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  4. A ball is held at rest at position A by two light strings. The horizontal string is cut and the ball starts swinging like a pendulum. Point B is the farthest to the right the ball goes as it swings back and forth. What is the ratio of the tension in the supporting string in position B to its value at A before the horizontal string was cut? <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-yvvfobdzfis9.png />

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Q. A ball is held at rest at position $A$ by two light strings. The horizontal string is cut and the ball starts swinging like a pendulum. Point $B$ is the farthest to the right the ball goes as it swings back and forth. What is the ratio of the tension in the supporting string in position $B$ to its value at $A$ before the horizontal string was cut?

Question

NTA AbhyasNTA Abhyas 2020Laws of Motion

Solution:

Solution
At A:
Solution

$T^{′}cos \beta =mg$
$T′=\frac{m g}{cos \beta }$
At: B

Solution
$T^{′ ′}- \, mg \, cos \beta = \, \frac{m v^{2}}{1}$
at $B \, ; \, v=0$
$T^{′ ′}=mg \, cos \beta $
$\frac{T ′ ′}{T ′}= \, \frac{m g \, cos \beta }{m g / cos ⁡ \beta }$
$\frac{T e n s i o n \, a t \, B}{T e n s i o n \, a t \, A}=\frac{T ′ ′}{T ′}=cos^{2}\beta $