# 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

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.

#### Solution

Given,
Length of the long, light suspension wire, l = 3 cm = 0.03 m
â€‹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]

#### APPEARS IN

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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