Find the period of a vertical spring - block system by both methods.

Explanation

To find the period of a vertical spring-block system using both the force method and the energy method, we can follow these steps:

### Method 1: Force Method

1. **Identify the System**:
- We have a spring with a spring constant \( k \) and a block of mass \( m \) attached to it. The spring is vertical.

2. **Establish Equilibrium**:
- When the block is at rest, the weight of the block \( mg \) is balanced by the spring force \( kx_0 \) at the equilibrium position. Here, \( x_0 \) is the displacement from the natural length of the spring.
- At equilibrium:
\[ mg = kx_0 \quad \text{(Equation 1)} \]

3. **Displacement from Equilibrium**:
- When the block is displaced further by a distance \( x \) from the equilibrium position, the net force acting on the block is:
\[ F = -k(x + x_0) + mg \]
- Substituting \( mg \) from Equation 1:
\[ F = -k(x + x_0) + kx_0 = -kx \]

4. **Simple Harmonic Motion**:
- The equation \( F = -kx \) indicates that the motion is simple harmonic. The general formula for the time period \( T \) of simple harmonic motion is:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

### Method 2: Energy Method

1. **Total Mechanical Energy**:
- At the mean position (equilibrium), let \( h = 0 \). The total mechanical energy when the block is displaced to position \( C \) is the sum of kinetic and potential energies:
\[ E = \frac{1}{2} mv^2 + \frac{1}{2} k(x + x_0)^2 - mgx \]

2. **Conservation of Energy**:
- The total mechanical energy remains constant. Thus, we can express the total energy at the displaced position:
\[ E = \frac{1}{2} mv^2 + \frac{1}{2} k(x + x_0)^2 - mgx \]

3. **Differentiate the Energy Equation**:
- To find the relationship between acceleration and displacement, differentiate the energy equation with respect to time:
\[ \frac{dE}{dt} = 0 \]
- This leads to:
\[ mv \frac{dv}{dt} + k(x + x_0) \frac{dx}{dt} - mg \frac{dx}{dt} = 0 \]

4. **Substituting Variables**:
- Replace \( \frac{dv}{dt} \) with acceleration \( a \) and \( \frac{dx}{dt} \) with velocity \( v \). Using \( kx_0 = mg \) from equilibrium:
\[ ma = -kx \]

5. **Simple Harmonic Motion**:
- This again indicates simple harmonic motion, leading us to the same time period:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

### Final Result
In both methods, we find that the period of the vertical spring-block system is:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]