A body of mass 2 kg suspended through a vertical spring executes simple harmonic motionof period 4s. If the oscillations are stopped and the body hangs in equillibrium, find the potential energy stored in the spring.

Explanation

To solve the problem step by step, we need to find the potential energy stored in the spring when the mass is in equilibrium. Here’s how to approach it:

### Step 1: Understand the equilibrium condition
When the mass is suspended and at rest, the forces acting on it must balance. The upward force exerted by the spring (spring force) equals the downward gravitational force (weight of the mass).

### Step 2: Write the equations for forces
The spring force can be expressed as:
\[ F_{\text{spring}} = k x_0 \]
where \( k \) is the spring constant and \( x_0 \) is the extension of the spring.

The weight of the mass is given by:
\[ F_{\text{weight}} = mg \]
where \( m = 2 \, \text{kg} \) and \( g \approx 10 \, \text{m/s}^2 \).

At equilibrium:
\[ k x_0 = mg \]

### Step 3: Solve for the extension \( x_0 \)
From the equilibrium condition, we can express \( x_0 \) as:
\[ x_0 = \frac{mg}{k} \]

### Step 4: Find the spring constant \( k \)
We can find \( k \) using the time period \( T \) of the oscillation. The formula for the time period of a mass-spring system is:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

Given \( T = 4 \, \text{s} \) and \( m = 2 \, \text{kg} \), we can rearrange this formula to solve for \( k \):
\[ T^2 = 4\pi^2 \frac{m}{k} \]
\[ k = \frac{4\pi^2 m}{T^2} \]

Substituting the values:
\[ k = \frac{4\pi^2 \cdot 2}{4^2} \]
\[ k = \frac{8\pi^2}{16} = \frac{\pi^2}{2} \]

Using \( \pi^2 \approx 10 \):
\[ k \approx \frac{10}{2} = 5 \, \text{N/m} \]

### Step 5: Substitute \( k \) into the extension formula
Now substituting \( k \) back into the equation for \( x_0 \):
\[ x_0 = \frac{mg}{k} = \frac{2 \cdot 10}{5} = 4 \, \text{m} \]

### Step 6: Calculate the potential energy stored in the spring
The potential energy \( PE \) stored in the spring is given by:
\[ PE = \frac{1}{2} k x_0^2 \]

Substituting the values of \( k \) and \( x_0 \):
\[ PE = \frac{1}{2} \cdot 5 \cdot (4)^2 \]
\[ PE = \frac{1}{2} \cdot 5 \cdot 16 = \frac{80}{2} = 40 \, \text{J} \]

### Final Answer
The potential energy stored in the spring is **40 Joules**.

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