A `1 m` long rope, having a mass of `40 g`, is fixed at one end and is tied to a light string at the other end. The tension in the string in `400 N`. Find the wavelength in second overtone (in `cm`).

Explanation

To solve the problem, we need to find the wavelength of the second overtone for a rope fixed at one end. Here’s a step-by-step solution:

### Step 1: Convert Mass to Kilograms
The mass of the rope is given as 40 g. We need to convert this to kilograms.

\[ \text{Mass} = 40 \, \text{g} = \frac{40}{1000} \, \text{kg} = 0.04 \, \text{kg} \]

**Hint:** Remember that 1 kg = 1000 g.

### Step 2: Calculate Linear Mass Density (μ)
The linear mass density (μ) is calculated as the mass per unit length of the rope.

\[ \mu = \frac{\text{mass}}{\text{length}} = \frac{0.04 \, \text{kg}}{1 \, \text{m}} = 0.04 \, \text{kg/m} \]

**Hint:** Linear mass density is crucial for wave speed calculations.

### Step 3: Calculate Wave Velocity (v)
The wave velocity (v) on the rope can be calculated using the formula:

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

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

Given \( T = 400 \, \text{N} \):

\[ v = \sqrt{\frac{400 \, \text{N}}{0.04 \, \text{kg/m}}} = \sqrt{10000} = 100 \, \text{m/s} \]

**Hint:** The tension in the string affects the speed of the wave on the rope.

### Step 4: Determine the Fundamental Wavelength (λ)
For a rope fixed at one end, the fundamental wavelength (λ) is given by:

\[ \frac{\lambda}{4} = L \]

Where \( L \) is the length of the rope (1 m). Thus,

\[ \lambda = 4L = 4 \times 1 \, \text{m} = 4 \, \text{m} \]

**Hint:** The fundamental mode has a specific relationship between wavelength and length for fixed ends.

### Step 5: Calculate the Frequency (f)
The frequency can be calculated using the wave velocity and wavelength:

\[ f = \frac{v}{\lambda} = \frac{100 \, \text{m/s}}{4 \, \text{m}} = 25 \, \text{Hz} \]

**Hint:** Frequency is inversely proportional to wavelength.

### Step 6: Find the Wavelength of the Second Overtone
In a rope fixed at one end, only odd harmonics are present. The second overtone corresponds to the fifth harmonic (n=5).

The relationship for the wavelength of the nth overtone is given by:

\[ \frac{n \lambda}{4} = L \]

For the second overtone (n=5):

\[ \frac{5 \lambda}{4} = L \implies \lambda = \frac{4L}{5} \]

Substituting \( L = 1 \, \text{m} \):

\[ \lambda = \frac{4 \times 1 \, \text{m}}{5} = \frac{4}{5} \, \text{m} = 0.8 \, \text{m} \]

### Step 7: Convert Wavelength to Centimeters
To convert the wavelength from meters to centimeters:

\[ \lambda = 0.8 \, \text{m} \times 100 = 80 \, \text{cm} \]

### Final Answer
The wavelength in the second overtone is **80 cm**.

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