The kinetic energy of a particle executing shm is 32 J when it passes through the mean position. If the mass of the particle is.4 kg and the amplitude is one metre, find its time period.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the given information
- Kinetic Energy (KE) at mean position = 32 J
- Mass (m) of the particle = 4 kg
- Amplitude (A) = 1 m

### Step 2: Write the formula for kinetic energy in SHM
The kinetic energy (KE) of a particle executing simple harmonic motion (SHM) at the mean position is given by the formula:
\[ KE = \frac{1}{2} m v_{\text{max}}^2 \]
where \( v_{\text{max}} \) is the maximum velocity of the particle.

### Step 3: Relate maximum velocity to amplitude and angular frequency
The maximum velocity \( v_{\text{max}} \) can be expressed in terms of amplitude (A) and angular frequency (ω):
\[ v_{\text{max}} = A \omega \]
Substituting this into the kinetic energy formula gives:
\[ KE = \frac{1}{2} m (A \omega)^2 = \frac{1}{2} m A^2 \omega^2 \]

### Step 4: Substitute the known values into the kinetic energy formula
We know:
- \( KE = 32 \, \text{J} \)
- \( m = 4 \, \text{kg} \)
- \( A = 1 \, \text{m} \)

Substituting these values into the equation:
\[ 32 = \frac{1}{2} \cdot 4 \cdot (1)^2 \cdot \omega^2 \]

### Step 5: Simplify the equation
This simplifies to:
\[ 32 = 2 \omega^2 \]
Now, divide both sides by 2:
\[ 16 = \omega^2 \]

### Step 6: Solve for angular frequency (ω)
Taking the square root of both sides gives:
\[ \omega = 4 \, \text{rad/s} \]

### Step 7: Calculate the time period (T)
The time period \( T \) of SHM is related to the angular frequency by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{s} \]

### Final Answer
The time period \( T \) of the particle executing SHM is:
\[ T = \frac{\pi}{2} \, \text{s} \]

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