JEE Main 2023PhysicsOscillationsHardNumerical

JEE Main 2023 — Oscillations Question with Solution

JEE Main 2023 (31 Jan Shift 1)

Question

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

 

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Show full solutionCorrect answer: 20
Correct answer
20

Step-by-step explanation

Both the springs are connected in parallel, therefore

Keq=K+K=2K

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

T=2πmKeq=2πm2K

Given here, m=490 gm=0.49 kg and K=2 N m-1

So, T=2π0.492×2

=2π49400=2π720=7π10

Now, number of oscillation in 14π is

N=timeT=14π7π10=20

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About this question

This is a previous-year question from JEE Main 2023, covering the Oscillations chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.