A body of mass 40g is executing linear simple harmonic motion and has a velocity of 2cm/s when its displacement is 3cm and a velocity of 3 cm/s when its displacement is 2cm . Then

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A. The angular frequency of oscillation is 1rad/s
B. The amplitude of oscillation is `3sqrt2cm`
C. The maximum kinetic energy of oscillator is `2.6xx10^(-5)J`
D. The maximum potential energy ofocillator is `2.6xx10^(-5)J`

Explanation

To solve the problem step by step, we will use the equations of motion for a body executing simple harmonic motion (SHM).

### Given:
- Mass of the body, \( m = 40 \, \text{g} = 0.04 \, \text{kg} \)
- Displacement \( x_1 = 3 \, \text{cm} = 0.03 \, \text{m} \) with velocity \( v_1 = 2 \, \text{cm/s} = 0.02 \, \text{m/s} \)
- Displacement \( x_2 = 2 \, \text{cm} = 0.02 \, \text{m} \) with velocity \( v_2 = 3 \, \text{cm/s} = 0.03 \, \text{m/s} \)

### Step 1: Use the equation for velocity in SHM
The velocity \( v \) of a particle in SHM is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
where:
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude,
- \( x \) is the displacement.

### Step 2: Set up equations for both conditions
For the first condition (at \( x_1 = 0.03 \, \text{m} \)):
\[ 0.02 = \omega \sqrt{A^2 - (0.03)^2} \]
Squaring both sides:
\[ (0.02)^2 = \omega^2 (A^2 - 0.0009) \]
\[ 0.0004 = \omega^2 (A^2 - 0.0009) \quad \text{(Equation 1)} \]

For the second condition (at \( x_2 = 0.02 \, \text{m} \)):
\[ 0.03 = \omega \sqrt{A^2 - (0.02)^2} \]
Squaring both sides:
\[ (0.03)^2 = \omega^2 (A^2 - 0.0004) \]
\[ 0.0009 = \omega^2 (A^2 - 0.0004) \quad \text{(Equation 2)} \]

### Step 3: Divide Equation 1 by Equation 2
\[ \frac{0.0004}{0.0009} = \frac{A^2 - 0.0009}{A^2 - 0.0004} \]
This simplifies to:
\[ \frac{4}{9} = \frac{A^2 - 0.0009}{A^2 - 0.0004} \]

### Step 4: Cross-multiply and simplify
Cross-multiplying gives:
\[ 4(A^2 - 0.0004) = 9(A^2 - 0.0009) \]
Expanding both sides:
\[ 4A^2 - 0.0016 = 9A^2 - 0.0081 \]
Rearranging terms:
\[ 5A^2 = 0.0065 \]
\[ A^2 = \frac{0.0065}{5} = 0.0013 \]
Taking the square root:
\[ A = \sqrt{0.0013} \approx 0.036 \, \text{m} \approx 3.6 \, \text{cm} \]

### Step 5: Find the angular frequency \( \omega \)
Using Equation 1:
\[ 0.0004 = \omega^2 (A^2 - 0.0009) \]
Substituting \( A^2 = 0.0013 \):
\[ 0.0004 = \omega^2 (0.0013 - 0.0009) = \omega^2 (0.0004) \]
Thus:
\[ \omega^2 = 1 \implies \omega = 1 \, \text{rad/s} \]

### Step 6: Calculate maximum kinetic and potential energy
The maximum kinetic energy \( K_{\text{max}} \) and maximum potential energy \( U_{\text{max}} \) in SHM are given by:
\[ K_{\text{max}} = U_{\text{max}} = \frac{1}{2} k A^2 \]
where \( k = m \omega^2 \):
\[ k = 0.04 \times 1^2 = 0.04 \, \text{N/m} \]
Thus:
\[ K_{\text{max}} = U_{\text{max}} = \frac{1}{2} \times 0.04 \times 0.0013 = 2.6 \times 10^{-5} \, \text{J} \]

### Final Results:
1. Amplitude \( A \approx 3.6 \, \text{cm} \)
2. Angular frequency \( \omega = 1 \, \text{rad/s} \)
3. Maximum kinetic energy \( K_{\text{max}} = 2.6 \times 10^{-5} \, \text{J} \)
4. Maximum potential energy \( U_{\text{max}} = 2.6 \times 10^{-5} \, \text{J} \)