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Oscillations

JEE Main level, single correct, medium

159 of 194

Question 159 of 194

In figure (A), mass $2m$ is fixed on mass $m$ which is attached to two springs of spring constant $k$ . In figure (B), mass $m$ is attached to two spring of spring constant $k$ and $2k$ . If mass $m$ in (A) and (B) are displaced by distance $x$ horizontally and then released, then time period ${T}_{1}$ and ${T}_{2}$ corresponding to (A) and (B) respectively follow the relation.

Show answer and solution

Answer: (A) $\dfrac{{T}_{1}}{{T}_{2}}=\dfrac{3}{\sqrt{2}}$

The time period of spring block system is given by

T=2\pi \sqrt{\dfrac{m}{k}}

.
In both the cases springs are connected in parallel,
In case

1

{k}_{eq}=2k

and total mass is

3m

. Therefore,

{T}_{1}=2\pi \sqrt{\dfrac{\left(3m\right)}{{k}_{eq}}}=2\pi \sqrt{\dfrac{\left(3m\right)}{2k}}

In case

2:

{k}_{eq}=3k

and total mass is

m

. Therefore,

{T}_{2}=2\pi \sqrt{\dfrac{\left(m\right)}{3k}}

\Rightarrow \dfrac{{T}_{1}}{{T}_{2}}=\sqrt{\dfrac{9}{2}}=\dfrac{3}{\sqrt{2}}

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Oscillations (Applications of SHM), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.