Answer:

(a) \(\mu_{\min}=\frac13\).

(b) The extension is \(\frac16a\).

The natural length is \(a\), and initially \(AP=\frac43a\). Thus the initial extension is

\(\frac43a-a=\frac13a.\)

The modulus of elasticity is \(2mg\), so the elastic potential energy initially is

\(\frac{2mg}{2a}\left(\frac13a\right)^2 =\frac{mga}{9}.\)

(a) For the string never to become slack, the particle must not pass the natural length point. The greatest distance it can move before reaching the natural length is \(\frac13a\). The work done against friction over this distance is

\(\mu mg\left(\frac13a\right).\)

For the minimum \(\mu\), this work equals the initial elastic potential energy:

\(\frac{mga}{9}=\mu mg\left(\frac13a\right).\)

Therefore

\(\mu=\frac13.\)

(b) Now \(\mu=\frac12\). Let \(x\) be the extension when the particle next comes to rest.

The particle moves a distance \(\frac13a-x\), so the work done against friction is

\(\frac12mg\left(\frac13a-x\right).\)

Using energy,

\(\frac{2mg}{2a}\left(\frac13a\right)^2 =\frac{2mg}{2a}x^2+\frac12mg\left(\frac13a-x\right).\)

Cancel \(mg\):

\(\frac{a}{9}=\frac{x^2}{a}+\frac{a}{6}-\frac{x}{2}.\)

Multiplying by \(18a\),

\(2a^2=18x^2+3a^2-9ax.\)

So

\(18x^2-9ax+a^2=0.\)

Factorising,

\((6x-a)(3x-a)=0.\)

The root \(x=\frac13a\) is the initial extension, so the extension when the particle next comes to rest is

\(x=\frac16a.\)