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. - Physics

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Sum

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, 

\[T = 2\pi\sqrt{\left( \frac{l}{A} \right)}\]

\[= 2\pi\sqrt{\left( \frac{0 . 03}{12 . 65} \right)}\] 

\[         = 0 . 30  \text { second }\]

  Is there an error in this question or solution?
Chapter 12: Simple Harmonics Motion - Exercise [Page 255]

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HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 12 Simple Harmonics Motion
Exercise | Q 47 | Page 255

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