A harmonic wave has been set up on a very long string which travels along the length of string. The wave has frequency of 50 Hz. Amplitude 1 cm and wavelength 0.5 m. for the above described wave.
Statement (i): time taken by a point on the string to travel a distance of 8 m along the length of string is 0.32 s.
Statement (ii): time taken by a point in the string to travel a distance of 8m, once the wave has reached at that point and sets it into motion is 0.32 s.

Options

A. Both the statement are correct
B. statement I is correct but statement II is incorect
C. statement I is incorrect but statement II is correct
D. both the statement are incorrect

Explanation

To solve the problem, we need to analyze both statements regarding the harmonic wave set up on a string.

### Given Data:
- Frequency (f) = 50 Hz
- Amplitude (A) = 1 cm = 0.01 m
- Wavelength (λ) = 0.5 m

### Step 1: Calculate the wave speed (v)
The speed of a wave on a string can be calculated using the formula:
\[ v = f \cdot \lambda \]

Substituting the values:
\[ v = 50 \, \text{Hz} \cdot 0.5 \, \text{m} = 25 \, \text{m/s} \]

### Step 2: Calculate the time taken to travel 8 m
Using the formula for time:
\[ t = \frac{d}{v} \]
where \( d \) is the distance traveled.

Substituting the values:
\[ t = \frac{8 \, \text{m}}{25 \, \text{m/s}} = 0.32 \, \text{s} \]

### Conclusion for Statement (i):
The time taken by a point on the string to travel a distance of 8 m along the length of the string is indeed 0.32 s. Therefore, **Statement (i) is true**.

### Step 3: Analyze Statement (ii)
Statement (ii) refers to the time taken by a point in the string to travel a distance of 8 m once the wave has reached that point and sets it into motion.

### Step 4: Calculate the oscillation of the point
When the wave reaches a point on the string, that point will oscillate up and down. The maximum distance it can move in one complete oscillation (one period) is twice the amplitude (up and down).

- Maximum displacement in one period = 2 * Amplitude = 2 * 0.01 m = 0.02 m (or 2 cm)

### Step 5: Calculate the number of oscillations needed to cover 8 m
To find out how many complete oscillations are needed to cover 8 m:
\[ \text{Number of oscillations} = \frac{8 \, \text{m}}{0.02 \, \text{m}} = 400 \]

### Step 6: Calculate the time taken for these oscillations
The time period (T) of the wave is given by:
\[ T = \frac{1}{f} = \frac{1}{50} = 0.02 \, \text{s} \]

The total time taken for 400 oscillations:
\[ \text{Total time} = \text{Number of oscillations} \cdot T = 400 \cdot 0.02 \, \text{s} = 8 \, \text{s} \]

### Conclusion for Statement (ii):
The time taken by a point in the string to travel a distance of 8 m, once the wave has reached that point and sets it into motion, is 8 s. Therefore, **Statement (ii) is false**.

### Final Conclusion:
- Statement (i) is true.
- Statement (ii) is false.

### Summary:
- **Correct Answer**: Statement (i) is true, Statement (ii) is false.