Question

A ring of mass $m$ can freely slide on a smooth vertical rod. The ring is symmetrically attached with two springs (as shown in the figure) each of spring constant $k$ . Each spring makes an angle $\theta$ with the horizontal. What will be the time period of the ring if it is slightly displaced vertically?

• A. $2\pi \sqrt{\frac{m}{2k}}$
• B. $2\pi \sqrt{\frac{m}{k}}$
• C. $2\pi \sqrt{\frac{2m}{ksi{n}^2\theta }}$
• D. $2\pi \sqrt{\frac{m}{2ksi{n}^2\theta }}$

Answer 1

Answer: (D)

If ring is pulled below the current position by displacement $y$ then elongation in the spring will be $x$ (for small displacement).
$y=xsin\theta$ ...(1)
If ring is displaced along $y$ ,
$ma=2kysin\theta$
$\Rightarrow ma=2kxsi{n}^2\theta$
$\Rightarrow a=\left(\frac{2k{\left(sin\right)}^2\theta }{m}\right)x$
$\Rightarrow {\left(\omega \right)}^2=\left(\frac{2k{\left(sin\right)}^2\theta }{m}\right)$
Therefore, $T=2\pi \sqrt{\frac{m}{2ksi{n}^2\theta }}$