Q.
A mass $m$ is suspended separately by two different springs in successive order then a time period is $t_{1}and\ t_{2}$ respectively. If $m$ is connected by both spring as shown in figure, then time period is $t_{0},\ \ the\ correct\ relation\ is\ $
A) $t_{0}^{2} = t_{1}^{2} + t_{2}^{2}$
B) $t_{0}^{- 2} = t_{1}^{- 2} + t_{2}^{- 2}$
C) $t_{0}^{- 1} = t_{1}^{- 1} + t_{2}^{- 1}$
D) $t_{0} = t_{1} + t_{2}$
Explanation
(b) $t_{1} = 2\pi\sqrt{\frac{m}{k_{1}}}\ \ \ or\ \ \ t_{1}^{2} = \frac{2\pi^{2}m}{k_{1}}$
or $k_{1} = \frac{4\pi^{2}m}{t_{1}^{2}}$
Similarly, $k_{2} = \frac{4\pi^{2}m}{t_{2}^{2}}$
And $\left( k_{1} + k_{2} \right) = \frac{4\pi^{2}m}{t_{0}^{2}}$
$\therefore\ \ \frac{4\pi^{2}m}{t_{0}^{2}} + \frac{4\pi^{2}m}{t_{1}^{2}} + \frac{4\pi^{2}m}{t_{2}^{2}}$
or $\frac{1}{t_{0}^{2}} = \frac{1}{t_{1}^{2}} + \frac{1}{t_{2}^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.