91Ó°ÊÓ

1. Block position of the block,$\mathrm{x}=(1.0\mathrm{cm})\cos \left(\mathrm{Ó\neg ³Ù}+\frac{\mathrm{Ï€}}{2}\right)$
2. From above equation amplitude,${\mathrm{x}}_{\max }=1.0\mathrm{cm}\mathrm{or}0.010\mathrm{m}$
3. Mass of block 1,${\mathrm{M}}_1=4.0\mathrm{kg}$
4. Mass of block 2,${\mathrm{M}}_2=2.0\mathrm{kg}$
5. Velocity of block 1,${\mathrm{M}}_1=6.0\mathrm{m}/\mathrm{s}$
6. Period of oscillation of block 2,$\mathrm{T}=20\mathrm{ms}\mathrm{or}20×{10}^{-3}\sec$
7. Time at collision,$t=5.0\mathrm{ms}\mathrm{or}5.0×{10}^{-3}\sec$

Using the concept of inelastic collision, we can find the final velocity of the system of two blocks. Next, we can use the concept of conservation of momentum to find the maximum amplitude of SHM.

$$\begin{gathered} \mathrm{Formula}: \\ \mathrm{The}\mathrm{general}\mathrm{expression}\mathrm{of}\mathrm{displacement}\mathrm{of}\mathrm{a}\mathrm{body}\mathrm{in}\mathrm{motion},x\mathit{=}a\mathit{cos}\mathit{(}\mathit{Ó\neg }\mathit{t}\mathit{+}\mathit{Ï•}\mathit{)}……\left(\mathrm{i}\right) \\ \mathrm{The}\mathrm{general}\mathrm{expression}\mathrm{of}\mathrm{velocity}\mathrm{of}\mathrm{a}\mathrm{body}\mathrm{in}\mathrm{motion},\mathrm{v}=-\mathrm{Ó\neg asin}\mathit{(}\mathrm{Ó\neg }t\mathit{+}\mathrm{Ï•}\mathit{)}.......\left(\mathrm{ii}\right) \\ \mathrm{Angular}\mathrm{velocity}\mathrm{of}\mathrm{a}\mathrm{body},\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{\mathrm{T}}…….\left(\mathrm{iii}\right) \\ \mathrm{Law}\mathrm{of}\mathrm{conservation}\mathrm{of}\mathrm{momentum},\mathrm{I}\mathrm{nitial}\mathrm{momentum}=\mathrm{Final}\mathrm{momentum}………\left(\mathrm{iv}\right) \\ \mathrm{The}\mathrm{energy}\mathrm{constant}\mathrm{of}\mathrm{a}\mathrm{spring},k\mathit{=}M{\mathrm{Ó\neg }}^{\mathit{2}}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{\left(}v\mathit{\right)} \end{gathered}$$

First, the angular frequency of block 2 using equation (iii) is given by:

$$\begin{gathered} \mathrm{Ó\neg }=\frac{2×3.14}{20×{10}^{-3}} \\ =3.14\mathrm{rad}/\sec \\ \mathrm{Then}\mathrm{velocity}\mathrm{of}\mathrm{SHM}\mathrm{at}5.0\text{'}\mathrm{ms}\mathrm{using}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{is}\mathrm{given}\mathrm{by}: \\ \mathrm{v}=(-314)(0.010)\sin \left(\left(314\right)\left(5.0×{10}^{-3}\right)+\frac{\mathrm{Ï€}}{2}\right) \\ =-3.14\sin \left(\mathrm{Ï€}\right) \\ =0\mathrm{m}/\mathrm{s} \\ \mathrm{In}\mathrm{this}\mathrm{case},\mathrm{there}\mathrm{is}\mathrm{inelastic}\mathrm{collision}\mathrm{so}\mathrm{using}\mathrm{equation}\left(\mathrm{iv}\right),\mathrm{we}\mathrm{get} \\ {M}_{\mathit{1}}v\mathit{+}{M}_{\mathit{2}}v\mathit{=}\mathit{(}{M}_{\mathit{1}}\mathit{+}{M}_{\mathit{2}}\mathit{)}\mathit{×}{V}_{final} \\ (4.0×6.0)+0=(4.0+2.0)×{\mathrm{V}}_{\mathrm{final}} \\ {\mathrm{V}}_{\mathrm{final}}=4\mathrm{m}/\mathrm{s} \\ \mathrm{Now},\mathrm{total}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{system}\mathrm{is}\mathrm{given}\mathrm{as}: \\ E=KE+PE \\ E=\left(\frac{1}{2}\left(M_1+M_2\right)V_{final}^2\right)+\left(\frac{1}{2}k{A}^2\right) \\ \frac{1}{2}×k×A_{\mathrm{maximum}}^2=\frac{1}{2}\left(6\right)\left(16\right)+\frac{1}{2}\left(k\right){(0.010)}^2{A}_{maximum}=\frac{\sqrt{\left(96\right)}}{k}+(1.0×{10}^{-4}) \\ \mathrm{Calculate}\mathrm{the}\mathrm{value}\mathrm{of}k\mathit{,} \\ \mathrm{k}={\mathrm{MÓ\neg }}^2 \\ =2×{\left(314\right)}^2=1.97×{10}^5\mathrm{N}/\mathrm{m} \\ \mathrm{Substitute}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{for}\mathrm{amplitude} \\ {\mathrm{A}}_{\mathrm{maximum}}=\sqrt{\frac{\left(96\right)}{1.98×{10}^5}}+\left(1.0×{10}^{-4}\right) \\ =\sqrt{4.84×{10}^{-4}+(1.0×{10}^{-4})} \\ =2.41×{10}^{-2} \\ =0.024\mathrm{m} \\ \mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{maximum}\mathrm{amplitude}\mathrm{is}0.024\mathrm{m}. \end{gathered}$$