A particle performs S.H.M. with amplitude 'A'. Its speed is tripled at the instant
when it is at a distance of \(\frac{2 \mathrm{~A}}{3}\) from the mean position. The new amplitude of the
motion is

Kinetic energy of a particle performing S.H.M. is given by
\(
k=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)
\)
When \(x=\frac{2}{3} A\),
\(
\mathrm{k}=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\frac{4}{9} \mathrm{~A}^{2}\right)=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2} \times \frac{5}{9}
\)
If the velocity is tripled, its kinetic energy will become 9 times. The new kinetic energy will be \(k^{\prime}=\frac{1}{2} m \omega^{2} A^{2} \times 5\)
The potential energy \(p=\frac{1}{2} m \omega^{2}\left(\frac{2}{3} A\right)^{2}=\frac{1}{2} m \omega^{2} A^{2} \cdot \frac{4}{9}\)
If \(A^{\prime}\) is the new amplitude then the total energy e is given by
\(E=\frac{1}{2} m \omega^{2} A^{\prime 2}\)
Also, \(E=P+k^{\prime}\)
\(\therefore \frac{1}{2} m \omega^{2} A^{\prime 2}=\frac{1}{2} m \omega^{2} A^{2} \cdot \frac{4}{9}+\frac{1}{2} m \omega^{2} A^{2} \cdot 5\)
\(\therefore A^{\prime 2}=\left(\frac{4}{9}+5\right) A^{2}\)
\(\therefore A^{\prime}=\frac{7}{3} A\)