To find the amplitude of oscillation, let's start by considering the given data and the formulae related to simple harmonic motion (SHM).

The total mechanical energy of a system in SHM is constant and is the sum of kinetic energy (KE) and potential energy (PE) at any point in its motion. We are given:

Kinetic Energy, $$ KE = 0.5 \, \text{J} $$

Potential Energy, $$ PE = 0.4 \, \text{J} $$

Therefore, the total energy (E) is:

$$ E = KE + PE = 0.5 + 0.4 = 0.9 \, \text{J} $$

In SHM, the total mechanical energy can also be expressed as:

$$ E = \frac{1}{2} k A^2 $$

where $ k $ is the spring constant and $ A $ is the amplitude.

We can also express kinetic and potential energy in SHM in terms of amplitude $ A $, position $ x $, and angular frequency $ \omega $:

Kinetic Energy: $$ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) $$

Potential Energy: $$ PE = \frac{1}{2} m \omega^2 x^2 $$

Given mass $ m = 0.2 \, \text{kg} $, position $ x = 0.04 \, \text{m} $.

The angular frequency $ \omega $ is related to the frequency $ f $ by:

$$ \omega = 2\pi f $$

Given frequency:

$$ f = \frac{25}{\pi} \, \text{Hz} $$

Thus,

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

Using the expression for the total energy in SHM:

$$ \frac{1}{2} k A^2 = \frac{1}{2} m \omega^2 A^2 $$

Since the total energy $ E = 0.9 \, \text{J} $, it follows:

$$ 0.9 = \frac{1}{2} \cdot 0.2 \cdot (50)^2 \cdot A^2 $$

Solving for $ A^2 $:

$$ 0.9 = \frac{1}{2} \cdot 0.2 \cdot 2500 \cdot A^2 $$

$$ 0.9 = \frac{1}{2} \cdot 500 \cdot A^2 $$

$$ 0.9 = 250 \cdot A^2 $$

$$ A^2 = \frac{0.9}{250} $$

$$ A^2 = 0.0036 $$

$$ A = \sqrt{0.0036} $$

$$ A = 0.06 \, \text{m} $$

The amplitude of the oscillations is 0.06 m.