The displacement equation of a spring block system is given by `y = A sin omegat` in air. It is completely immersed in water. If `A^(1)` and `omega^(1)` be new amplitude and angular freuency then

Options

A. `omega = omega^(1)`
B. `A lt A^(1)`
C. both
D. none

Answer

To solve the problem, we need to analyze how the immersion of a spring-block system in water affects its amplitude and angular frequency.

### Step-by-step Solution:

1. **Understanding the Initial Conditions**:
- The displacement equation of the spring-block system in air is given by \( y = A \sin(\omega t) \).
- Here, \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Angular Frequency**:
- The angular frequency \( \omega \) of a spring-block system is given by the formula:
\[ \omega = \sqrt{\frac{k}{m}} \]
- Where \( k \) is the spring constant and \( m \) is the mass of the block.

3. **Effect of Immersion in Water**:
- When the spring-block system is immersed in water, the buoyant force acts on the block, effectively reducing the net force acting on it.
- However, the spring constant \( k \) and the mass \( m \) remain unchanged. Therefore, the angular frequency \( \omega \) does not change:
\[ \omega' = \omega \]

4. **Amplitude**:
- The amplitude \( A \) is related to the maximum displacement of the system. When the system is immersed in water, the buoyant force reduces the effective weight of the block.
- As a result, the maximum acceleration of the system decreases, which is directly proportional to the amplitude. Thus, the new amplitude \( A' \) will be less than the original amplitude \( A \):
\[ A' < A \]

5. **Conclusion**:
- After immersion in water, the angular frequency remains the same, \( \omega' = \omega \).
- The amplitude decreases, \( A' < A \).

### Final Results:
- New amplitude: \( A' < A \)
- New angular frequency: \( \omega' = \omega \)