An object of mass `0.2 kg` executes SHm along the X-axis with frequency of `(25//pi) Hz`. At the point `X=0.4 m` the object has `KE 0.5 J` and `PE 0.4 J`. The amplitude of oscilation is-

Answer

To find the amplitude of the oscillation for the object executing Simple Harmonic Motion (SHM), we can follow these steps:

### Step 1: Calculate Total Energy
The total energy (E) in SHM is the sum of kinetic energy (KE) and potential energy (PE):
\[ E = KE + PE \]
Given:
- KE = 0.5 J
- PE = 0.4 J

Calculating total energy:
\[ E = 0.5 \, \text{J} + 0.4 \, \text{J} = 0.9 \, \text{J} \]

### Step 2: Relate Total Energy to Amplitude
The total energy in SHM can also be expressed in terms of mass (m), angular frequency (ω), and amplitude (A):
\[ E = \frac{1}{2} m \omega^2 A^2 \]
We need to rearrange this equation to find the amplitude (A):
\[ A^2 = \frac{2E}{m \omega^2} \]
Thus,
\[ A = \sqrt{\frac{2E}{m \omega^2}} \]

### Step 3: Calculate Angular Frequency (ω)
The angular frequency (ω) is related to the frequency (f) by the formula:
\[ \omega = 2 \pi f \]
Given:
- Frequency \( f = \frac{25}{\pi} \, \text{Hz} \)

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

### Step 4: Substitute Values into Amplitude Formula
Now we can substitute the values of E, m, and ω into the amplitude formula:
- E = 0.9 J
- m = 0.2 kg
- ω = 50 rad/s

Calculating A:
\[ A = \sqrt{\frac{2 \times 0.9 \, \text{J}}{0.2 \, \text{kg} \times (50 \, \text{rad/s})^2}} \]
Calculating the denominator:
\[ (50 \, \text{rad/s})^2 = 2500 \, \text{rad}^2/\text{s}^2 \]
Now substituting:
\[ A = \sqrt{\frac{1.8}{0.2 \times 2500}} = \sqrt{\frac{1.8}{500}} = \sqrt{0.0036} \]
Calculating the square root:
\[ A = 0.06 \, \text{m} = 6 \, \text{cm} \]

### Final Answer
The amplitude of oscillation is:
\[ \boxed{0.06 \, \text{m} \, (6 \, \text{cm})} \]

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