Answer:

Answer: (a) \(\lambda=\dfrac{10}{3}mg\). (b) \(v=\dfrac{4}{3}\sqrt{\dfrac{ag}{3}}\).

(a) In the equilibrium position, each half of the string has horizontal component \(2a\) and vertical component \(\dfrac32a\). Hence each half has length

The natural length of each half is \(2a\), so the extension of each half is \(\dfrac12a\). Therefore the tension in each half is

If \(\phi\) is the angle that either half-string makes with the vertical, then

Resolving vertically for equilibrium,

Thus

so

\(\lambda=\dfrac{10}{3}mg\).

(b) At the release position, each half of the string has length

So the extension of each half is

The total elastic potential energy at release is therefore

When \(P\) passes through \(M\), the string is unstretched, so the elastic potential energy is zero. The particle has risen by \(\dfrac83a\), so its gravitational potential energy has increased by

Using conservation of energy,

Hence

so

\(v=\dfrac{4}{3}\sqrt{\dfrac{ag}{3}}\).