A man generates a symmetrical pulse in a string by moving his hand up and down. At `t=0` the point in his hand moves downward. The pulse travels with speed of `3 m//s` on the string & his hands passes `6` times in each second from the mean position. Then the point on the string at a distance `3m` will reach its upper extreme first time at time `t=`

Options

A. 1.25 s
B. 1s
C. `11//12 s `
D. 23/24 s

Explanation

To solve the problem step by step, we will analyze the motion of the pulse generated by the man's hand and determine when a point on the string at a distance of 3 meters will reach its upper extreme for the first time.

### Step-by-Step Solution:

1. **Understanding the Motion of the Hand:**
- The man moves his hand up and down, crossing the mean position 6 times per second.
- Since he crosses the mean position twice in one complete cycle (once going up and once going down), the frequency \( f \) of the motion is:
\[ f = \frac{6 \text{ crossings}}{2} = 3 \text{ Hz} \]

2. **Calculating Angular Frequency (\( \omega \)):**
- The angular frequency \( \omega \) is given by the formula:
\[ \omega = 2 \pi f \]
- Substituting the value of \( f \):
\[ \omega = 2 \pi \times 3 = 6 \pi \text{ radians/second} \]

3. **Calculating Wave Number (\( k \)):**
- The wave travels with a speed \( v = 3 \text{ m/s} \).
- The wave number \( k \) is calculated using the formula:
\[ k = \frac{\omega}{v} \]
- Substituting the values of \( \omega \) and \( v \):
\[ k = \frac{6 \pi}{3} = 2 \pi \text{ radians/meter} \]

4. **Using the Wave Equation:**
- The wave equation can be expressed as:
\[ y(x, t) = A \sin(kx - \omega t) \]
- For the upper extreme, \( y \) must equal the amplitude \( A \). Thus, we set:
\[ A \sin(kx - \omega t) = A \]
- This implies:
\[ \sin(kx - \omega t) = 1 \]
- The sine function equals 1 at \( kx - \omega t = \frac{\pi}{2} + 2n\pi \) for any integer \( n \). We will consider the first occurrence, so we set:
\[ kx - \omega t = \frac{\pi}{2} \]

5. **Substituting Values:**
- For \( x = 3 \text{ m} \):
\[ k \cdot 3 - \omega t = \frac{\pi}{2} \]
- Substituting \( k = 2\pi \) and \( \omega = 6\pi \):
\[ 2\pi \cdot 3 - 6\pi t = \frac{\pi}{2} \]
- Simplifying:
\[ 6\pi - 6\pi t = \frac{\pi}{2} \]

6. **Solving for \( t \):**
- Rearranging gives:
\[ 6\pi - \frac{\pi}{2} = 6\pi t \]
- Converting \( 6\pi \) to a fraction:
\[ 6\pi = \frac{12\pi}{2} \]
- Thus:
\[ \frac{12\pi}{2} - \frac{\pi}{2} = 6\pi t \]
\[ \frac{11\pi}{2} = 6\pi t \]
- Dividing both sides by \( 6\pi \):
\[ t = \frac{11}{12} \text{ seconds} \]

### Final Answer:
The point on the string at a distance of 3 meters will reach its upper extreme for the first time at time \( t = \frac{11}{12} \) seconds.