Along the principal axis of a convex lens having a focal length of...

Given the focal length of the convex lens, $f = +12 \text{ cm}$.

The mean position of the particle is at $u = -20 \text{ cm}$ and its amplitude of oscillation is $A_o = 1.0 \text{ cm}$.

The extreme positions of the oscillating particle on the principal axis are $u_1 = -20 - 1 = -21 \text{ cm}$ and $u_2 = -20 + 1 = -19 \text{ cm}$.

Using the lens formula, $v = \dfrac{uf}{u+f}$, we find the corresponding extreme positions of the image.

For $u_1 = -21 \text{ cm}$:
$v_1 = \dfrac{-21 \times 12}{-21 + 12} = \dfrac{-252}{-9} = 28 \text{ cm}$

For $u_2 = -19 \text{ cm}$:
$v_2 = \dfrac{-19 \times 12}{-19 + 12} = \dfrac{-228}{-7} = \dfrac{228}{7} \text{ cm}$

The path length of the image's oscillation is the distance between these two extreme positions:
Path length $= v_2 - v_1 = \dfrac{228}{7} - 28 = \dfrac{228 - 196}{7} = \dfrac{32}{7} \text{ cm}$

The amplitude of oscillation for the image is half of its path length:
$A_i = \dfrac{1}{2} \times \dfrac{32}{7} = \dfrac{16}{7} \approx 2.28 \text{ cm} \approx 2.3 \text{ cm}$