A string of length l is fixed at both ends and is vibrating in second harmonic. The amplitude at anti-node is 5 mm. The amplitude of a particle at distance `l//8` from the fixed end is

Options

A. `5sqrt(2)`mm
B. `(5)/(sqrt(2))mm`
C. 5mm
D. `(10)/(sqrt(2))mm`

Explanation

To solve the problem step by step, we need to determine the amplitude of a particle at a distance \( \frac{l}{8} \) from one end of a string vibrating in its second harmonic.

### Step 1: Understand the Harmonic Vibration
In the second harmonic, the string has one node at each end and one antinode in the middle. The length of the string \( l \) is equal to one full wavelength \( \lambda \).

### Step 2: Determine the Wavelength
For the second harmonic:
\[ \lambda = l \]

### Step 3: Write the Equation for Amplitude
The general equation for the displacement \( Y \) of a particle in a stationary wave is given by:
\[ Y = 2a \sin(kx) \cos(\omega t) \]
where \( a \) is the amplitude at the antinode, \( k \) is the wave number, and \( x \) is the position along the string.

### Step 4: Find the Wave Number \( k \)
The wave number \( k \) is defined as:
\[ k = \frac{2\pi}{\lambda} \]
Substituting \( \lambda = l \):
\[ k = \frac{2\pi}{l} \]

### Step 5: Substitute \( k \) into the Amplitude Equation
The amplitude at any point \( x \) along the string can be expressed as:
\[ Y = 2a \sin\left(\frac{2\pi}{l} x\right) \]

### Step 6: Substitute Known Values
Given that the amplitude at the antinode (which is at the midpoint of the string) is \( 5 \, \text{mm} \), we have:
\[ 2a = 5 \, \text{mm} \implies a = \frac{5}{2} \, \text{mm} \]

### Step 7: Calculate the Amplitude at \( x = \frac{l}{8} \)
Now, we need to find the amplitude at the position \( x = \frac{l}{8} \):
\[ Y\left(\frac{l}{8}\right) = 2a \sin\left(\frac{2\pi}{l} \cdot \frac{l}{8}\right) \]
This simplifies to:
\[ Y\left(\frac{l}{8}\right) = 2a \sin\left(\frac{\pi}{4}\right) \]

### Step 8: Substitute \( a \) and Calculate
Substituting \( a = \frac{5}{2} \):
\[ Y\left(\frac{l}{8}\right) = 2 \left(\frac{5}{2}\right) \sin\left(\frac{\pi}{4}\right) \]
Since \( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):
\[ Y\left(\frac{l}{8}\right) = 5 \cdot \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}} \, \text{mm} \]

### Final Answer
The amplitude of the particle at a distance \( \frac{l}{8} \) from the fixed end is:
\[ \frac{5}{\sqrt{2}} \, \text{mm} \approx 3.54 \, \text{mm} \]