A body of mass 0.2 kg performing SHM has a velocity of `0.6 ms^(-1)` after one second of its starting from the mean position. If the time period is 6 seconds, find the kinetic energy, potential energy and total energy.

Explanation

To solve the problem step by step, let's summarize the given information and then calculate the kinetic energy, potential energy, and total energy of the body performing simple harmonic motion (SHM).

### Given:
- Mass of the body, \( m = 0.2 \, \text{kg} \)
- Velocity after 1 second, \( v = 0.6 \, \text{m/s} \)
- Time period, \( T = 6 \, \text{s} \)

### Step 1: Calculate Kinetic Energy (KE)
The formula for kinetic energy is:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times 0.2 \times (0.6)^2 \]
Calculating:
\[ KE = 0.1 \times 0.36 = 0.036 \, \text{J} \]
Thus, the kinetic energy is:
\[ KE = 3.6 \times 10^{-2} \, \text{J} \]

### Step 2: Calculate Angular Frequency (\( \omega \))
The angular frequency is given by:
\[ \omega = \frac{2\pi}{T} \]
Substituting the time period:
\[ \omega = \frac{2\pi}{6} = \frac{\pi}{3} \, \text{rad/s} \]

### Step 3: Calculate Amplitude (A)
Using the relationship between velocity, amplitude, and angular frequency:
\[ v = A \omega \cos(\omega t) \]
At \( t = 1 \, \text{s} \):
\[ 0.6 = A \left(\frac{\pi}{3}\right) \cos\left(\frac{\pi}{3}\right) \]
Since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ 0.6 = A \left(\frac{\pi}{3}\right) \left(\frac{1}{2}\right) \]
Solving for \( A \):
\[ 0.6 = \frac{A \pi}{6} \implies A = \frac{0.6 \times 6}{\pi} = \frac{3.6}{\pi} \]

### Step 4: Calculate Displacement (x)
Using the displacement formula:
\[ x = A \sin(\omega t) \]
Substituting the values:
\[ x = \left(\frac{3.6}{\pi}\right) \sin\left(\frac{\pi}{3}\right) \]
Since \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \):
\[ x = \left(\frac{3.6}{\pi}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{1.8 \sqrt{3}}{\pi} \]

### Step 5: Calculate Potential Energy (PE)
The potential energy in SHM is given by:
\[ PE = \frac{1}{2} k x^2 \]
Where \( k = m \omega^2 \):
\[ k = 0.2 \left(\frac{\pi}{3}\right)^2 = 0.2 \cdot \frac{\pi^2}{9} = \frac{0.2 \pi^2}{9} \]
Now substituting \( x \):
\[ PE = \frac{1}{2} \cdot \frac{0.2 \pi^2}{9} \cdot \left(\frac{1.8 \sqrt{3}}{\pi}\right)^2 \]
Calculating \( x^2 \):
\[ x^2 = \left(\frac{1.8 \sqrt{3}}{\pi}\right)^2 = \frac{3.24 \cdot 3}{\pi^2} = \frac{9.72}{\pi^2} \]
Thus:
\[ PE = \frac{1}{2} \cdot \frac{0.2 \pi^2}{9} \cdot \frac{9.72}{\pi^2} = \frac{0.1 \cdot 9.72}{9} = \frac{0.972}{9} = 0.108 \, \text{J} \]
So, the potential energy is:
\[ PE = 10.8 \times 10^{-2} \, \text{J} \]

### Step 6: Calculate Total Energy (TE)
The total energy in SHM is the sum of kinetic and potential energy:
\[ TE = KE + PE \]
Substituting the values:
\[ TE = 3.6 \times 10^{-2} + 10.8 \times 10^{-2} = 14.4 \times 10^{-2} \, \text{J} \]

### Final Answers:
- Kinetic Energy, \( KE = 3.6 \times 10^{-2} \, \text{J} \)
- Potential Energy, \( PE = 10.8 \times 10^{-2} \, \text{J} \)
- Total Energy, \( TE = 14.4 \times 10^{-2} \, \text{J} \)