A mass $$0.9 \text{ kg}$$, attached to a horizontal spring, executes SHM with an amplitude $$A_1$$. When this mass passes through its mean position, then a smaller mass of $$124 \text{ g}$$ is placed over it and both masses move together with amplitude $$A_2$$. If the ratio $$\dfrac{A_1}{A_2}$$ is $$\dfrac{\alpha}{\alpha - 1}$$, then the value of $$\alpha$$ will be ______.

We are given a mass $$m_1 = 0.9 \text{ kg}$$ executing SHM with amplitude $$A_1$$. When it passes through the mean position, a mass $$m_2 = 124 \text{ g} = 0.124 \text{ kg}$$ is placed on it, and they move together with amplitude $$A_2$$. We need to find $$\alpha$$ where $$\dfrac{A_1}{A_2} = \dfrac{\alpha}{\alpha - 1}$$.

At the mean position, the velocity of the first mass is maximum:

$$v_1 = A_1 \omega_1$$

where $$\omega_1 = \sqrt{\dfrac{k}{m_1}}$$ is the angular frequency.

When the smaller mass is placed on the first mass at the mean position, by conservation of linear momentum:

$$m_1 v_1 = (m_1 + m_2) v_2$$

$$v_2 = \frac{m_1 v_1}{m_1 + m_2}$$

After the mass is added, the new angular frequency is $$\omega_2 = \sqrt{\dfrac{k}{m_1 + m_2}}$$.

At the mean position, the velocity equals the maximum velocity:

$$v_2 = A_2 \omega_2$$

$$\frac{A_1}{A_2} = \frac{v_1 / \omega_1}{v_2 / \omega_2} = \frac{v_1}{v_2} \times \frac{\omega_2}{\omega_1}$$

$$\frac{v_1}{v_2} = \frac{m_1 + m_2}{m_1}$$

$$\frac{\omega_2}{\omega_1} = \sqrt{\frac{k/(m_1 + m_2)}{k/m_1}} = \sqrt{\frac{m_1}{m_1 + m_2}}$$

$$\frac{A_1}{A_2} = \frac{m_1 + m_2}{m_1} \times \sqrt{\frac{m_1}{m_1 + m_2}} = \sqrt{\frac{m_1 + m_2}{m_1}}$$

$$\frac{A_1}{A_2} = \sqrt{\frac{0.9 + 0.124}{0.9}} = \sqrt{\frac{1.024}{0.9}} = \sqrt{\frac{1024}{900}} = \frac{32}{30} = \frac{16}{15}$$

Comparing with $$\dfrac{\alpha}{\alpha - 1} = \dfrac{16}{15}$$:

$$\alpha - 1 = 15$$ and $$\alpha = 16$$.

Therefore, $$\alpha = 16$$.