A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is 4 mm. Minimum distance between the two points having amplitude 2 mm is:

Options

A. 1m
B. 75 cm
C. 60 cm
D. 50 cm

Explanation

To solve the problem step by step, we will analyze the fundamental mode of vibration of a string and use the properties of standing waves.

### Step 1: Understand the fundamental mode of vibration
In the fundamental mode, a string clamped at both ends vibrates with a single antinode at the center and nodes at the ends. The amplitude at the center is maximum, and it decreases towards the ends.

### Step 2: Identify the given parameters
- Length of the string \( L = 1.5 \, \text{m} \)
- Maximum amplitude at the center \( A = 4 \, \text{mm} \)
- We need to find the minimum distance between two points where the amplitude is \( 2 \, \text{mm} \).

### Step 3: Write the equation for the standing wave
The equation of the standing wave can be expressed as:
\[ y(x, t) = A \sin(kx) \cos(\omega t) \]
At \( t = 0 \), this simplifies to:
\[ y(x) = A \sin(kx) \]

### Step 4: Set up the equation for amplitude
For points where the amplitude is \( 2 \, \text{mm} \):
\[ 2 = 4 \sin(kx) \]
This simplifies to:
\[ \sin(kx) = \frac{1}{2} \]

### Step 5: Solve for \( kx \)
The solution for \( \sin(kx) = \frac{1}{2} \) gives:
\[ kx = \frac{\pi}{6} \quad \text{and} \quad kx = \frac{5\pi}{6} \]

### Step 6: Relate \( k \) to the wavelength
The wave number \( k \) is related to the wavelength \( \lambda \) by:
\[ k = \frac{2\pi}{\lambda} \]
For a string of length \( L \) vibrating in the fundamental mode, the wavelength is:
\[ \lambda = 2L = 2 \times 1.5 \, \text{m} = 3 \, \text{m} \]
Thus,
\[ k = \frac{2\pi}{3} \]

### Step 7: Substitute \( k \) into the equation
Substituting \( k \) into the equation:
\[ \frac{2\pi}{3} x = \frac{\pi}{6} \]
Solving for \( x \):
\[ x = \frac{\pi/6}{2\pi/3} = \frac{1}{4} \, \text{m} = 0.25 \, \text{m} \]

### Step 8: Calculate the second point
The second point where the amplitude is \( 2 \, \text{mm} \) corresponds to:
\[ \frac{2\pi}{3} x = \frac{5\pi}{6} \]
Solving for \( x \):
\[ x = \frac{5\pi/6}{2\pi/3} = \frac{5}{4} \, \text{m} = 1.25 \, \text{m} \]

### Step 9: Find the distance between the two points
The minimum distance \( d \) between the two points with amplitude \( 2 \, \text{mm} \) is:
\[ d = x_2 - x_1 = 1.25 \, \text{m} - 0.25 \, \text{m} = 1.0 \, \text{m} \]

### Final Answer
The minimum distance between the two points having amplitude \( 2 \, \text{mm} \) is:
\[ \boxed{1 \, \text{m}} \]