Q135 / 180 · AIIMS 2010 · Physics
Let $T_1$ and $T_2$ be the time periods of springs $A$ and $B$ when mass $M$ is suspended from one end of each spring. If both springs are taken in series and the same mass $M$ is suspended from the series combination, the time period is $T$, then
- A$T_1+T_2+T_3$
- B$\frac{1}{T}=\frac{1}{T_1}+\frac{1}{T_2}$
- Correct answerC$T^2=T_1^2+T_2^2$
- D$\frac{1}{T^2}=\frac{1}{T_1^2}+\frac{1}{T^2}$
Solution
$T_1=2 \pi \sqrt{\frac{M}{k_1}}$ or $k_1=\frac{4 \pi^2 M}{T_1^2}$ and $k_2=\frac{4 \pi^2 M}{T_2^2}$
In series combination, $k_{\text {eff }}=\frac{k_1 k_2}{k_1+k_2}=\frac{4 \pi^2 M}{T_1^2+T_2^2}$
$$
\therefore T=2 \pi \sqrt{\frac{M}{k_{\mathrm{eff}}}}=\sqrt{T_1^2+T_2^2} \text {. }
$$
Hence, the correct answer is option: 'C'.