An object of mass `0.2 kg` exectes simple harmonic oscillations alongs `x-`axis with a freqency of `(25//pi) Hz`. At the positions `x = 0.04m`, the object has kinetic energy of `0.5 J` and potential energy `0.4 J`. Find the amplitude of oscillaties

Answer

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

### Step 1: Calculate Total Energy
The total energy \( E \) in simple harmonic motion (SHM) is the sum of kinetic energy \( KE \) and potential energy \( PE \).

Given:
- Kinetic Energy \( KE = 0.5 \, J \)
- Potential Energy \( PE = 0.4 \, J \)

Total Energy \( E \) can be calculated as:
\[ E = KE + PE = 0.5 \, J + 0.4 \, J = 0.9 \, J \]

### Step 2: Find Angular Frequency \( \omega \)
The angular frequency \( \omega \) can be calculated using the formula:
\[ \omega = 2\pi f \]

Given frequency \( f = \frac{25}{\pi} \, Hz \):
\[ \omega = 2\pi \left(\frac{25}{\pi}\right) = 50 \, \text{rad/s} \]

### Step 3: Use Total Energy Formula
The total energy in SHM is also given by the formula:
\[ E = \frac{1}{2} m \omega^2 A^2 \]

Where:
- \( m \) is the mass of the object
- \( A \) is the amplitude

Given:
- Mass \( m = 0.2 \, kg \)
- Total Energy \( E = 0.9 \, J \)
- \( \omega = 50 \, \text{rad/s} \)

Substituting the known values into the energy formula:
\[ 0.9 = \frac{1}{2} \times 0.2 \times (50)^2 \times A^2 \]

### Step 4: Simplify and Solve for Amplitude \( A \)
First, calculate \( \frac{1}{2} \times 0.2 \):
\[ \frac{1}{2} \times 0.2 = 0.1 \]

Now substitute this back into the equation:
\[ 0.9 = 0.1 \times 2500 \times A^2 \]
\[ 0.9 = 250 \times A^2 \]

Now, divide both sides by 250:
\[ A^2 = \frac{0.9}{250} = 0.0036 \]

Taking the square root to find \( A \):
\[ A = \sqrt{0.0036} = 0.06 \, m \]

### Final Answer
The amplitude of the oscillations is:
\[ A = 0.06 \, m \]

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