\(\therefore\) Total Energy \(=\frac{1}{2} KA ^{2}\)

By energy conservation we can final \(v\) at \(x =5\)

\(\frac{1}{2} K (10)^{2}=\frac{1}{2} K (5)^{2}+\frac{1}{2} mv ^{2}\)

\(V =\sqrt{\frac{75 K }{ m }}\)

Now, velocity is tripled through external mean so the amplitude of \(SHM\) will charge and so the total energy, (but potential) energy at this moment will remain same)

\(\therefore \frac{1}{2} K (5)^{2}+\frac{1}{2} m \left(3 \sqrt{\frac{75 K }{ m }}\right)^{2}=\frac{1}{2} KA ^{2}\)

\(\Rightarrow 25 K +675 K = KA ^{2}\)

\(\therefore A =\sqrt{700}\)

\(\therefore x =700\)