1. Given data

• Initial mass: $m_1 = 0.9\,\text{kg}$
• Added mass: $m_2 = 124\,\text{g} = 0.124\,\text{kg}$
• Initial amplitude: $A_1$
• New amplitude after masses stick together: $A_2$

The smaller mass is placed when the block is at the mean position.

2. Speed at mean position before adding the second mass

For SHM, the speed at the mean position is maximum:

$v_1 = \omega_1 A_1$

where

$\omega_1 = \sqrt{\frac{k}{m_1}}$

So,

$v_1 = A_1\sqrt{\frac{k}{m_1}}$

3. Velocity just after the second mass is placed

At the mean position, displacement is zero, so spring potential energy is zero. The added mass is gently placed and both move together, so use conservation of linear momentum at that instant:

$m_1 v_1 = (m_1+m_2)v_2$

Hence,

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

4. New angular frequency

After the masses move together, total mass becomes

$M = m_1+m_2$

So the new angular frequency is

$\omega_2 = \sqrt{\frac{k}{M}} = \sqrt{\frac{k}{m_1+m_2}}$

5. Find new amplitude $A_2$

Just after collision, the system is at mean position with speed $v_2$. For SHM, if a particle starts from mean position with speed $v_2$, then amplitude is

$A_2 = \frac{v_2}{\omega_2}$

Substitute $v_2$:

$A_2 = \frac{1}{\omega_2}\cdot \frac{m_1}{m_1+m_2}v_1$

Now put $v_1 = \omega_1 A_1$:

$A_2 = \frac{m_1}{m_1+m_2}\cdot \frac{\omega_1}{\omega_2} A_1$

Using

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

therefore,