A simple harmonic oscillator of period 6 second has 6 joule potential energy when its displacement is 3 cm. Calculate (i) force constant and (ii) average kinetic energy when the amplitude is 5 cm.

Explanation

To solve the problem step by step, we will break it down into two parts: (i) calculating the force constant \( k \) and (ii) calculating the average kinetic energy when the amplitude is 5 cm.

### Given Data:
- Time period \( T = 6 \) seconds
- Potential energy \( U = 6 \) Joules
- Displacement \( x = 3 \) cm = \( 3 \times 10^{-2} \) m
- Amplitude \( A = 5 \) cm = \( 5 \times 10^{-2} \) m

### Step 1: Calculate the Force Constant \( k \)

The potential energy \( U \) in a simple harmonic oscillator is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
Rearranging this formula to solve for \( k \):
\[ k = \frac{2U}{x^2} \]

Substituting the known values:
\[ k = \frac{2 \times 6 \, \text{J}}{(3 \times 10^{-2} \, \text{m})^2} \]

Calculating \( x^2 \):
\[ x^2 = (3 \times 10^{-2})^2 = 9 \times 10^{-4} \, \text{m}^2 \]

Now substituting back:
\[ k = \frac{12}{9 \times 10^{-4}} = \frac{12}{0.0009} = 13333.33 \, \text{N/m} \approx 1.33 \times 10^4 \, \text{N/m} \]

### Step 2: Calculate the Average Kinetic Energy

The average kinetic energy \( K \) in a simple harmonic motion can be calculated using the formula:
\[ K_{\text{avg}} = \frac{1}{2} k A^2 \]

Substituting the values:
\[ K_{\text{avg}} = \frac{1}{2} \times 1.33 \times 10^4 \, \text{N/m} \times (5 \times 10^{-2} \, \text{m})^2 \]

Calculating \( A^2 \):
\[ A^2 = (5 \times 10^{-2})^2 = 25 \times 10^{-4} \, \text{m}^2 \]

Now substituting back:
\[ K_{\text{avg}} = \frac{1}{2} \times 1.33 \times 10^4 \times 25 \times 10^{-4} \]
\[ K_{\text{avg}} = \frac{1}{2} \times 1.33 \times 25 \times 10^0 \]
\[ K_{\text{avg}} = \frac{1}{2} \times 33.25 = 16.625 \, \text{J} \]

### Final Answers:
(i) Force constant \( k \approx 1.33 \times 10^4 \, \text{N/m} \)

(ii) Average kinetic energy \( K_{\text{avg}} \approx 16.625 \, \text{J} \)