Question

The ear-ring of a lady shown in figure (12−E18) 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 s⁻¹ in a circle of radius 2 m. Find the time period of small oscillations of the ear-ring.

Figure

Answer 1

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
$$\begin{gathered} \mathrm{T}=2\mathrm{\pi }\sqrt{\left(\frac{l}{g}\right)} \\ =2\mathrm{\pi }\sqrt{\left(\frac{0.03}{9.8}\right)} \\ =0.34\mathrm{second} \end{gathered}$$

(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 $\left(a\right)$ is given by,
$a=\frac{v^2}{r}=\frac{4^2}{2}=8\mathrm{m}/{\mathrm{s}}^2$
Resultant acceleration $\left(A\right)$ is given by,
$$\begin{gathered} A=\sqrt{\left(g^2+a^2\right)} \\ =\sqrt{\left(96.04+64\right)} \\ =12.65\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$
Time period, $T=2\mathrm{\pi }\sqrt{\left(\frac{l}{A}\right)}$
$$\begin{gathered} =2\mathrm{\pi }\sqrt{\left(\frac{0.03}{12.65}\right)} \\ =0.30\mathrm{second} \end{gathered}$$