Find the amplitude of oscillation of a 2 kg mass which is executing a harmonic motion with a frequency `25//pi Hz` along x-axis, also it is given that the kinetic energy and potential energy at x= 0.04 m are 0.5 J, 0.4 J:

Options

A. 0.05
B. 0.06
C. 0.01
D. none of thers

Explanation

To find the amplitude of oscillation of a 2 kg mass executing harmonic motion with a frequency of \( \frac{25}{\pi} \) Hz, given the kinetic energy and potential energy at \( x = 0.04 \) m are 0.5 J and 0.4 J respectively, we can follow these steps:

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

\[ E = KE + PE \]

Substituting the given values:

\[ E = 0.5 \, \text{J} + 0.4 \, \text{J} = 0.9 \, \text{J} \]

### Step 2: Relate Total Energy to Amplitude
The total energy in simple harmonic motion can also be expressed in terms of mass \( m \), angular frequency \( \omega \), and amplitude \( A \):

\[ E = \frac{1}{2} m \omega^2 A^2 \]

Where \( \omega = 2 \pi f \) and \( f \) is the frequency.

### Step 3: Calculate Angular Frequency
Given the frequency \( f = \frac{25}{\pi} \) Hz, we can calculate \( \omega \):

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

### Step 4: Substitute Values into Energy Equation
Now substituting \( E \), \( m \), and \( \omega \) into the energy equation:

\[ 0.9 = \frac{1}{2} \cdot 2 \cdot (50)^2 \cdot A^2 \]

### Step 5: Simplify the Equation
This simplifies to:

\[ 0.9 = 1 \cdot 2500 \cdot A^2 \]

\[ 0.9 = 2500 A^2 \]

### Step 6: Solve for Amplitude \( A \)
Now, solving for \( A^2 \):

\[ A^2 = \frac{0.9}{2500} \]

\[ A^2 = 0.00036 \]

Taking the square root to find \( A \):

\[ A = \sqrt{0.00036} = 0.018 \, \text{m} \]

### Conclusion
Thus, the amplitude of oscillation is:

\[ \boxed{0.018 \, \text{m}} \]