A 2.00 m-long rope, having a mass of 80 g, is fixed at one end and is tied to a light string at the other end. The tension in the string is 256 N. (a) Find the frequencies of the fundamental and the first two overtones. (b) Find the wavelength in the fundamental and the first two overtones.

Explanation

To solve the problem, we will follow these steps:

### Given Data:
- Length of the rope (L) = 2.00 m
- Mass of the rope (m) = 80 g = 0.08 kg
- Tension in the string (T) = 256 N

### Step 1: Calculate the mass per unit length (μ)
The mass per unit length (μ) is given by the formula:
\[ \mu = \frac{m}{L} \]
Substituting the values:
\[ \mu = \frac{0.08 \, \text{kg}}{2.00 \, \text{m}} = 0.04 \, \text{kg/m} \]

### Step 2: Calculate the wave velocity (v)
The velocity of the wave on the string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the values:
\[ v = \sqrt{\frac{256 \, \text{N}}{0.04 \, \text{kg/m}}} = \sqrt{6400} = 80 \, \text{m/s} \]

### Step 3: Determine the frequencies of the fundamental and the first two overtones
For a string fixed at one end, the frequencies are given by:
\[ f_n = \frac{(2n + 1)v}{4L} \]
Where \( n \) is the harmonic number (0 for fundamental, 1 for first overtone, 2 for second overtone).

#### Fundamental Frequency (n = 0):
\[ f_0 = \frac{(2 \cdot 0 + 1) \cdot 80 \, \text{m/s}}{4 \cdot 2.00 \, \text{m}} = \frac{80}{8} = 10 \, \text{Hz} \]

#### First Overtone (n = 1):
\[ f_1 = \frac{(2 \cdot 1 + 1) \cdot 80 \, \text{m/s}}{4 \cdot 2.00 \, \text{m}} = \frac{3 \cdot 80}{8} = 30 \, \text{Hz} \]

#### Second Overtone (n = 2):
\[ f_2 = \frac{(2 \cdot 2 + 1) \cdot 80 \, \text{m/s}}{4 \cdot 2.00 \, \text{m}} = \frac{5 \cdot 80}{8} = 50 \, \text{Hz} \]

### Step 4: Calculate the wavelengths for the fundamental and the first two overtones
The wavelength for each mode is given by:
\[ \lambda_n = \frac{4L}{2n + 1} \]

#### Wavelength for Fundamental (n = 0):
\[ \lambda_0 = \frac{4 \cdot 2.00 \, \text{m}}{2 \cdot 0 + 1} = \frac{8}{1} = 8 \, \text{m} \]

#### Wavelength for First Overtone (n = 1):
\[ \lambda_1 = \frac{4 \cdot 2.00 \, \text{m}}{2 \cdot 1 + 1} = \frac{8}{3} \approx 2.67 \, \text{m} \]

#### Wavelength for Second Overtone (n = 2):
\[ \lambda_2 = \frac{4 \cdot 2.00 \, \text{m}}{2 \cdot 2 + 1} = \frac{8}{5} = 1.6 \, \text{m} \]

### Final Results:
- Frequencies:
- Fundamental Frequency: \( f_0 = 10 \, \text{Hz} \)
- First Overtone: \( f_1 = 30 \, \text{Hz} \)
- Second Overtone: \( f_2 = 50 \, \text{Hz} \)

- Wavelengths:
- Wavelength for Fundamental: \( \lambda_0 = 8 \, \text{m} \)
- Wavelength for First Overtone: \( \lambda_1 \approx 2.67 \, \text{m} \)
- Wavelength for Second Overtone: \( \lambda_2 = 1.6 \, \text{m} \)