The vertical motion of a ship at sea is described by the equation `(d^2(x))/(dt^(2))=-4x` , where x is the vertical height of the ship (in meter) above its mean position. If it oscillates through a height of 1 m

Options

A. its maximum vertical speed will be `1 ms^(-1)`
B. its maximum vertical speed will be `2 ms^(-1)`
C. its greater vertical acceleration is `2 ms^(-2)`
D. its greater vertical acceleration is `1 ms^(-2)`

Explanation

To solve the problem, we need to analyze the given equation of motion and determine the maximum vertical speed and acceleration of the ship.

### Step-by-Step Solution:

1. **Identify the Equation of Motion**:
The equation given is:
\[ \frac{d^2x}{dt^2} = -4x \]
This represents simple harmonic motion (SHM), where \( x \) is the vertical height of the ship above its mean position.

2. **Compare with Standard SHM Equation**:
The standard form of the SHM equation is:
\[ \frac{d^2x}{dt^2} = -\omega^2 x \]
By comparing both equations, we can identify:
\[ \omega^2 = 4 \]
Therefore, we find:
\[ \omega = \sqrt{4} = 2 \, \text{rad/s} \]

3. **Determine the Amplitude**:
The problem states that the ship oscillates through a height of 1 meter. This means the amplitude \( A \) is:
\[ A = 1 \, \text{m} \]

4. **Calculate Maximum Velocity**:
The maximum velocity \( V_{\text{max}} \) in SHM is given by the formula:
\[ V_{\text{max}} = A \cdot \omega \]
Substituting the values of \( A \) and \( \omega \):
\[ V_{\text{max}} = 1 \cdot 2 = 2 \, \text{m/s} \]

5. **Calculate Maximum Acceleration**:
The maximum acceleration \( a_{\text{max}} \) in SHM is given by:
\[ a_{\text{max}} = A \cdot \omega^2 \]
Substituting the values:
\[ a_{\text{max}} = 1 \cdot 4 = 4 \, \text{m/s}^2 \]

6. **Conclusion**:
From the calculations, we find:
- Maximum vertical speed: \( 2 \, \text{m/s} \)
- Maximum vertical acceleration: \( 4 \, \text{m/s}^2 \)

### Final Answer:
The correct option is that the maximum vertical speed will be \( 2 \, \text{m/s} \).