MHT CET · Physics · Oscillations
All the springs in fig. (a), (b) and (c) are identical, each having force constant K. Mass attached to each system is ' \(m\) '. If \(T_a, T_b\) and \(T_c\) are the time periods of oscillations of the three systems respectively, then

- A \(\mathrm{T}_{\mathrm{a}}=\sqrt{2} \mathrm{~T}_{\mathrm{b}}\)
- B \(T_a=\frac{T_c}{\sqrt{2}}\)
- C \(\mathrm{T}_{\mathrm{b}}=2 \mathrm{~T}_{\mathrm{a}}\)
- D \(T_b=2 T_c\)
Answer & Solution
Correct Answer
(D) \(T_b=2 T_c\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
T_a & =2 \pi \sqrt{\frac{m}{K}} \\
T_b & =2 \pi \sqrt{\frac{m}{\frac{K}{2}}} \\
\therefore \quad T_b & =2 \pi \sqrt{\frac{2 m}{K}}=\sqrt{2} T_a \Rightarrow T_a=\frac{T_b}{\sqrt{2}}...(i) \\
T_c & =2 \pi \sqrt{\frac{m}{2 K}}=\frac{T_a}{\sqrt{2}} \Rightarrow T_a=\sqrt{2} T_c
...(ii)\end{aligned}\)
From (i) and (ii),
\(\frac{\mathrm{T}_{\mathrm{b}}}{\sqrt{2}}=\sqrt{2} \mathrm{~T}_{\mathrm{c}} \Rightarrow \mathrm{~T}_{\mathrm{b}}=2 \mathrm{~T}_{\mathrm{c}}\)
T_a & =2 \pi \sqrt{\frac{m}{K}} \\
T_b & =2 \pi \sqrt{\frac{m}{\frac{K}{2}}} \\
\therefore \quad T_b & =2 \pi \sqrt{\frac{2 m}{K}}=\sqrt{2} T_a \Rightarrow T_a=\frac{T_b}{\sqrt{2}}...(i) \\
T_c & =2 \pi \sqrt{\frac{m}{2 K}}=\frac{T_a}{\sqrt{2}} \Rightarrow T_a=\sqrt{2} T_c
...(ii)\end{aligned}\)
From (i) and (ii),
\(\frac{\mathrm{T}_{\mathrm{b}}}{\sqrt{2}}=\sqrt{2} \mathrm{~T}_{\mathrm{c}} \Rightarrow \mathrm{~T}_{\mathrm{b}}=2 \mathrm{~T}_{\mathrm{c}}\)
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