Q. Two springs are joined and attached to a mass of 16 kg. This system is then suspended vertically from a rigid support. The spring constant of the two springs are $k_{1}\ and\ k_{2}$ respectively. The period of vertical oscillations of the system will be
A) $\frac{1}{8\pi}\sqrt{k_{1} + k_{2}}$
B) $8\pi\sqrt{\frac{k_{1} + k_{2}}{k_{1}k_{2}}}$
C) $\frac{\pi}{2}\sqrt{k_{1} - k_{2}}$
D) $\frac{\pi}{2}\sqrt{\frac{k_{1}}{k_{2}}}$
Explanation
(b) The two springs are in series. Therefore, the time period is
$T = 2\pi\sqrt{\frac{m}{k}}$
$= \ 2\pi\sqrt{m\left( \frac{k_{1} + k_{2}}{k_{1}k_{2}} \right)}$
As $m = 16$ kg;
$T = 8\pi\sqrt{\frac{k_{1} + k_{2}}{k_{1}k_{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.