91Ó°ÊÓ

If a particle of mass (m) experience a nonzero net force (F), its acceleration (a) is related to the net force by Newton’s second law.

$\underset{\mathbf{}}{\mathbf{∑}}\mathbf{F}\mathbf{=}\mathbf{ma}$

Consider the following free-body diagram of ball A.

At point A the sting is in equilibrium so according to Newton’s second law the force is each direction is equal to zero.

$$\begin{gathered} ∑F_y=0 \\ {T}_{A}cos\mathrm{β}=mg \end{gathered}$$

${\mathrm{T}}_{\mathrm{A}}=\frac{\mathrm{mg}}{\mathrm{³¦´Ç²õβ}}$ ….. (1)

After the rope is cut the ball will move in simple harmonic motion and one type of simple harmonic motion is the uniform circular motion so focus in radial direction were the tension force equalizes the weight, so at point B the ball is not in equilibrium but the acceleration is zero in radial direction so the component of weight is in equilibrium with tension at (b).

Consider the following free-body diagram of ball B.

Apply the Newton’s second law the force.

$∑F_R=0$

$T_B=mgcos\mathrm{β}$ ….. (2)

Divide equations (1) and (2) as below.

$$\begin{gathered} \frac{{\mathrm{T}}_{\mathrm{B}}}{{\mathrm{T}}_{\mathrm{A}}}=\frac{\mathrm{mg³¦´Ç²õβ}}{{\displaystyle \frac{\mathrm{mg}}{\mathrm{³¦´Ç²õβ}}}} \\ ={\cos }^2\mathrm{β} \end{gathered}$$

Therefore, the required ratio is $\frac{{\mathrm{T}}_{\mathrm{B}}}{{\mathrm{T}}_{\mathrm{A}}}={\cos }^2\mathrm{β}$.