Question

Question forJEE Aspirant​

Answer 1

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,

$\longrightarrow\sf{E_P=U_P+K_P}$

$\longrightarrow\sf{E_P=\dfrac{1}{2}k(2A)^2+0}$

$\longrightarrow\sf{E_P=2kA^2}$

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,

$\longrightarrow\sf{E_Q=U_Q+K_Q}$

$\longrightarrow\sf{E_Q=\dfrac{1}{2}kA^2+K_Q}$

By energy conservation,

$\longrightarrow\sf{E_P=E_Q}$

$\longrightarrow\sf{2kA^2=\dfrac{1}{2}kA^2+K_Q}$

$\longrightarrow\sf{K_Q=\dfrac{3}{2}kA^2}$

Here the mass undergoes SHM with O as the mean position and P as extreme. So let its displacement be,

$\longrightarrow\sf{x=2A\sin(\omega t+\phi)\quad\quad\dots(1)}$

$\sf{2A}$ is because the maximum amplitude is the same as maximum compression.

The mass has a displacement $\sf{x=-2A}$ at time $\sf{t=0.}$

$\longrightarrow\sf{-2A=2A\sin(\omega(0)+\phi)}$

$\longrightarrow\sf{-2A=2A\sin\phi}$

$\longrightarrow\sf{\sin\phi=-1}$

$\longrightarrow\sf{\phi=-\dfrac{\pi}{2}}$

Also,

$\longrightarrow\sf{\omega=\sqrt{\dfrac{k}{m}}}$

From the question,

$\longrightarrow\sf{\omega=\dfrac{\pi}{12}}$

Hence (1) becomes,

$\longrightarrow\sf{x=2A\sin\left(\dfrac{\pi t}{12}-\dfrac{\pi}{2}\right)}$

$\longrightarrow\sf{x=2A\sin\left[\dfrac{\pi}{12}\left(t-6\right)\right]}$

We've to find the time at which mass will be at Q having displacement $\sf{x=A.}$

$\longrightarrow\sf{A=2A\sin\left[\dfrac{\pi}{12}\left(t-6\right)\right]}$

$\longrightarrow\sf{\sin\left[\dfrac{\pi}{12}\left(t-6\right)\right]=\dfrac{1}{2}}$

$\longrightarrow\sf{\dfrac{\pi}{12}\left(t-6\right)=\dfrac{\pi}{6}}$

$\longrightarrow\sf{t=8\ s}$

The mass loses two - third of its kinetic energy at Q, so its new kinetic energy at Q will be,

$\longrightarrow\sf{K_Q\,\!'=\dfrac{1}{3}K_Q}$

$\longrightarrow\sf{K_Q\,\!'=\dfrac{1}{2}kA^2}$

Thus the total energy of the mass at Q becomes,

$\longrightarrow\sf{E_Q\,\!'=U_Q+K_Q\,\!'}$

$\longrightarrow\sf{E_Q\,\!'=\dfrac{1}{2}kA^2+\dfrac{1}{2}kA^2}$

$\longrightarrow\sf{E_Q\,\!'=kA^2}$

Now the mass gets compressed to its maximum at R at a distance $\sf{A'}$ from mean position.

The mass has potential energy only at R. So its total energy at R will be,

$\longrightarrow\sf{E_R=U_R+K_R}$

$\longrightarrow\sf{E_R=\dfrac{1}{2}k(A')^2}$

By energy conservation.

$\longrightarrow\sf{E_Q\,\!'=E_R}$

$\longrightarrow\sf{kA^2=\dfrac{1}{2}k(A')^2}$

$\longrightarrow\sf{A'=A\sqrt2}$

The displacement of the mass will be,

$\longrightarrow\sf{x'=A\sqrt2\,\sin(\omega t+\phi')}$

$\longrightarrow\sf{x'=A\sqrt2\,\sin\left(\dfrac{\pi t}{12}+\phi'\right)\quad\quad\dots(2)}$

so that it has a displacement $\sf{x=A}$ at $\sf{t=8\ s}$ at Q.

$\longrightarrow\sf{A=A\sqrt2\,\sin\left(\dfrac{8\pi}{12}+\phi'\right)}$

$\longrightarrow\sf{\sin\left(\dfrac{2\pi}{3}+\phi'\right)=\dfrac{1}{\sqrt2}}$

$\longrightarrow\sf{\dfrac{2\pi}{3}+\phi'=\dfrac{3\pi}{4}\quad\quad\dots(i)}$

$\longrightarrow\sf{\phi'=\dfrac{\pi}{12}}$

Hence (2) becomes,

$\longrightarrow\sf{x'=A\sqrt2\,\sin\left(\dfrac{\pi t}{12}+\dfrac{\pi}{12}\right)}$

$\longrightarrow\sf{x'=A\sqrt2\,\sin\left[\dfrac{\pi}{12}\left(t+1\right)\right]}$

Let's find time at which the mass reaches R with displacement $\sf{x'=-A\sqrt2.}$

$\longrightarrow\sf{-A\sqrt2=A\sqrt2\,\sin\left[\dfrac{\pi}{12}\left(t+1\right)\right]}$

$\longrightarrow\sf{\sin\left[\dfrac{\pi}{12}\left(t+1\right)\right]=-1}$

$\longrightarrow\sf{\dfrac{\pi}{12}\left(t+1\right)=\dfrac{3\pi}{2}\quad\quad\dots(ii)}$

$\longrightarrow\underline{\underline{\sf{t=17\ s}}}$

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.

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