Q 11-13-114JEE MainJEE Main 2021 (26 Feb, Shift 1)Easy
If two similar springs each of spring constant $K_1$ are joined in series, the new spring constant and time period would be changed by a factor:
Answer: (D) $\frac{1}{2},\ \sqrt{2}$
For two equal springs in series, $\dfrac{1}{K} = \dfrac{1}{K_1}+\dfrac{1}{K_1} \Rightarrow K = \dfrac{K_1}{2}$.
Time period $T = 2\pi\sqrt{m/K}$, so $T \propto K^{-1/2}$. Halving $K$ multiplies $T$ by $\sqrt{2}$.
Factors: $\frac{1}{2}$ and $\sqrt{2}$.
Solution by Sreeraj P, M.Sc Physics