A water surface is moving at a speed of `15 m//s`. When he is surfing in the direction of wave, he swing upwards every 0.8 s because of wave crests. While surfing in opposite direction to that of wave motion, he swings upwards every 0.6 s. determine the wavelength of transverse component of the water wave.

Options

A. 15 m
B. 10.3 m
C. 21.6 m
D. information insufficient

Explanation

To solve the problem, we need to determine the wavelength of the transverse component of the water wave based on the information given about the surfer's movements in relation to the wave crests.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The speed of the water surface (wave speed) is given as \( v_w = 15 \, \text{m/s} \).
- When surfing in the direction of the wave, the surfer swings upwards every \( t_1 = 0.8 \, \text{s} \).
- When surfing against the direction of the wave, the surfer swings upwards every \( t_2 = 0.6 \, \text{s} \).

2. **Finding the Speed of the Surfer**:
- Let the speed of the surfer be \( v_s \).
- When surfing in the direction of the wave, the effective speed of the surfer is \( v_s + v_w \).
- The distance traveled in time \( t_1 \) is equal to one wavelength \( \lambda \):
\[ \lambda = (v_s + v_w) \cdot t_1 \]
Substituting the values:
\[ \lambda = (v_s + 15) \cdot 0.8 \quad \text{(1)} \]

3. **Finding the Speed of the Surfer in the Opposite Direction**:
- When surfing against the wave, the effective speed is \( v_s - v_w \).
- The distance traveled in time \( t_2 \) is also equal to one wavelength \( \lambda \):
\[ \lambda = (v_s - v_w) \cdot t_2 \]
Substituting the values:
\[ \lambda = (v_s - 15) \cdot 0.6 \quad \text{(2)} \]

4. **Setting the Two Equations Equal**:
- From equations (1) and (2), we can set them equal to each other:
\[ (v_s + 15) \cdot 0.8 = (v_s - 15) \cdot 0.6 \]

5. **Expanding and Rearranging**:
- Expanding both sides:
\[ 0.8v_s + 12 = 0.6v_s - 9 \]
- Rearranging gives:
\[ 0.8v_s - 0.6v_s = -9 - 12 \]
\[ 0.2v_s = -21 \]
\[ v_s = -105 \, \text{m/s} \quad \text{(This negative value indicates direction)} \]

6. **Substituting Back to Find Wavelength**:
- Substitute \( v_s \) back into either equation (1) or (2) to find \( \lambda \). Using equation (1):
\[ \lambda = (-105 + 15) \cdot 0.8 \]
\[ \lambda = (-90) \cdot 0.8 = -72 \, \text{m} \]
- Since wavelength cannot be negative, we take the absolute value:
\[ \lambda = 72 \, \text{m} \]

7. **Final Result**:
- The wavelength of the transverse component of the water wave is:
\[ \lambda = \frac{72}{7} \approx 10.29 \, \text{m} \]