A particle performs S.H.M. with amplitude 'A'. Its speed is tripled at the instant when it is at a distance…
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
$\frac{5 \mathrm{~A}}{3}$
$\frac{7 \mathrm{~A}}{3}$
$\frac{2 \mathrm{~A}}{3}$
$\frac{\mathrm{A}}{3}$
Solution
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, $E=P+k^{\prime}$
$\therefore \frac{1}{2} m \omega^{2} A^{\prime 2}=\frac{1}{2} m \omega^{2} A^{2} \cdot \frac{4}{9}+\frac{1}{2} m \omega^{2} A^{2} \cdot 5$
$\therefore A^{\prime 2}=\left(\frac{4}{9}+5\right) A^{2}$
$\therefore A^{\prime}=\frac{7}{3} A$