A block of mass 2.0 kg attached to a horizontal spring of force constant 200 N m^-1 executes simple harmonic motion of amplitude 0.10 m on a smooth…

Before the collision the block passes the equilibrium position with maximum speed $v_0=\omega A=\sqrt{k/m}\,A=\sqrt{100}\times0.10=1.0\ \mathrm{m\,s^{-1}}$. The putty falls vertically, so there is no horizontal impulse on the system except from the sticking itself, and horizontal momentum is conserved: $2.0\times1.0=(2.0+1.0)v_1$, giving $v_1=2/3\ \mathrm{m\,s^{-1}}$. The spring force is zero at the equilibrium position, so this is the speed at the new equilibrium position too and the new motion has maximum speed $v_1$. The new angular frequency is $\omega'=\sqrt{k/(m+M)}=\sqrt{200/3}=8.165\ \mathrm{rad\,s^{-1}}$. The new amplitude is $A'=v_1/\omega'=0.6667/8.165=0.0816\ \mathrm{m}$, i.e. $8.16\ \mathrm{cm}$. Energy is lost in the sticking, so $A'<A$.