A pendulum of length $28 c m$ oscillates such that its string makes an angles of $30^{o}$ from the vertical when it is at one of the extreme positions. Find the ratio of the distance to displacement of the bob of the pendulum when it moves from one extreme position to the other.

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Solution

Angle made with vertical at extreme position $= 30 °$
When it goes from one extreme position to the other extreme position, angle traversed $= 2 \times 30 = 60 °$
Radius $= 28 c m$
Distance travelled $= \frac{60}{360} \times 2 Π r = \frac{176}{6} c m$
Displacement $= 2 \times r sin 30 = 2 \times 28 \times \frac{1}{2} = 28 c m$
Distance to displacement ratio $= \frac{176}{(6 \times 28)} = 22 : 21$

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A pendulum of length 28cm oscillates such that its string makes an angles of 30o from the vertical when it is at one of the extreme positions. Find the ratio of the distance to displacement of the bob of the pendulum when it moves from one extreme position to the other.

Angle made with vertical at extreme position=30°When it goes from one extreme position to the other extreme position, angle traversed =2×30=60°Radius =28cmDistance travelled =60360×2Πr=1766cmDisplacement =2×rsin30=2×28×12=28cm Distance to displacement ratio =176(6×28)=22:21

A pendulum of length $28 c m$ oscillates such that its string makes an angles of $30^{o}$ from the vertical when it is at one of the extreme positions. Find the ratio of the distance to displacement of the bob of the pendulum when it moves from one extreme position to the other.