Can I use energy conservation to find out the new amplitude of SHM? 
A mass $m_1$ connected to a horizontal spring performs S.H.M. with amplitude $A$. While mass $m_1$ is passing through its mean position another mass $m_2$ is placed on it so that both the masses move together with amplitude $A_1$. The ratio $A_1/A$ is? ($m_2<m_1$).

I know that I can use conservation of momentum to solve this question. I just wanted to know why I can't use conservation of energy to do the same.
I want to equate the energy in SHM for the first SHM at mean position and the energy in SHM for the second SHM with two blocks at the mean position.
 A: The question is ambiguous, as the answer depends on how $m_2$ is placed on $m_1$. If it is moving with $m_1$, then you're adding kinetic energy to the system:
$$ \frac 1 2 m_1 v^2 \rightarrow \frac 1 2  (m_1+m_2) v^2$$
and if it's stationary, then you are taking energy out (but conserving momentum):
$$ \frac{p^2}{2m_1} \rightarrow \frac{p^2}{2(m_1+m_2)}$$
The question doesn't specify other than to say it is done such that the amplitude is $A_1$.
Since potential energy goes as $\frac 1 2 kA^2$, and the question was multiple choice, you should be able to pick the correct answer without doing any serious calculation.
A: The total kinetic energy is not conserved when $m_2$ is placed on $m_1$. You may think of this as an inelastic collision, since the masses stick together after coming into contact. Thus, it would be invalid to use conservation of energy to solve the problem.
For an elastic collision, you would then be able to use conservation of energy. This case would correspond to $m_2$ bouncing off $m_1$ after colliding with it. In this case, both conservation of energy and conservation of momentum are needed to solve the problem.
You might be wondering why in the former case 1 equation is required, while in the latter case 2 equations are required. This is because when the blocks stick together, the number of degrees of freedom in the system decrease from 2 to 1, since we know their velocities become equal. This additional constraint balances out the fact that we didn't use conservation of energy.
