These springs again!

Let $x(t)$ be the length of the unwrapped part of the spring, $\theta (t)$ be the angle through which the spring moves, $v(t)$ is the speed of the mass and $k(t)$ be the spring constant of the unwrapped part of the spring.

$\frac { m{ v }^{ 2 } }{ x } =kx$

Using this value of v, the frequency of the circular motion is given by:

$\omega =\frac { \Delta \theta }{ \Delta t } =\frac { v }{ x } =\sqrt { \frac { k }{ m } } \\$

Note: $k$ is inversely proportional to the equilibrium length, the change in angle of the contact point on the pole equals the change in angle of the mass around the pole which is 
$\Delta \theta$ and the fractional decrease in the equilibrium length of the unwrapped part is $\Delta \theta =\frac { a\Delta \theta }{ x }$ so the spring constant is:

${ k }_{ new }=\frac { { k }_{ old } }{ 1-\frac { a\Delta \theta }{ x } } \approx { k }_{ old }=1+\frac { a\Delta \theta }{ x } \\ \therefore \Delta k=\frac { ka\Delta \theta }{ x }$ dividing be $\Delta t$ $\Rightarrow$ $\overset { . }{ k } =\frac { ka\omega }{ x }$

The final equation we need is the conservation of energy:

$\frac { 1 }{ 2 } k{ x }^{ 2 }+\frac { 1 }{ 2 } m{ v }^{ 2 }=\frac { 1 }{ 2 } k^{ ' }x^{ '2 }+\frac { 1 }{ 2 } m^{ ' }v^{ '2 }+(\frac { 1 }{ 2 } k{ x }^{ 2 })(\frac { a\Delta \theta }{ x } )$

By using these equations:

$k{ x }^{ 2 }=k^{ ' }x^{ '2 }+\frac { 1 }{ 2 } m^{ ' }v^{ '2 }+\frac { 1 }{ 2 } k{ x }a\Delta \theta \\ -\frac { 1 }{ 2 } Kxa\omega =\frac { \Delta (K{ x }^{ 2 }) }{ \Delta t } \\ \quad \quad \quad \quad \quad =\overset { . }{ k } { x }^{ 2 }+2Lx\overset { . }{ x } \\ \quad \quad \quad \quad \quad =(\frac { ka\omega }{ x } ){ x }^{ 2 }+2kx\overset { . }{ x } \\ \overset { . }{ x } =-\frac { 3 }{ 4 } a\omega \\ \frac { \overset { . }{ x } }{ x } =-\frac { 3\overset { . }{ k } }{ 4k } \\ k=\frac { K{ L }^{ \frac { 4 }{ 3 } } }{ { x }^{ \frac { 4 }{ 3 } } } \\ { x }^{ \frac { 2 }{ 3 } }\overset { . }{ x } =-\frac { 3a{ K }^{ \frac { 1 }{ 2 } }{ L }^{ \frac { 2 }{ 3 } } }{ a{ m }^{ \frac { 1 }{ 2 } } } \\ { x }^{ \frac { 5 }{ 3 } }={ L }^{ \frac { 5 }{ 3 } }-(\frac { 5a{ K }^{ \frac { 1 }{ 2 } }{ L }^{ \frac { 2 }{ 3 } } }{ 4{ m }^{ \frac { 1 }{ 2 } } } )t\\ x(t)=L{ (1-\frac { t }{ T } ) }^{ \frac { 3 }{ 5 } }\\where \quad T=\frac { 4 }{ 5 } \frac { L }{ a } \sqrt { \frac { m }{ K } }$

is the time for which $x(t)=0$ and the mass hits the pole.

\\ T=\frac { \Delta }{ \Delta +1 } \frac { L }{ a } \sqrt { \frac { m }{ K } } =\frac { 4 }{ 5 } \frac { L }{ a } \sqrt { \frac { m }{ K } } \\ so\quad \Delta =4\\