Answer
$v=\omega\sqrt{A^2-x^2}$
Work Step by Step
According to given condition
$K.E+U=E$
$\implies \frac{1}{2}mv^2+\frac{1}{2}Kx^2=\frac{1}{2}KA^2$
This can be rearranged as:
$mv^2=K(A^2-x^2)$
This simplifies to:
$v=\sqrt{\frac{K}{m}(A^2-x^2)}$
As $\omega=\sqrt{\frac{K}{m}}$
$\implies v=\omega\sqrt{A^2-x^2}$
