Question

The earring of a lady shown in the figure (12-E18) has a three 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/s in a circle of radius 2m. Find the time period of small oscillations of the ear-ring.

Answer 1

Given conditions,

length of the pendulum = 3 cm

= 0.03 m.

   

(a) When the lady is standing on the ground, the time period of small oscillations will be given by,

$T = 2\pi \sqrt{\frac{l}{g} }$

T = 2 × 22/7 × √(0.03/9.8)

∴ T = 0.347 seconds.

Hence, the time period is 0.347 seconds.

_______________________________

(b) When the lady is in a merry go round the velocity v = 4 m/s, r = 2 m.

Then there will be the centrifugal force on the bob of the ring.

Refer the attachment for this part of the Question.

∴ Net Force = √[(mv²/r)² + (mg)²]

∴ a = √[v⁴/r² + g²]

Now, using the formula,

$T = 2\pi \sqrt{\frac{l}{a _e_f_f} }$

$T = 2\pi \sqrt{\frac{l}{\frac{v^4}{r^2} + g^2 } }$

∴ T = 2 × 22/7 × √(0.03/a)

Let us first find a.

a =√(4⁴/2² + 9.8²)

∴ a = √(64 + 96.04)

∴ a = √160.04

∴ a = 12.65 m/s²

∴ Time period = T

T = 2 × 22/7 × √(0.03/12.65)

∴ T = 0.306 seconds.

Hence, the time period of girl's earing is is 0.306 seconds.

Hope it helps.