$\therefore v_{2}=v_{1} \pm 4=512 \mathrm{Hz} \pm 4=516 \mathrm{Hz}$ or $508 \mathrm{Hz}$

Increase in the tension of a piano string increases its frequency.

If $v_{2}=516 \mathrm{Hz}$, further increase in $v_{2},$ resulted in an increase in the beat frequency. But this is not given in the question.

If $v_{2}=508 \mathrm{Hz}$, further increase in $v_{2}$ resulted in decrease in the beat frequency. This is given in the question. When the beat frequency decreases to $2$ beats per second. Therefore, the frequency of the piano string before increasing the tension was $508\,\mathrm{Hz}$.

$\mathop {512\,Hz}\limits_{({v_1})} \,\,\xrightarrow{{ + \,\,4\,\,Hz}}\,\mathop {516\,Hz}\limits_{({v_2})} $

$\mathop {512\,Hz}\limits_{({v_1})} \,\,\xrightarrow{{ - \,\,4\,\,Hz}}\,\mathop {508\,Hz}\limits_{({v_2})} $