Question

A bob is suspended from a ideal string of length l pulled a. Side through 60 degree to vertical whirled along a horizontal circle.Then it’s period of revolution to is

Answer 1

The time period is π/√5.

Explanation:

We need to resolve tension in the string in the vertical and the horizontal direction.

Let T be the tension in the string.

Therefore, Tcos(60°) = mg

Tcos(30°) = Centripetal force = m ω^2lsin (60°) = √3 / 2 mω^2

cos(60°) / cos(30°) = mg / √3 / 2 mω^2

ω^2 = 2g / √3  x cos(30°) / cos(60°) = 2g = 20

Time period T = 2π / 2√5 = π/√5

Thus the time period is  π/√5.

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