And that is Mechanics

To find the tension in the string at position A, identify all forces acting on the object. There is a tension force ${T_1}$ acting at 60 degree angle to the horizontal. A second tension force ${T_2}$ acts directly to the left of the object, and weight ${mg}$ acts directly downward. Because the object is in static equilibrium, the sum of all forces in the x and y directions must equal zero. Therefore, summing forces in y direction results in:

${T_1}{sin(60)} - mg = 0$

Solving for ${T_1}$ ,

${T_1} = \frac{mg}{sin(60)} = \frac{2mg}{\sqrt{3}}$

Now for Point B, two forces act on the object: a tension force ${T_3}$ again at 60 degree angle and weight ${mg}$ downward. Because Point B is the farthest the pendulum can go, the object is at maximum gravitational potential and has no kinetic energy. Therefore, radial acceleration, which depends on velocity, is zero as well. Summing forces along the direction of the radial acceleration vector (pointing toward the center of the pendulum) gives:

${T_3} - {mg cos(30)} = 0$

Solving for ${T_3}$ ,

${T_3} = {mg cos(30)} N = \frac{\sqrt(3)mg}{2}$

$\frac{T_3}{T_1} = \frac{\frac{\sqrt(3)mg}{2}}{\frac{2mg}{\sqrt{3}}} = (\frac{\sqrt(3)mg}{2})(\frac{\sqrt(3)}{2mg}) = 0.75$

At position B, the ball is momentary stop. There is no centripetal force and the forces involve are the weight of the ball ( $mg$ ) and the tension in the pendulum string ( $T_B$ ). By the Newton's third law of motion, we have $T_B = mg \cos 30^\circ = \dfrac {\sqrt 3mg}2$ .

At position A, there are three forces acting: the weight of the ball ( $W=mg$ ), the tension in the pendulum string ( $T_A$ ) and the tension in the horizontal string ( $T_s$ ). By the Newton's third law of motion, we have $\vec{T_A} = \vec W + \vec {T_s}$ (vector addition) $\implies T_A = \dfrac {mg}{\cos 30^\circ} = \dfrac {2mg}{\sqrt 3}$ .

Therefore, $\dfrac {T_B}{T_A} = \dfrac {\frac {\sqrt 3mg}2}{\frac {2mg}{\sqrt 3}} = \dfrac 34 = \boxed{0.75}$ .