In Figure 15-41, block 2 of massoscillates on the end of a spring in SHM with a period of$\mathbf{20}\mathbf{}\mathbf{ms}$.The block’s position is given by$\mathbf{x}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{cm}\mathbf{)}\mathbf{cos}\mathbf{(}\mathbf{\omega t}\mathbf{+}\mathbf{\pi }\mathbf{/}\mathbf{2}\mathbf{)}$Block 1 of mass$\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}$slides toward block 2with a velocity of magnitude$\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$, directed along the spring’s length. The two blocks undergo a completely inelastic collision at time$\mathbf{t}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{ms}$. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?

1. Block position of the block,$\mathrm{x}=(1.0\mathrm{cm})\cos \left(\mathrm{\omega t}+\frac{\mathrm{\pi }}{2}\right)$
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,$\mathrm{T}=20\mathrm{ms}\mathrm{or}20\times {10}^{-3}\sec$
7. Time at collision,$t=5.0\mathrm{ms}\mathrm{or}5.0\times {10}^{-3}\sec$

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.

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

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

$$\begin{gathered} \mathrm{\omega }=\frac{2\times 3.14}{20\times {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\times {10}^{-3}\right)+\frac{\mathrm{\pi }}{2}\right) \\ =-3.14\sin \left(\mathrm{\pi }\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{\times }{V}_{final} \\ (4.0\times 6.0)+0=(4.0+2.0)\times {\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}\times k\times 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\times {10}^{-4}) \\ \mathrm{Calculate}\mathrm{the}\mathrm{value}\mathrm{of}k\mathit{,} \\ \mathrm{k}={\mathrm{M\omega }}^2 \\ =2\times {\left(314\right)}^2=1.97\times {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\times {10}^5}}+\left(1.0\times {10}^{-4}\right) \\ =\sqrt{4.84\times {10}^{-4}+(1.0\times {10}^{-4})} \\ =2.41\times {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}$$