A 1 kg block executes simple harmonic motion with an amplitude of...

Initial mass $m_1 = 1 \text{ kg}$, amplitude $A_1 = 0.1 \text{ m}$, and spring constant $k = 100 \text{ N/m}$.

The angular frequency of the initial block is:
$\omega_1 = \sqrt{\dfrac{k}{m_1}} = \sqrt{\dfrac{100}{1}} = 10 \text{ rad/s}$

The velocity of the block at the mean position is:
$v_1 = A_1 \omega_1 = 0.1 \times 10 = 1 \text{ m/s}$

When the $3 \text{ kg}$ block is gently placed on the $1 \text{ kg}$ block at the mean position, the horizontal momentum of the system is conserved. Let $v_2$ be the new velocity at the mean position:
$m_1 v_1 = (m_1 + m_2) v_2$
$1 \times 1 = (1 + 3) v_2 \Rightarrow v_2 = 0.25 \text{ m/s}$

The new angular frequency of the system is:
$\omega_2 = \sqrt{\dfrac{k}{m_1 + m_2}} = \sqrt{\dfrac{100}{4}} = 5 \text{ rad/s}$

The new frequency is:
$f_2 = \dfrac{\omega_2}{2\pi} = \dfrac{5}{2\pi} \text{ Hz}$

The new amplitude $A_2$ can be found using the relation $v_2 = A_2 \omega_2$ at the mean position:
$0.25 = A_2 \times 5 \Rightarrow A_2 = 0.05 \text{ m} = 5 \text{ cm}$