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

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" /> (A) sin2 β  (B) cos2 β  (C) tan2 β  (D) cos β

Tardigrade Admin · Accepted Answer

At A:

Tâ²cos β =mg Tâ²=(m g/cos β ) At: B Solution

Tâ² â²- mg cos β = (m v2/1) at B ; v=0 Tâ² â²=mg cos β (T â² â²/T â²)= (m g cos β /m g / cos â¡ β ) (T e n s i o n a t B/T e n s i o n a t A)=(T â² â²/T â²)=cos2β

$s i n^{2} β$

$co s^{2} β$

$t a n^{2} β$

$cos β$

At A:

$T^{'} cos β = m g$
$T' = \frac{m g}{cos β}$
At: B

$T^{''} - m g cos β = \frac{m v ^{2}}{1}$
at $B; v = 0$
$T^{''} = m g cos β$
$\frac{T ''}{T '} = \frac{m g cos β}{m g / cos β}$
$\frac{T e n s i o n a t B}{T e n s i o n a t A} = \frac{T ''}{T '} = co s^{2} β$

sin2β

cos2β

tan2β

cosβ

At A: T′cosβ=mg T′=cosβmg​ At: B T′′−mgcosβ=1mv2​ at B;v=0 T′′=mgcosβ T′T′′​=mg/cos⁡βmgcosβ​ TensionatATensionatB​=T′T′′​=cos2β

$s i n^{2} β$

$co s^{2} β$

$t a n^{2} β$

$cos β$

At A:

$T^{'} cos β = m g$
$T' = \frac{m g}{cos β}$
At: B

$T^{''} - m g cos β = \frac{m v ^{2}}{1}$
at $B; v = 0$
$T^{''} = m g cos β$
$\frac{T ''}{T '} = \frac{m g cos β}{m g / cos β}$
$\frac{T e n s i o n a t B}{T e n s i o n a t A} = \frac{T ''}{T '} = co s^{2} β$