \( m \) is the mass of the wire and \( L \) is its length.

Given:
- Mass \( m = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Length \( L = 10.0 \, \text{m} \)

Calculating \( \mu \):
\[ \mu = \frac{0.1 \, \text{kg}}{10.0 \, \text{m}} = 0.01 \, \text{kg/m} \]

### Step 2: Calculate the wave speed (v)
The wave speed (v) in a stretched string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire.

Given:
- Tension \( T = 275 \, \text{N} \)

Calculating \( v \):
\[ v = \sqrt{\frac{275 \, \text{N}}{0.01 \, \text{kg/m}}} = \sqrt{27500} \approx 165.83 \, \text{m/s} \]

### Step 3: Calculate the fundamental frequency (f1)
The fundamental frequency (f1) for standing waves on a wire is given by:
\[ f_1 = \frac{v}{2L} \]

Calculating \( f_1 \):
\[ f_1 = \frac{165.83 \, \text{m/s}}{2 \times 10.0 \, \text{m}} = \frac{165.83}{20} \approx 8.29 \, \text{Hz} \]

### Step 4: Calculate the second lowest frequency (f2)
The second harmonic frequency (f2) is given by:
\[ f_2 = 2f_1 \]

Calculating \( f_2 \):
\[ f_2 = 2 \times 8.29 \approx 16.58 \, \text{Hz} \]

### Step 5: Calculate the third lowest frequency (f3)
The third harmonic frequency (f3) is given by:
\[ f_3 = 3f_1 \]

Calculating \( f_3 \):
\[ f_3 = 3 \times 8.29 \approx 24.87 \, \text{Hz} \]

### Final Answers:
(a) The lowest frequency \( f_1 \) is approximately **8.29 Hz**.
(b) The second lowest frequency \( f_2 \) is approximately **16.58 Hz**.
(c) The third lowest frequency \( f_3 \) is approximately **24.87 Hz**.

---