Angle at Which Bob Speed is Half Initial Velocity in Vertical Circular Motion

To find the angle $\theta$ at which the speed of the bob is half of its initial speed at $A$, we analyze the motion using the principle of conservation of mechanical energy.

Step 1: Determine the velocity at $A$

For the bob suspended by a string to just reach the top point $B$ of the vertical circular path, the tension in the string at point $B$ must be zero ($T_B = 0$).

At point $B$, gravity provides the required centripetal acceleration:
$\frac{M v_B^2}{L} = M g \implies v_B^2 = gL$

Using the conservation of mechanical energy between the lowest point $A$ (taken as the reference level for potential energy, $U_A = 0$) and the highest point $B$ (at a height $2L$ above $A$):
$\frac{1}{2} M V^2 = \frac{1}{2} M v_B^2 + Mg(2L)$
$V^2 = v_B^2 + 4gL = gL + 4gL = 5gL$
$V = \sqrt{5gL}$

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Step 2: Relate speed $v$ at angle $\theta$ to the initial speed $V$

Let $\theta$ be the angle made by the string with the downward vertical line passing through $A$. The height $h$ of the bob above point $A$ at angle $\theta$ is:
$h = L(1 - \cos\theta)$

Applying the conservation of mechanical energy between point $A$ and the point at angle $\theta$:
$\frac{1}{2} M V^2 = \frac{1}{2} M v^2 + M g L (1 - \cos\theta)$
$v^2 = V^2 - 2gL(1 - \cos\theta)$

Substituting $V^2 = 5gL$:
$v^2 = 5gL - 2gL(1 - \cos\theta) = gL(3 + 2\cos\theta)$

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Step 3: Find $\theta$ when $v = \frac{V}{2}$

Given that the speed of the bob is half of its velocity at $A$, $v = \frac{V}{2}$:
$v^2 = \left(\frac{V}{2}\right)^2 = \frac{V^2}{4} = \frac{5gL}{4}$

Equating this to the energy expression for $v^2$:
$gL(3 + 2\cos\theta) = \frac{5gL}{4}$
$3 + 2\cos\theta = \frac{5}{4}$
$2\cos\theta = \frac{5}{4} - 3 = -\frac{7}{4}$
$\cos\theta = -\frac{7}{8} = -0.875$

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Step 4: Determine the interval for $\theta$

We know the standard values of cosine in the second quadrant:
$\cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}} \approx -0.707$
$\cos(\pi) = -1$

Since $-1 < -0.875 < -0.707$, it follows that:
$\cos(\pi) < \cos\theta < \cos\left(\frac{3\pi}{4}\right)$

Since $\cos\theta$ is a strictly decreasing function in the range $[0, \pi]$, reversing the inequalities gives:
$\frac{3\pi}{4} < \theta < \pi$

Thus, the correct option is D.