Suppose i have a block of mass $M $ performing simple harmonic motion under a spring. Now suppose i gently place a particle of mass $m $ on top of it.
Case 1
The mass $m$ is placed when block of mass $M$ is passing through its equilibrium position with a speed, say $v_°$
The new angular frequency $\omega_{new}$ will become $$\omega_{new} = \sqrt{\frac{k}{m+M}} = \omega_{old}.\sqrt{\frac{M}{m+M}}$$ By momentum conservation $$v_{new} = \frac{M v_°}{m + M}$$ so $$\text{Energy}_{\text{oscillation}} = \text{Kinetic Energy}_{max} = \frac12 m u_{_{new}}^2$$
Case 2
the mass $m$ is placed on the block when it is at its extreme position.
again $$\omega_{new} = \omega_{old}.\sqrt{\frac{M}{m+M}}$$
but this time since amplitude is same (unlike case 1 where it was changed), $v_{max} = \omega_{new}.A = \sqrt{\frac{M}{m+M}} v$ clearly here $v_{max}$ is not like before ($v/2$). so total oscillation energy must be also different.
Both cases have similar physical setup. so where is this difference in energy going in one of them ?