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When the mass is compressed to the maximum, it experiences potential energy but no kinetic energy. So the total energy of the mass at P is,
[tex]\longrightarrow\sf{E_P=U_P+K_P}[/tex]
[tex]\longrightarrow\sf{E_P=\dfrac{1}{2}k(2A)^2+0}[/tex]
[tex]\longrightarrow\sf{E_P=2kA^2}[/tex]
When the mass collides with right wall, i.e., when it reaches Q, it has kinetic energy as well as potential energy.
So the total energy of the mass at Q is,
[tex]\longrightarrow\sf{E_Q=U_Q+K_Q}[/tex]
[tex]\longrightarrow\sf{E_Q=\dfrac{1}{2}kA^2+K_Q}[/tex]
By energy conservation,
[tex]\longrightarrow\sf{E_P=E_Q}[/tex]
[tex]\longrightarrow\sf{2kA^2=\dfrac{1}{2}kA^2+K_Q}[/tex]
[tex]\longrightarrow\sf{K_Q=\dfrac{3}{2}kA^2}[/tex]
Here the mass undergoes SHM with O as the mean position and P as extreme. So let its displacement be,
[tex]\longrightarrow\sf{x=2A\sin(\omega t+\phi)\quad\quad\dots(1)}[/tex]
[tex]\sf{2A}[/tex] is because the maximum amplitude is the same as maximum compression.
The mass has a displacement [tex]\sf{x=-2A}[/tex] at time [tex]\sf{t=0.}[/tex]
[tex]\longrightarrow\sf{-2A=2A\sin(\omega(0)+\phi)}[/tex]
[tex]\longrightarrow\sf{-2A=2A\sin\phi}[/tex]
[tex]\longrightarrow\sf{\sin\phi=-1}[/tex]
[tex]\longrightarrow\sf{\phi=-\dfrac{\pi}{2}}[/tex]
Also,
[tex]\longrightarrow\sf{\omega=\sqrt{\dfrac{k}{m}}}[/tex]
From the question,
[tex]\longrightarrow\sf{\omega=\dfrac{\pi}{12}}[/tex]
Hence (1) becomes,
[tex]\longrightarrow\sf{x=2A\sin\left(\dfrac{\pi t}{12}-\dfrac{\pi}{2}\right)}[/tex]
[tex]\longrightarrow\sf{x=2A\sin\left[\dfrac{\pi}{12}\left(t-6\right)\right]}[/tex]
We've to find the time at which mass will be at Q having displacement [tex]\sf{x=A.}[/tex]
[tex]\longrightarrow\sf{A=2A\sin\left[\dfrac{\pi}{12}\left(t-6\right)\right]}[/tex]
[tex]\longrightarrow\sf{\sin\left[\dfrac{\pi}{12}\left(t-6\right)\right]=\dfrac{1}{2}}[/tex]
[tex]\longrightarrow\sf{\dfrac{\pi}{12}\left(t-6\right)=\dfrac{\pi}{6}}[/tex]
[tex]\longrightarrow\sf{t=8\ s}[/tex]
The mass loses two - third of its kinetic energy at Q, so its new kinetic energy at Q will be,
[tex]\longrightarrow\sf{K_Q\,\!'=\dfrac{1}{3}K_Q}[/tex]
[tex]\longrightarrow\sf{K_Q\,\!'=\dfrac{1}{2}kA^2}[/tex]
Thus the total energy of the mass at Q becomes,
[tex]\longrightarrow\sf{E_Q\,\!'=U_Q+K_Q\,\!'}[/tex]
[tex]\longrightarrow\sf{E_Q\,\!'=\dfrac{1}{2}kA^2+\dfrac{1}{2}kA^2}[/tex]
[tex]\longrightarrow\sf{E_Q\,\!'=kA^2}[/tex]
Now the mass gets compressed to its maximum at R at a distance [tex]\sf{A'}[/tex] from mean position.
The mass has potential energy only at R. So its total energy at R will be,
[tex]\longrightarrow\sf{E_R=U_R+K_R}[/tex]
[tex]\longrightarrow\sf{E_R=\dfrac{1}{2}k(A')^2}[/tex]
By energy conservation.
[tex]\longrightarrow\sf{E_Q\,\!'=E_R}[/tex]
[tex]\longrightarrow\sf{kA^2=\dfrac{1}{2}k(A')^2}[/tex]
[tex]\longrightarrow\sf{A'=A\sqrt2}[/tex]
The displacement of the mass will be,
[tex]\longrightarrow\sf{x'=A\sqrt2\,\sin(\omega t+\phi')}[/tex]
[tex]\longrightarrow\sf{x'=A\sqrt2\,\sin\left(\dfrac{\pi t}{12}+\phi'\right)\quad\quad\dots(2)}[/tex]
so that it has a displacement [tex]\sf{x=A}[/tex] at [tex]\sf{t=8\ s}[/tex] at Q.
[tex]\longrightarrow\sf{A=A\sqrt2\,\sin\left(\dfrac{8\pi}{12}+\phi'\right)}[/tex]
[tex]\longrightarrow\sf{\sin\left(\dfrac{2\pi}{3}+\phi'\right)=\dfrac{1}{\sqrt2}}[/tex]
[tex]\longrightarrow\sf{\dfrac{2\pi}{3}+\phi'=\dfrac{3\pi}{4}\quad\quad\dots(i)}[/tex]
[tex]\longrightarrow\sf{\phi'=\dfrac{\pi}{12}}[/tex]
Hence (2) becomes,
[tex]\longrightarrow\sf{x'=A\sqrt2\,\sin\left(\dfrac{\pi t}{12}+\dfrac{\pi}{12}\right)}[/tex]
[tex]\longrightarrow\sf{x'=A\sqrt2\,\sin\left[\dfrac{\pi}{12}\left(t+1\right)\right]}[/tex]
Let's find time at which the mass reaches R with displacement [tex]\sf{x'=-A\sqrt2.}[/tex]
[tex]\longrightarrow\sf{-A\sqrt2=A\sqrt2\,\sin\left[\dfrac{\pi}{12}\left(t+1\right)\right]}[/tex]
[tex]\longrightarrow\sf{\sin\left[\dfrac{\pi}{12}\left(t+1\right)\right]=-1}[/tex]
[tex]\longrightarrow\sf{\dfrac{\pi}{12}\left(t+1\right)=\dfrac{3\pi}{2}\quad\quad\dots(ii)}[/tex]
[tex]\longrightarrow\underline{\underline{\sf{t=17\ s}}}[/tex]
Hence the mass comes back to rest again after 17 seconds of releasing.
Equations (i) and (ii) are made in such a way that the right hand sides should have minimum difference.
[tex]\setlength{\unitlength}{1.5mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){50}}\multiput(0,0)(50,0){2}{\line(0,1){10}}\put(17.5,0.25){\framebox(5,5){\sf{m}}}\put(14,3){\line(1,0){3.5}}\put(0,3){\line(1,0){3}}\multiput(3,3)(1.2,0){8}{\qbezier(0,0)(1,3)(2,0)\qbezier(2,0)(1.5,-3)(1.2,0)}\qbezier(12.5,3)(13,6)(14,3)\multiput(20,0)(10,0){4}{\line(0,-1){1}}\put(40,-5){\line(0,-1){4}}\put(19.3,-4){\sf{P}}\put(29.3,-4){\sf{R}}\put(39.3,-4){\sf{O}}\put(49.3,-4){\sf{Q}}\put(40,-6){\vector(-1,0){20}}\put(40,-8){\vector(-1,0){10}}\put(40,-7){\vector(1,0){10}}\put(19,-9.2){\sf{2A}}\put(29,-11.2){\sf{A'}}\put(49,-10.2){\sf{A}}\end{picture}[/tex]