A particle executes linear S.H.M. along the principal axis of a convex lens of focal length \(8 \mathrm{~cm}\). The mean position of oscillation is at \(14 \mathrm{~cm}\) from the lens with amplitude \(1 \mathrm{~cm}\). The amplitude of oscillating image of the particle is nearly

\(\mathrm{f}=8 \mathrm{~cm}\), when the particle is at mean position, \(\mathrm{u}=-14 \mathrm{~cm}\)
\(
\begin{aligned}
& \frac{1}{\mathrm{~V}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{u}}=\frac{1}{8}-\frac{1}{14}=\frac{3}{56} \\
& \therefore \mathrm{V}=\frac{56}{3} \approx 19 \mathrm{~cm}
\end{aligned}
\)
When the particle is at one of the extreme positions its distance from the lens is \(14+1=15 \mathrm{~cm}\)
\(
\therefore \mathrm{u}=-15 \mathrm{~cm}
\)
Again, \(\frac{1}{\mathrm{~V}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{u}}=\frac{1}{8}-\frac{1}{15}=\frac{7}{120}\)
\(
\therefore \mathrm{v}=\frac{120}{7} \approx 17 \mathrm{~cm}
\)
Amplitude of the image \(=19-17=2 \mathrm{~cm}\)