Question

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?

• A. $sin^{2} \beta$
• B. $cos^{2} \beta$
• C. $tan^{2} \beta$
• D. $cos \beta$

Answer 1

Answer: (B)

At A:


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


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