Step 1: Energy Conservation in SHM
The total mechanical energy in simple harmonic motion is:
\[
E = \frac{1}{2} m \omega^2 A^2
\]
At displacement \( x \), the energy distribution is:
\[
E_x = \frac{1}{2} m \omega^2 x^2 + \frac{1}{2} m \omega^2 (A^2 - x^2)
\]
Since the velocity is doubled instantly, kinetic energy increases by a factor of 4:
\[
\text{New total energy} = 4 \times \text{Original kinetic energy} + \text{Potential energy}
\]
Equating energies:
\[
\frac{1}{2} m \omega^2 A_{\text{new}}^2 = \frac{1}{2} m \omega^2 (4A^2 - 3x^2)
\]
Solving for \( A_{\text{new}} \):
\[
A_{\text{new}} = \sqrt{4A^2 - 3x^2}
\]
Conclusion
Thus, the correct answer is:
\[
\sqrt{4A^2 - 3x^2}
\]