Question

Consider the situation of the previous problem. (a) Find the tension in the string in equilibrium. (b) Suppose the ball is slightly pushed aside and released. Find the time period of the small oscillations.

Answer 1

(a) The tension is the string in equilibrium is 0.20N.

(b) 0.45 seconds of time is consumed when the ball is slightly pushed aside and released.

Explanation:

(a) Once we resolve the tension in the string along horizontal and vertical directions, we get:

T cos 60°=mg

= (10 $\times$ 10-3 $\times$ 10)/(1/2) = 0.20 N

(b) When the same figure is straightened, the resultant ‘R’ in the pendulum, the acceleration is straightened.

$T=2 \times \pi \sqrt{\frac{\ell}{g}}=2 \pi \sqrt{\frac{\ell}{\left[g^{2}+\left(\frac{\sigma q}{2 \varepsilon_{0} m}\right)^{2}\right]^{1 / 2}}}$

$=2 \times 3.1416 \times \sqrt{\frac{10 \times 10^{-2}}{20}}=0.45 \mathrm{sec}$