A block of mass 0.5 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.1 m and time period 0.314s. Find the maximum fore exerted by the spring on the blockl.

Explanation

To solve the problem step-by-step, we need to find the maximum force exerted by the spring on the block. Here's how we can do it:

### Step 1: Identify the given values
- Mass of the block (m) = 0.5 kg
- Amplitude (A) = 0.1 m
- Time period (T) = 0.314 s

### Step 2: Write the expression for the maximum force
The maximum force exerted by the spring on the block can be expressed as:
\[ F_{\text{max}} = kA + mg \]
where:
- \( k \) is the spring constant,
- \( A \) is the amplitude,
- \( m \) is the mass of the block,
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 3: Find the spring constant \( k \)
The time period \( T \) of a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
We can rearrange this formula to solve for \( k \):
\[ k = \frac{4\pi^2 m}{T^2} \]

### Step 4: Substitute the values into the equation for \( k \)
Substituting the known values into the equation:
- \( m = 0.5 \, \text{kg} \)
- \( T = 0.314 \, \text{s} \)

Calculating \( k \):
\[ k = \frac{4\pi^2 \times 0.5}{(0.314)^2} \]
Calculating \( \pi^2 \approx 9.87 \):
\[ k = \frac{4 \times 9.87 \times 0.5}{0.098596} \approx \frac{19.74}{0.098596} \approx 200 \, \text{N/m} \]

### Step 5: Calculate the maximum force \( F_{\text{max}} \)
Now that we have \( k \), we can calculate the maximum force:
\[ F_{\text{max}} = kA + mg \]
Substituting the values:
- \( k = 200 \, \text{N/m} \)
- \( A = 0.1 \, \text{m} \)
- \( m = 0.5 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)

Calculating:
\[ F_{\text{max}} = 200 \times 0.1 + 0.5 \times 10 \]
\[ F_{\text{max}} = 20 + 5 = 25 \, \text{N} \]

### Final Answer
The maximum force exerted by the spring on the block is **25 N**.

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