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Question
The ear-ring of a lady shown in figure has a 3 cm long light suspension wire. (a) Find the time period of small oscillations if the lady is standing on the ground. (b) The lady now sits in a merry-go-round moving at 4 m/s1 in a circle of radius 2 m. Find the time period of small oscillations of the ear-ring.
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Solution
Given,
Length of the long, light suspension wire, l = 3 cm = 0.03 m
Acceleration due to gravity, `g = 9.8 "ms"^(- 2)`
(a)Time Period \[\left( T \right)\] is given by ,
\[T = 2\pi\sqrt{\left( \frac{l}{g} \right)}\]
\[ = 2\pi\sqrt{\left( \frac{0 . 03}{9 . 8} \right)}\]
\[ = 0 . 34 \text { second}\]
(b) Velocity of merry-go-round, v = 4 `"ms"^(- 1)`
Radius of circle, r = 2 m
As the lady sits on the merry-go-round, her earring experiences centripetal acceleration.
Centripetal acceleration (a) is given by,
\[a = \frac{v^2}{r} = \frac{4^2}{2} = 8 m/ s^2\]
Resultant acceleration (A) is given by ,
\[A = \sqrt{\left( g^2 + a^2 \right)}\]
\[ = \sqrt{\left( 96 . 04 + 64 \right)}\]
\[ = 12 . 65 m/ s^2\]
Time Period,
\[T = 2\pi\sqrt{\left( \frac{l}{A} \right)}\]
\[= 2\pi\sqrt{\left( \frac{0 . 03}{12 . 65} \right)}\]
\[ = 0 . 30 \text { second }\]
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