A horizontal coiled spring is found to stretch 3 cm by a force of `6 xx 10^(-5)` N. A bob of `2 xx 10^(-3)` kg is attached to the end of the spring and the spring is pulled by 4 cm along a horizontal frictionless table and then released. Find the force constant, period, potential energy and kinetic energy when the displacement is 2 cm.

Explanation

To solve the problem step by step, we will find the force constant (k), the period of oscillation (T), the potential energy (PE) at a displacement of 2 cm, and the kinetic energy (KE) at the same displacement.

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

We are given that a force of \( F = 6 \times 10^{-5} \) N stretches the spring by \( x = 3 \) cm. We can convert the displacement into meters:

\[ x = 3 \text{ cm} = 3 \times 10^{-2} \text{ m} \]

Using Hooke's Law, which states that \( F = kx \), we can rearrange to find \( k \):

\[ k = \frac{F}{x} \]

Substituting the values:

\[ k = \frac{6 \times 10^{-5}}{3 \times 10^{-2}} = 2 \times 10^{-3} \text{ N/m} \]

### Step 2: Calculate the Period of Oscillation (T)

The formula for the period of a mass-spring system is given by:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

Where \( m = 2 \times 10^{-3} \) kg (mass of the bob) and \( k = 2 \times 10^{-3} \) N/m (spring constant). Substituting the values:

\[ T = 2\pi \sqrt{\frac{2 \times 10^{-3}}{2 \times 10^{-3}}} = 2\pi \sqrt{1} = 2\pi \text{ seconds} \]

### Step 3: Calculate the Potential Energy (PE) at 2 cm Displacement

The potential energy stored in the spring when it is displaced by \( x = 2 \) cm (which is \( 2 \times 10^{-2} \) m) is given by:

\[ PE = \frac{1}{2} k x^2 \]

Substituting the values:

\[ PE = \frac{1}{2} \times 2 \times 10^{-3} \times (2 \times 10^{-2})^2 \]

Calculating:

\[ PE = \frac{1}{2} \times 2 \times 10^{-3} \times 4 \times 10^{-4} = 4 \times 10^{-7} \text{ Joules} \]

### Step 4: Calculate the Kinetic Energy (KE) at 2 cm Displacement

The kinetic energy at a displacement \( x \) can be calculated using the formula:

\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]

Where:
- \( A = 4 \) cm = \( 4 \times 10^{-2} \) m (amplitude)
- \( \omega = \sqrt{\frac{k}{m}} \)

First, calculate \( \omega \):

\[ \omega = \sqrt{\frac{2 \times 10^{-3}}{2 \times 10^{-3}}} = 1 \text{ rad/s} \]

Now substituting into the KE formula:

\[ KE = \frac{1}{2} \times 2 \times 10^{-3} \times (1^2) \times ((4 \times 10^{-2})^2 - (2 \times 10^{-2})^2) \]

Calculating:

\[ KE = \frac{1}{2} \times 2 \times 10^{-3} \times (16 \times 10^{-4} - 4 \times 10^{-4}) = \frac{1}{2} \times 2 \times 10^{-3} \times 12 \times 10^{-4} \]

\[ KE = 12 \times 10^{-7} \text{ Joules} \]

### Summary of Results
- Force Constant \( k = 2 \times 10^{-3} \) N/m
- Period \( T = 2\pi \) seconds
- Potential Energy at 2 cm \( PE = 4 \times 10^{-7} \) Joules
- Kinetic Energy at 2 cm \( KE = 12 \times 10^{-7} \) Joules

---