Length of a stretched wire is 2m. It is oscillating in its fourth overtone mode. Maximum amplitude of oscillations is 2mm. Find amplitude of oscillation at a distance of 0.2m from one fixed end.

Explanation

To solve the problem step by step, we will follow the reasoning and calculations outlined in the video transcript.

### Step 1: Understand the Problem
We have a stretched wire of length \( L = 2 \, \text{m} \) oscillating in its fourth overtone mode. The maximum amplitude of oscillation is given as \( A = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \). We need to find the amplitude of oscillation at a distance \( x = 0.2 \, \text{m} \) from one fixed end.

### Step 2: Determine the Number of Loops
In the fourth overtone mode, the number of loops (or segments) formed in the wire can be calculated. The number of loops \( n \) in the \( n^{th} \) overtone is given by:
\[ n = 4 \quad (\text{since it's the fourth overtone}) \]
Thus, there are \( n + 1 = 5 \) loops in total.

### Step 3: Relate Length and Wavelength
The length of the wire \( L \) is related to the wavelength \( \lambda \) by the formula:
\[ L = \frac{n \lambda}{2} \]
Substituting \( n = 5 \) and \( L = 2 \, \text{m} \):
\[ 2 = \frac{5 \lambda}{2} \]
From this, we can solve for \( \lambda \):
\[ \lambda = \frac{2 \times 2}{5} = \frac{4}{5} \, \text{m} \]

### Step 4: Write the Wave Equation
The displacement \( y \) of the wave at a distance \( x \) can be expressed using the wave equation:
\[ y = A \sin(kx) \]
where \( k \) is the wave number given by:
\[ k = \frac{2\pi}{\lambda} \]
Substituting \( \lambda = \frac{4}{5} \):
\[ k = \frac{2\pi}{\frac{4}{5}} = \frac{10\pi}{4} = \frac{5\pi}{2} \]

### Step 5: Calculate the Amplitude at \( x = 0.2 \, \text{m} \)
Now we substitute \( A = 2 \times 10^{-3} \, \text{m} \) and \( x = 0.2 \, \text{m} \) into the wave equation:
\[ y = 2 \times 10^{-3} \sin\left(\frac{5\pi}{2} \times 0.2\right) \]
Calculating the argument of the sine function:
\[ \frac{5\pi}{2} \times 0.2 = \frac{5\pi}{10} = \frac{\pi}{2} \]
Thus, we have:
\[ y = 2 \times 10^{-3} \sin\left(\frac{\pi}{2}\right) \]
Since \( \sin\left(\frac{\pi}{2}\right) = 1 \):
\[ y = 2 \times 10^{-3} \times 1 = 2 \times 10^{-3} \, \text{m} = 2 \, \text{mm} \]

### Final Answer
The amplitude of oscillation at a distance of \( 0.2 \, \text{m} \) from one fixed end is \( 2 \, \text{mm} \).

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