In the figure given below. a block of mass $M=490g$ placed on a frictionless table is connected with two springs having same spring constant ($K=2N{m}^{-1}$). If the block is horizontally displaced through $X$ $m$ then the number of complete oscillations it will make in $14\pi$ seconds will be ______. ![JEE Main 2023 Physics, Oscillations — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Oscillations/648b5a6f417cc3fb48d6828c/question_1__q_648b5a6f417cc3fb48d6828c__cdn-question-pool.getmarks.app__def3d0ba-9e6a-4709-805f-bdce7f881a7a-9896057__b5ea1fcc34_final_ppt_sync.png)

![JEE Main 2023 Physics, Oscillations — solution figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Oscillations/648b5a6f417cc3fb48d6828c/solution_1__q_648b5a6f417cc3fb48d6828c__cdn-question-pool.getmarks.app__b1fba8f3-a42b-4208-93ad-459d2ef403a8-9896057_1__3a4fe5f84c_final_ppt_sync.png) Both the springs are connected in parallel, therefore ${K}_{eq}=K+K=2K$ Now, time period of spring-block system is given by, $T=2\pi \sqrt{\frac{m}{{K}_{eq}}} =2\pi \sqrt{\frac{m}{2K}}$ Given here, $m=490\mathrm{gm}$$=0.49\mathrm{kg}$ and $K=2N{m}^{-1}$ So, $T=2\pi \sqrt{\frac{0.49}{2\times 2}}$ $=2\pi \sqrt{\frac{49}{400}}=2\pi \frac{7}{20}=\frac{7\pi }{10}$ Now, number of oscillation in $14\pi$ is $N=\frac{\mathrm{time}}{T}=\frac{14\pi }{\frac{7\pi }{10}}=20$

In the figure given below. a block of mass… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Waves & Oscillations

2023integerhard

In the figure given below. a block of mass $M=490g$ placed on a frictionless table is connected with two springs having same spring constant ( $K=2N{m}^{-1}$ ). If the block is horizontally displaced through $X$ $m$ then the number of complete oscillations it will make in $14\pi$ seconds will be ______.

JEE Main 2023 Physics, Oscillations — question figure

Official previous-year question

Held on 31 Jan 2023 · Verified 6 Jul 2026.

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Solution

JEE Main 2023 Physics, Oscillations — solution figure

Both the springs are connected in parallel, therefore

${K}_{eq}=K+K=2K$

Now, time period of spring-block system is given by,

$T=2\pi \sqrt{\frac{m}{{K}_{eq}}} =2\pi \sqrt{\frac{m}{2K}}$

Given here, $m=490\mathrm{gm}$ $=0.49\mathrm{kg}$ and $K=2N{m}^{-1}$

So, $T=2\pi \sqrt{\frac{0.49}{2\times 2}}$

$=2\pi \sqrt{\frac{49}{400}}=2\pi \frac{7}{20}=\frac{7\pi }{10}$

Now, number of oscillation in $14\pi$ is

$N=\frac{\mathrm{time}}{T}=\frac{14\pi }{\frac{7\pi }{10}}=20$

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