A particle executes shm of period.8 s. After what time of its passing through the mean position the energy will be half kinetic and half potential.

Explanation

To solve the problem step by step, we need to determine the time after a particle passes through the mean position when the energy is equally divided between kinetic and potential energy.

### Step 1: Understand the Energy Distribution in SHM
In simple harmonic motion (SHM), the total mechanical energy (E) is constant and is the sum of kinetic energy (KE) and potential energy (PE). The condition given is that the potential energy is half of the total energy, which means:
\[ PE = \frac{1}{2} E \]
This implies that:
\[ KE = \frac{1}{2} E \]

### Step 2: Relate Potential Energy to Displacement
The potential energy in SHM is given by:
\[ PE = \frac{1}{2} k x^2 \]
Where \( k \) is the spring constant and \( x \) is the displacement from the mean position. The total energy in SHM is given by:
\[ E = \frac{1}{2} k A^2 \]
Where \( A \) is the amplitude.

Given that \( PE = \frac{1}{2} E \), we can substitute:
\[ \frac{1}{2} k x^2 = \frac{1}{2} \left( \frac{1}{2} k A^2 \right) \]
This simplifies to:
\[ k x^2 = \frac{1}{2} k A^2 \]
Dividing both sides by \( k \) (assuming \( k \neq 0 \)):
\[ x^2 = \frac{1}{2} A^2 \]

### Step 3: Solve for Displacement
Taking the square root of both sides, we find:
\[ x = \frac{A}{\sqrt{2}} \]

### Step 4: Write the Equation of Motion
The displacement \( x \) in SHM is described by the equation:
\[ x = A \sin(\omega t) \]
Where \( \omega = \frac{2\pi}{T} \) and \( T \) is the period. Given that the period \( T = 8 \) seconds, we find:
\[ \omega = \frac{2\pi}{8} = \frac{\pi}{4} \]

### Step 5: Set Up the Equation
Now we set up the equation using the displacement we found:
\[ \frac{A}{\sqrt{2}} = A \sin\left(\frac{\pi}{4} t\right) \]
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{\sqrt{2}} = \sin\left(\frac{\pi}{4} t\right) \]

### Step 6: Solve for Time
We know that:
\[ \sin\left(\frac{\pi}{4} t\right) = \frac{1}{\sqrt{2}} \]
This occurs at:
\[ \frac{\pi}{4} t = \frac{\pi}{4} \]
Thus:
\[ t = 1 \text{ second} \]

### Final Answer
The time after passing through the mean position when the energy is half kinetic and half potential is:
\[ t = 1 \text{ second} \]