Answer:

Answer: The maximum extension of the spring is \(\dfrac{3a}{10}\).

When the spring has length \(\dfrac{3a}{5}\), its compression is

\(a-\dfrac{3a}{5}=\dfrac{2a}{5}\).

Let the maximum extension after release be \(x\). At this instant the particle is momentarily at rest, and the spring has length \(a+x\).

The elastic potential energy in a spring of modulus \(\lambda\), natural length \(a\), and extension or compression \(e\), is

\(\dfrac{\lambda e^2}{2a}\).

Use conservation of mechanical energy between the release position and the position of maximum extension:

Divide by \(mg\):

This simplifies to

Factorising or using the quadratic formula gives

So

\(x=\dfrac{3a}{10}\) or \(x=-\dfrac{2a}{5}\).

The negative value corresponds to the initial compression, so the maximum extension is

\(\dfrac{3a}{10}\).