**1. The problem statement, all variables and given/known data**

A light elastic string on a smooth horizontal table has one of its ends fastened. The other end is attached to a particle of mass m. The string of force constant ##k## is stretched to twice its natural length of ##l_0## and the particle is projected along the table at right angles with a speed ##v_0##.

i)Show that in the subsequent motion, the string will attain its natural length again if its initial kinetic energy is less than a critical fraction of its initial total energy. Find this critical fraction.

ii)If one-fifth of its initial energy is kinetic, and if the string attains its natural length at some instant, describe the motion when the string becomes slack. Show that it will remain slack for duration ##\tau_0##. Determine ##\tau_0##.

**2. Relevant equations**

**3. The attempt at a solution**

Let ##r## be the distance of mass from the fixed end at any point of time. Then,

$$m\frac{d^2r}{dt^2}=-k(r-l_0)$$

$$\Rightarrow r(t)=A\sin(\omega t)+B\cos(\omega t)+l_0$$

where ##\omega=\sqrt{k/m}##.

From the initial conditions, ##r(0)=2l_0## and ##r'(0)=0##, I get ##A=0## and ##B=l_0## i.e

$$r(t)=l_0(1+\cos(\omega t))$$

When the string attains its natural length, ##r(t)=l_0 \Rightarrow \sin (\omega t)=1##. Hence, radial speed at that instant is ##v_r=l_0\omega##.

There is also a component of velocity perpendicular to radial velocity at any instant. I denote its magnitude by ##v_{\perp}##. This can be obtained from conservation of angular momentum about the fixed end of string.

When the string is at its natural length, ##v_{\perp}=2v_0##.

Now I am not sure how to proceed, I tried energy conservation. LHS of the following equation is energy at t=0 and RHS is energy when the string is relaxed.

$$\frac{1}{2}kl_0^2+\frac{1}{2}mv_0^2=\frac{1}{2}m\left(v_r^2+v_{\perp} ^2\right)$$

$$\Rightarrow kl_0+mv_0^2=m\left( l_0^2\frac{k}{m}+4v_0^2\right)$$

As it can be seen, both sides are not equal so I guess I went wrong somewhere. π

Any help is appreciated. Thanks!

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