The gravitational potential energy of the spring is equal to the spring potential energy. The energy can be expressed as:
\(\begin{array}{c}{E_g} = {E_s}\\mg\left( {{y_2} - {y_1}} \right) = \frac{1}{2}k{\left( {{y_2} - {y_1}} \right)^2}\\mg = \frac{1}{2}k\left( {{y_2} - {y_1}} \right)\\\left( {{y_2} - {y_1}} \right) = \frac{{2mg}}{k}\end{array}\)
Here,\({E_g}\)is the gravitational potential energy and\({E_s}\)is the spring’s potential energy.
Substitute the values in the above expression.
\(\begin{array}{c}\left( {{y_2} - 15{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)} \right) = \frac{{2 \times {\rm{2}}{\rm{.5 kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2}}}{{83{\bf{ }}{\rm{N/m}}\left( {\frac{{1{\rm{ kg/}}{{\rm{s}}^2}}}{{1{\rm{ N/m}}}}} \right)}}\\{y_2} - 0.15{\rm{ m}} = 0.59{\rm{ m}}\\{y_2} = 0.74{\rm{ m}}\left( {\frac{{100{\rm{ cm}}}}{{1{\rm{ m}}}}} \right)\\{y_2} = 74\;{\rm{cm}}\end{array}\)
Thus, the end of the spring will line up with the ruler marks at 74 cm when the mass is at its lowest position.