\[ x= A sin\left( \omega t \right) \]

\[ \frac{A}{2}= A sin\left( \omega t \right) \]

\[ \frac{1}{2}= sin\left( \omega t \right) \]

\[ \frac{\Pi}{6}= \omega t =\frac{2\Pi}{T}t = \frac{2\Pi}{3}t \]

\[ t= \frac{1}{4}s \]

Equation of displacement x = A sin (ωt) so, acceleration a= -Aω sin(ωt) force will have same nature so,

At the mean position, tension is \(T = mg + \frac{mv^2}{L}\). Given \(T = 3mg ⇒ \frac{mv^2}{L} = 2mg ⇒ v^2 = 2gL\). Using conservation of energy, \(mgL(1 - \cos\theta) = \frac{1}{2}mv^2 = mgL ⇒\cos\theta = 0 ⇒ \theta = 90^\circ\).

Using standard SHM formula \(x = A sin(\omega t + phi)\), where \(A = 0.1 \text{m}\), \(\omega = 2\pi f = 2\pi(25) = 50\pi \text{rad/s}\), and \(\phi = \frac{\pi}{4}\). Substituting gives \(x = 0.1 sin \left(50\pi t + \frac{\pi}{4}\right)\).

The maximum acceleration in SHM is given by \(a_{\text{max}} = \omega^2 A = (2\pi f)^2 A\). Substituting \(f = 4\text{ Hz}\) and \(A = 0.15\text{ m}\), we find \(a_{\text{max}} = (8\pi)^2 \times 0.15 \approx 94.65 \text{ m s}^{-2}\).

The expression represents two perpendicular SHMs of the same frequency with a phase difference of \(\frac{\pi}{2}\). The resultant amplitude is \(A = \sqrt{\alpha^2 + \beta^2}\).

Separation occurs when the maximum downward acceleration of the piston equals \(g\). Thus, \(\omega^2 A = g \implies \left(\frac{2\pi}{T}\right)^2 A = g \implies 4\pi^2 A = 10 \implies 40 A = 10 \implies A = 0.25\text{ m}\).

Comparing the given equation with the standard SHM equation \( x = A\sin(\omega t) \), we get \( A = 2\text{ m} \) and \( \omega = \frac{\pi}{2}\text{ rad/s} \). The maximum speed is \( v_{\max} = A\omega = 2 \times \frac{\pi}{2} = \pi\text{ m/s} \).