A body weighing 0.02 kg has a velocity of `0.06 ms^(-1)` after one second of start from the mean position. If the time period is 6 s find the K.E., P.E. and the total energy.

Explanation

To solve the problem step by step, we will find the kinetic energy (K.E.), potential energy (P.E.), and total energy (T.E.) of a body in simple harmonic motion (SHM) given its mass, velocity, and time period.

### Given Data:
- Mass (m) = 0.02 kg
- Velocity (V) = 0.06 m/s (after 1 second)
- Time Period (T) = 6 s

### Step 1: Calculate Angular Frequency (ω)
The angular frequency (ω) can be calculated using the formula for the time period:
\[ T = \frac{2\pi}{\omega} \]
Rearranging gives:
\[ \omega = \frac{2\pi}{T} \]
Substituting the given time period:
\[ \omega = \frac{2\pi}{6} = \frac{\pi}{3} \text{ rad/s} \]

### Step 2: Calculate Spring Constant (k)
The angular frequency is also related to the spring constant (k) and mass (m) by the formula:
\[ \omega = \sqrt{\frac{k}{m}} \]
Squaring both sides gives:
\[ \omega^2 = \frac{k}{m} \]
Substituting for ω and m:
\[ \left(\frac{\pi}{3}\right)^2 = \frac{k}{0.02} \]
Calculating \( \left(\frac{\pi}{3}\right)^2 \):
\[ \frac{\pi^2}{9} = \frac{k}{0.02} \]
Thus,
\[ k = 0.02 \cdot \frac{\pi^2}{9} \approx 0.00219 \text{ N/m} \]

### Step 3: Calculate Amplitude (A)
Using the velocity formula in SHM:
\[ V = A \omega \cos(\omega t) \]
At \( t = 1 \) second, we have:
\[ 0.06 = A \cdot \frac{\pi}{3} \cdot \cos\left(\frac{\pi}{3}\right) \]
Since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ 0.06 = A \cdot \frac{\pi}{3} \cdot \frac{1}{2} \]
Solving for A:
\[ A = \frac{0.06 \cdot 3}{\pi \cdot \frac{1}{2}} = \frac{0.18}{\pi} \approx 0.0573 \text{ m} \]

### Step 4: Calculate Displacement (x) at t = 1 s
Using the displacement formula:
\[ x = A \sin(\omega t) \]
Substituting values:
\[ x = \frac{0.18}{\pi} \sin\left(\frac{\pi}{3}\right) \]
Since \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \):
\[ x = \frac{0.18}{\pi} \cdot \frac{\sqrt{3}}{2} \approx 0.0992 \text{ m} \]

### Step 5: Calculate Kinetic Energy (K.E.)
Using the formula for kinetic energy:
\[ K.E. = \frac{1}{2} m V^2 \]
Substituting values:
\[ K.E. = \frac{1}{2} \cdot 0.02 \cdot (0.06)^2 = \frac{1}{2} \cdot 0.02 \cdot 0.0036 \approx 3.6 \times 10^{-5} \text{ J} \]

### Step 6: Calculate Potential Energy (P.E.)
Using the formula for potential energy:
\[ P.E. = \frac{1}{2} k x^2 \]
Substituting values:
\[ P.E. = \frac{1}{2} \cdot 0.00219 \cdot (0.0992)^2 \approx 10.8 \times 10^{-5} \text{ J} \]

### Step 7: Calculate Total Energy (T.E.)
Total energy in SHM is the sum of kinetic and potential energy:
\[ T.E. = K.E. + P.E. \]
Substituting the values:
\[ T.E. = 3.6 \times 10^{-5} + 10.8 \times 10^{-5} = 14.4 \times 10^{-5} \text{ J} \]

### Final Answers:
- Kinetic Energy (K.E.) = \( 3.6 \times 10^{-5} \) J
- Potential Energy (P.E.) = \( 10.8 \times 10^{-5} \) J
- Total Energy (T.E.) = \( 14.4 \times 10^{-5} \) J