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 ?

  • 1
    $\begingroup$ I searched for a duplicate to this question because I expected that this question had been asked before, but I couldn't find one. That might suggest that I lack the proper search skills. $\endgroup$
    – march
    Commented Apr 20, 2022 at 17:48

1 Answer 1


In case 1, the collision between the two masses is inelastic (because they stick together), and so some of the energy is "lost", or rather, some of the energy is converted to thermal energy of the two masses. This doesn't happen in the case where you attach $m$ to $M$ when it's stationary, and hence the mechanical energy of the system is smaller in the first case then in the second case.

Quantitative analysis

In the first case, we have an inelastic collision, the result of which is that the new (maximum) speed of the oscillator is $$ v_{\textrm{new}} = \frac{M}{m+M}v\,, $$ and hence the new maximum kinetic energy (and hence total mechanical energy) is $$ K_{\textrm{max,new}} = \frac{1}{2}(M+m)\left(\frac{M}{m+M}v\right)^2 = \frac{M}{m+M} \frac{1}{2}Mv^2 <\frac{1}{2}Mv^2 = K_{\textrm{max,old}}\,. $$ In the second case, when $m$ is attached to $M$ when it is at its full amplitude, the total mechanical energy doesn't change, because the energy only depends on the amplitude, i.e., $$ U_{\textrm{max,old}} = \frac{1}{2}kA_{\textrm{old}}^2 = \frac{1}{2}kA_{\textrm{new}}^2 = U_{\textrm{max,new}}\,. $$


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