A wooden block of mass 0.5 kg and density 800 `kgm^-3` is fastened to the free end of a vertical spring of spring constant 50 `Nm^-1` fixed at the bottom. If the entire system is completely immersed in water, find a. the elongation (or compression) of the spring in equilibrium and b. the time period of vertical oscillations of the block when it is slightly depressed and released.

Explanation

To solve the problem step by step, we will break it down into two parts: (a) finding the elongation (or compression) of the spring in equilibrium, and (b) finding the time period of vertical oscillations of the block when it is slightly depressed and released.

### Part (a): Finding the Elongation (or Compression) of the Spring in Equilibrium

1. **Identify the Forces Acting on the Block**:
- The weight of the block \( W = mg \) acting downwards.
- The buoyant force \( F_b \) acting upwards.
- The spring force \( F_s = kx \) acting upwards.

2. **Write the Equation for Equilibrium**:
At equilibrium, the sum of the upward forces (buoyant force + spring force) equals the downward force (weight of the block):
\[ F_b + F_s = W \]

3. **Calculate the Weight of the Block**:
Given:
- Mass of the block \( m = 0.5 \, \text{kg} \)
- Acceleration due to gravity \( g = 9.81 \, \text{m/s}^2 \)
\[ W = mg = 0.5 \times 9.81 = 4.905 \, \text{N} \]

4. **Calculate the Buoyant Force**:
The buoyant force can be calculated using the formula:
\[ F_b = \text{density of water} \times g \times V \]
where \( V \) is the volume of the block. The volume of the block can be calculated as:
\[ V = \frac{m}{\text{density of block}} = \frac{0.5 \, \text{kg}}{800 \, \text{kg/m}^3} = 0.000625 \, \text{m}^3 \]
Thus,
\[ F_b = 1000 \times 9.81 \times 0.000625 = 6.13125 \, \text{N} \]

5. **Substitute the Forces into the Equilibrium Equation**:
\[ F_b + kx = W \]
\[ 6.13125 + 50x = 4.905 \]

6. **Solve for \( x \)**:
Rearranging the equation gives:
\[ 50x = 4.905 - 6.13125 \]
\[ 50x = -1.22625 \]
\[ x = \frac{-1.22625}{50} = -0.024525 \, \text{m} = -2.45 \, \text{cm} \]
The negative sign indicates that the spring is compressed by approximately 2.45 cm.

### Part (b): Finding the Time Period of Vertical Oscillations

1. **Identify the System Behavior**:
When the block is slightly depressed and released, the buoyant force remains constant, and the restoring force is provided solely by the spring.

2. **Write the Equation for the Restoring Force**:
The restoring force is given by Hooke's law:
\[ F = -kx \]

3. **Relate the Force to Mass and Acceleration**:
According to Newton's second law:
\[ F = ma \]
Therefore,
\[ ma = -kx \]
This can be rearranged to:
\[ a = -\frac{k}{m}x \]

4. **Identify the Angular Frequency**:
The equation resembles the form of simple harmonic motion (SHM):
\[ a = -\omega^2 x \]
Hence,
\[ \omega = \sqrt{\frac{k}{m}} \]

5. **Calculate the Time Period**:
The time period \( T \) of SHM is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Substituting the values:
- \( m = 0.5 \, \text{kg} \)
- \( k = 50 \, \text{N/m} \)
\[ T = 2\pi \sqrt{\frac{0.5}{50}} = 2\pi \sqrt{0.01} = 2\pi \times 0.1 = 0.2\pi \, \text{s} \]

### Final Answers:
- (a) The compression in the spring is approximately **2.45 cm**.
- (b) The time period of vertical oscillations is approximately **0.2π seconds**.