Question

a sphere of mass 0.1 kg is attached to an inextensible string of length 1.3 whose upper end is fix to the ceiling the sphere is made to describe a horizontal circle of radius 0.5m. calculate time period of recoup revolution and tension in the string​

Answer 1

Answer:

Time period is $2.2$ seconds

Tension in the string is $\frac{13}{12}$ N

Explanation:

This is a case of conical pendulum

Given length of the string $L=1.3\text{ m}$

Radius of the horizontal circle $r=0.5\text{ m}$

If the angle from the vertical is θ

Then

$\sin\theta=\frac{r}{L}$

$\implies \sin\theta=\frac{0.5}{1.3}$

$\sin\theta=\frac{5}{13}$

Therefore, we can also find the value of cosθ

$\cos\theta=\frac{12}{13}$

The time period of a conical pendulum is given by

$\boxed{t=2\pi\sqrt{\frac{L\cos\theta}{g}}}$

$\implies t=2\pi\sqrt{\frac{1.3\times (12/13)}{10}}$

$\implies t=2\times 3.14\sqrt{\frac{12}{100}}$

$\implies t=2\times 3.14\times \frac{2}{10}\times \sqrt{3}$

$\implies t=2\times 3.14\frac{\sqrt{3}}{5}$ seconds

$\implies t=2.17$ seconds

or, t ≈ 2.2 s

If the tension is T then

$T\cos\theta=mg$

$\implies T\times\frac{12}{13}=0.1\times 10$

$\implies T=\frac{13}{12}$ N