A body is performing simple harmonic with an amplitude of \(10 \,cm\). The velocity of the body was tripled by air Jet when it is at \(5 \,cm\) from its mean position. The new amplitude of vibration is \(\sqrt{ x } \,cm\). The value of \(x\) is________

\(\therefore\) 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)

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

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

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

\(\therefore x =700\)