A mass m = 50 g is released from a height h = 10 cm above a...

Given mass $m = 50 \text{ g} = 0.05 \text{ kg}$, height $h = 10 \text{ cm} = 0.1 \text{ m}$, and spring constant $k = 500 \text{ N m}^{-1}$.

The velocity of the mass just before striking the spring is $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.1} = \sqrt{2} \text{ m s}^{-1}$.

Let the uncompressed free end of the spring be the origin. When the mass sticks to the spring, it executes simple harmonic motion (SHM).

The mean position of the oscillation is at a compression $x_0$ where the spring force balances the gravitational force:

$kx_0 = mg \Rightarrow x_0 = \dfrac{0.05 \times 10}{500} = 0.001 \text{ m} = 0.1 \text{ cm}$.

The angular frequency of the SHM is $\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{500}{0.05}} = 100 \text{ rad s}^{-1}$.

Using the velocity-displacement relation for SHM at the uncompressed position (displacement from mean position is $x_0 = 0.1 \text{ cm}$):

$v^2 = \omega^2 (A^2 - x_0^2)$

Substituting the values in standard SI units:

$(\sqrt{2})^2 = 100^2 (A^2 - 0.001^2)$

$2 = 10000 A^2 - 0.01 \Rightarrow 10000 A^2 = 2.01 \Rightarrow A^2 = 2.01 \times 10^{-4} \text{ m}^2$.

Approximating $A^2 \approx 2 \times 10^{-4} \text{ m}^2$, the amplitude is $A \approx \sqrt{2} \times 10^{-2} \text{ m} = \sqrt{2} \text{ cm}$.

The length of the oscillation path of the mass is $L_o = 2A = 2\sqrt{2} \approx 2.828 \text{ cm}$.

The pole of the concave mirror is at a distance of $30 \text{ cm}$ from the uncompressed free end. The object distance of the mean position is approximately $u = -30 \text{ cm}$, and the focal length is $f = -12 \text{ cm}$.

The longitudinal magnification $m_L$ of the mirror is given by:

$m_L = \left( \dfrac{f}{u - f} \right)^2 = \left( \dfrac{-12}{-30 - (-12)} \right)^2 = \left( \dfrac{-12}{-18} \right)^2 = \dfrac{4}{9}$

The length of the image over which it oscillates is:

$L_i = m_L \times L_o = \dfrac{4}{9} \times 2\sqrt{2} = \dfrac{8 \times 1.414}{9} \approx 1.25 \text{ cm}$.

The closest provided option is $1.2 \text{ cm}$.