Velocity in SHM is given by $v = \omega\sqrt{A^2 - x^2}$. Here $A = a$, $x = a/2$, and $\omega = \frac{2\pi}{T}$. So, $v = \frac{2\pi}{T}\sqrt{a^2 - \frac{a^2}{4}} = \frac{2\pi}{T} \times \frac{a\sqrt{3}}{2} = \frac{\pi a \sqrt{3}}{T}$.

Velocity $v = \frac{dx}{dt} = a\omega\cos(\omega t + \pi/6)$. Maximum velocity is $a\omega$. Given $v = \frac{a\omega}{2}$, so $\cos(\omega t + \pi/6) = 1/2$. This gives $\omega t + \pi/6 = \pi/3 \Rightarrow \omega t = \pi/6$. Since $\omega = 2\pi/T$, we get $\frac{2\pi t}{T} = \frac{\pi}{6} \Rightarrow t = \frac{T}{12}$.

Maximum acceleration in SHM is given by $a_{\text{max}} = \omega^2 A$. Since amplitude $A$ is the same, $a_{\text{max}} \propto \omega^2$. The ratio is $a_1 / a_2 = (\omega_1 / \omega_2)^2 = (100 / 1000)^2 = (1/10)^2 = 1 : 100 = 1 : 10^2$.

Equation from equilibrium is $x = a\sin(\omega t)$. For $x = a/2$, $a/2 = a\sin(\omega t) \Rightarrow \sin(\omega t) = 1/2$. This gives $\omega t = \pi/6$. Substituting $\omega = 2\pi/T$, we get $\frac{2\pi}{T}t = \pi/6 \Rightarrow t = T/12$.

In SHM, velocity leads displacement by a phase of $\pi/2$, and acceleration leads velocity by a phase of $\pi/2$ (or $0.5\pi$). Thus, the phase difference between velocity and acceleration is $0.5\pi$.

Restoring force on the block is $F = -(\rho A g)y$. Acceleration $a = F/m = -(\frac{\rho A g}{m})y$. This is SHM with $\omega^2 = \frac{\rho A g}{m}$. Time period $T = 2\pi\sqrt{\frac{m}{\rho A g}}$. Therefore, $T \propto \frac{1}{\sqrt{A}}$.

In simple harmonic motion, speed is maximum at the mean position.\nAt this mean position, the displacement is zero, causing the restoring force and acceleration to be zero.

Substitute $\delta$: $x = a\cos(\omega t + \alpha + \pi/2) = -a\sin(\omega t + \alpha)$.\nCombining with $y = a\cos(\omega t + \alpha)$ yields $x^2 + y^2 = a^2$, a circle.\nThe direction of motion traces anti-clockwise as time progresses.

The time taken to travel from the extreme position to the mean position is $T/4$.\nThus, $T/4 = 6 \implies T = 24 \text{ s}$.\nFrequency $f = 1/T = 1/24 \approx 0.04 \text{ Hz}$.

Superposition of two mutually perpendicular SHMs of equal amplitude and a phase difference of $\pi/2$ produces a circular trajectory.\n$x = A\sin(\omega t)$ and $y = A\sin(\omega t + \pi/2) = A\cos(\omega t)$ gives $x^2 + y^2 = A^2$.