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JEE Main 202224 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
24 June
Shift
Evening Shift
Subject
Physics
Chapter
Oscillations

JEE Main 2022 · 24 June · Evening Shift · Physics · Oscillations

Two massless springs with spring constants 2k and 9k , carry 50g...

Two massless springs with spring constants 2k2k and 9k9k, carry 50g50g and 100g100g masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be :Two massless springs with spring constants 2k and 9k , carry 50g and 100g masses at their free ends. These two masses

Options

  1. A3:23: 2
  2. B1:21: 2
  3. C3:13: 1
  4. D2:32: 3

Correct answer

A. 3:23: 2

Step-by-step solution

Maximum velocity is given by {v}_\max = \omegaA =kmA= \sqrt{\dfrac{k}{m}}A

Two massless springs with spring constants 2k and 9k , carry 50g and 100g masses at their free ends. These two masses

Here, k1m1A1=k2m2A2\sqrt{\dfrac{{k}_{1}}{{m}_{1}}}{A}_{1} = \sqrt{\dfrac{{k}_{2}}{{m}_{2}}}{A}_{2}

⇒2k50×10−3A1=9k100×10−3A2⇒ \sqrt{\dfrac{2k}{50 \times {10}^{- 3}}}{A}_{1} = \sqrt{\dfrac{9k}{100 \times {10}^{- 3}}}{A}_{2}

⇒2A1=3A2⇒ 2{A}_{1} = 3{A}_{2}

Thus, ratio of amplitude is A1A2=32=3:2\dfrac{{A}_{1}}{{A}_{2}} = \dfrac{3}{2} = 3: 2.

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