Why Does Cutting the String Change the Tension Ratio in a Pendulum?

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Cutting the string attached to the wall changes the forces acting on the ball, leading to different tension ratios in the supporting string. Initially, the ball is in equilibrium, and the tension is calculated as T1 = mg/cos(theta). Once the wall string is cut, the ball swings as a pendulum, and the tension during this motion is T2 = mg cos(theta). The ratio of the tensions before and after cutting the string is T1/T2 = 1/cos^2(theta), indicating that the tension increases significantly as the pendulum swings. This change is due to the different forces acting on the ball in equilibrium versus during its motion.
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SO.

a ball is suspended by a string from the ceiling, and held at angle theta to the vertical by another string that is attached to the wall nearby... the string attched to the wall is cut and the ball swings like a pendulum, at the farthest it is theta from the vertical on both sides.

so. the book asks what is the ratio of the tension in the rope holding the rope to the ceiling before the other rope(attached to the wall) was cut, to when the ball is swinging and is at its farthest from the vertical(forming angle theta to it)

i thought it should be one to one, because the tension*sin(theta) would be supporting the ball in both cases, but it turns out the answer is cos^2(theta)

any ideas why?
 
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Check your vector diagram.
 
\begin{align*}\\<br /> Case\ 1 :\\<br /> Particle\ is\ in\ equilibrium. \\<br /> \sum \vec{F}=0\\<br /> T_1cos\theta + (-mg) = 0\\<br /> T_1=\frac{mg}{cos\theta}\\<br /> Case\ 2:\\<br /> Centripedal\ acceleration :\\<br /> T_2+(-mgcos\theta)=0\ since\ v=0\\<br /> T_2=mgcos\theta\\<br /> \frac{T_1}{T_2}=\frac{1}{cos^2\theta}\\<br /> Why\ the\ reciprocal\ ?\\<br /> \end{align}
 

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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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