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 find the maximum force exerted by the spring on the block, we can follow these steps:

### Step 1: Calculate Angular Frequency (ω)
The angular frequency \( \omega \) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the time period.

Given:
- \( T = 0.314 \, \text{s} \)

Calculating \( \omega \):
\[ \omega = \frac{2 \times 3.14}{0.314} \approx 20 \, \text{rad/s} \]

### Step 2: Calculate Maximum Acceleration (a_max)
The maximum acceleration \( a_{\text{max}} \) in simple harmonic motion is given by:
\[ a_{\text{max}} = \omega^2 \cdot A \]
where \( A \) is the amplitude.

Given:
- \( A = 0.1 \, \text{m} \)

Calculating \( a_{\text{max}} \):
\[ a_{\text{max}} = (20)^2 \cdot 0.1 = 400 \cdot 0.1 = 40 \, \text{m/s}^2 \]

### Step 3: Calculate the Weight of the Block (W)
The weight \( W \) of the block is given by:
\[ W = m \cdot g \]
where \( m \) is the mass and \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

Given:
- \( m = 0.5 \, \text{kg} \)

Calculating \( W \):
\[ W = 0.5 \cdot 10 = 5 \, \text{N} \]

### Step 4: Calculate Maximum Force (F_max)
The maximum force exerted by the spring on the block is the sum of the maximum force due to acceleration and the weight of the block:
\[ F_{\text{max}} = m \cdot a_{\text{max}} + W \]

Calculating \( F_{\text{max}} \):
\[ F_{\text{max}} = 0.5 \cdot 40 + 5 = 20 + 5 = 25 \, \text{N} \]

### Final Answer
The maximum force exerted by the spring on the block is:
\[ \boxed{25 \, \text{N}} \]