Q170 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Physics
A body is suspended from the two light springs separately. The periods of vertical oscillations are T and 3T respectively. The same body is suspended from two springs connected first in parallel and then in series. The periods of oscillation in both cases, respectively, are
- A$\dfrac{2T}{\sqrt{5}},\ \sqrt{5}T$
- Correct answerB$\dfrac{3T}{\sqrt{10}},\ \sqrt{10}T$
- C$\dfrac{4T}{\sqrt{15}},\ \sqrt{15}T$
- D$\sqrt{5}T,\ \dfrac{2T}{\sqrt{5}}$
Solution
The time period of a spring-mass system is given by $T = 2\pi \sqrt{\dfrac{m}{k}}$, which implies $k = \dfrac{4\pi^2 m}{T^2}$.
For the first spring:
$k_1 = \dfrac{4\pi^2 m}{T^2}$
For the second spring:
$k_2 = \dfrac{4\pi^2 m}{(3T)^2} = \dfrac{4\pi^2 m}{9T^2}$
When the springs are connected in parallel, the equivalent spring constant is $k_p = k_1 + k_2$.
$\dfrac{1}{T_p^2} = \dfrac{k_p}{4\pi^2 m} = \dfrac{k_1 + k_2}{4\pi^2 m} = \dfrac{1}{T_1^2} + \dfrac{1}{T_2^2}$
$\dfrac{1}{T_p^2} = \dfrac{1}{T^2} + \dfrac{1}{9T^2} = \dfrac{10}{9T^2}$
$T_p = \dfrac{3T}{\sqrt{10}}$
When the springs are connected in series, the equivalent spring constant is given by $\dfrac{1}{k_s} = \dfrac{1}{k_1} + \dfrac{1}{k_2}$.
$T_s^2 = \dfrac{4\pi^2 m}{k_s} = 4\pi^2 m \left(\dfrac{1}{k_1} + \dfrac{1}{k_2}\right) = T_1^2 + T_2^2$
$T_s^2 = T^2 + (3T)^2 = 10T^2$
$T_s = \sqrt{10}T$
The periods of oscillation in parallel and series are $\dfrac{3T}{\sqrt{10}}$ and $\sqrt{10}T$ respectively.
Answer: $\dfrac{3T}{\sqrt{10}},\ \sqrt{10}T$
For the first spring:
$k_1 = \dfrac{4\pi^2 m}{T^2}$
For the second spring:
$k_2 = \dfrac{4\pi^2 m}{(3T)^2} = \dfrac{4\pi^2 m}{9T^2}$
When the springs are connected in parallel, the equivalent spring constant is $k_p = k_1 + k_2$.
$\dfrac{1}{T_p^2} = \dfrac{k_p}{4\pi^2 m} = \dfrac{k_1 + k_2}{4\pi^2 m} = \dfrac{1}{T_1^2} + \dfrac{1}{T_2^2}$
$\dfrac{1}{T_p^2} = \dfrac{1}{T^2} + \dfrac{1}{9T^2} = \dfrac{10}{9T^2}$
$T_p = \dfrac{3T}{\sqrt{10}}$
When the springs are connected in series, the equivalent spring constant is given by $\dfrac{1}{k_s} = \dfrac{1}{k_1} + \dfrac{1}{k_2}$.
$T_s^2 = \dfrac{4\pi^2 m}{k_s} = 4\pi^2 m \left(\dfrac{1}{k_1} + \dfrac{1}{k_2}\right) = T_1^2 + T_2^2$
$T_s^2 = T^2 + (3T)^2 = 10T^2$
$T_s = \sqrt{10}T$
The periods of oscillation in parallel and series are $\dfrac{3T}{\sqrt{10}}$ and $\sqrt{10}T$ respectively.
Answer: $\dfrac{3T}{\sqrt{10}},\ \sqrt{10}T$
Hence, the correct answer is option: 'B'.