Question

Consider a simple pendulum of length 4 m. Its bob performs a circular motion in horizontal plane with its string making an angle 60° with the vertical. The Period of rotation of the bob is ____________.(Take g = 10 m/s²)

Options

• 2 s

• 2.8 s

• 1.98 s

• 2.4 s

Answer 1

Consider a simple pendulum of length 4 m. Its bob performs a circular motion in horizontal plane with its string making an angle 60° with the vertical. The Period of rotation of the bob is 2.8 s.

Explanation:

Using,

`"T sin"  theta = "m"omega^2 "r" = "m"omega^2  l "sin"  theta`     ....(i)

`"T cos"  theta = "mg"`     .....(ii)

`"from (i) and (ii),"  omega^2 = "g"/(l  "cos" theta)`

`therefore  omega = sqrt("g"/(l  "cos" theta))`

`therefore "Time period, T" = (2pi)/omega`

` = 2 pi sqrt ((l cos theta)/"g")`

`= 2 xx 3.14 xx sqrt ((4 xx "cos" 60°)/10)`

` = 2.8  "s"`