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Simple Harmonic Motion question

2023 · 30 Jan · Shift 2 · Q61
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  5. /2023 · 30 Jan · Shift 2 · Q61

Simple Harmonic Motion question

2023 · 30 Jan · Shift 2 · Q61

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is 1 kg1 \mathrm{~kg}1 kg, the angular frequency is ω1\omega_{1}ω1​. When the mass block is 2 kg2 \mathrm{~kg}2 kg the angular frequency is ω2\omega_{2}ω2​. The ratio ω2/ω1\omega_{2} / \omega_{1}ω2​/ω1​ is JEE Main 2023 (Online) 30th January Evening Shift Physics - Simple Harmonic Motion Question 45 English
  1. A
    1/21 / \sqrt{2}1/2​
  2. B
    1/21 / 21/2
  3. C
    2
  4. D
    2\sqrt{2}2​
View written solutionFree

Correct answer: A

  1. Angular frequency of a mass-spring system

For a block of mass mmm attached to a spring of spring constant kkk on a frictionless surface, the angular frequency is

ω=km.\omega = \sqrt{\frac{k}{m}}.ω=mk​​.
  1. Case 1: mass =1 kg=1\,\text{kg}=1kg
ω1=k1=k.\omega_1 = \sqrt{\frac{k}{1}} = \sqrt{k}.ω1​=1k​​=k​.
  1. Case 2: mass =2 kg=2\,\text{kg}=2kg
ω2=k2.\omega_2 = \sqrt{\frac{k}{2}}.ω2​=2k​​.
  1. Find the ratio
ω2ω1=k/2k=12=12.\frac{\omega_2}{\omega_1} = \frac{\sqrt{k/2}}{\sqrt{k}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}.ω1​ω2​​=k​k/2​​=21​​=2​1​.
  1. Match with the options
ω2ω1=12\frac{\omega_2}{\omega_1} = \frac{1}{\sqrt{2}}ω1​ω2​​=2​1​

So, the correct option is A.

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