Physics - SHM Question with Solution | TestHub

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, $E=P+k^{\prime }$ $\therefore \frac{1}{2}m{\omega }^2{A}^{\prime 2}=\frac{1}{2}m{\omega }^2{A}^2\cdot \frac{4}{9}+\frac{1}{2}m{\omega }^2{A}^2\cdot 5$ $\therefore A^{\prime 2}=\left(\frac{4}{9}+5\right)A^2$ $\therefore A^{\prime }=\frac{7}{3}A$