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Simple Harmonic Motion question

2021 · 17 Mar · Shift 2 · Q53
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  5. /2021 · 17 Mar · Shift 2 · Q53

Simple Harmonic Motion question

2021 · 17 Mar · Shift 2 · Q53

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
Two particles A and B of equal masses are suspended from two massless springs of spring constants K1 and K2 respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of A and B is
  1. A
    K1K2{{{K_1}} \over {{K_2}}}K2​K1​​
  2. B
    K1K2\sqrt {{{{K_1}} \over {{K_2}}}}K2​K1​​​
  3. C
    K2K1{{{K_2}} \over {{K_1}}}K1​K2​​
  4. D
    K2K1\sqrt {{{{K_2}} \over {{K_1}}}}K1​K2​​​
View written solutionFree

Correct answer: D

  1. For a mass-spring system executing SHM, the maximum speed is vmax⁡=Aωv_{\max}=A\omegavmax​=Aω where AAA is amplitude and ω\omegaω is angular frequency.

  2. For a spring of spring constant KKK with mass mmm, the angular frequency is ω=Km\omega=\sqrt{\frac{K}{m}}ω=mK​​

  3. Therefore, for particle AAA: vmax⁡,A=AAK1mv_{\max,A}=A_A\sqrt{\frac{K_1}{m}}vmax,A​=AA​mK1​​​

    And for particle BBB: vmax⁡,B=ABK2mv_{\max,B}=A_B\sqrt{\frac{K_2}{m}}vmax,B​=AB​mK2​​​

  4. Given that the maximum velocities are equal, AAK1m=ABK2mA_A\sqrt{\frac{K_1}{m}}=A_B\sqrt{\frac{K_2}{m}}AA​mK1​​​=AB​mK2​​​

  5. Since the masses are equal, mmm cancels out: AAK1=ABK2A_A\sqrt{K_1}=A_B\sqrt{K_2}AA​K1​​=AB​K2​​

  6. Hence, AAAB=K2K1\frac{A_A}{A_B}=\sqrt{\frac{K_2}{K_1}}AB​AA​​=K1​K2​​​

  7. Comparing with the options, this is Option D.

Final Answer: AAAB=K2K1\boxed{\frac{A_A}{A_B}=\sqrt{\frac{K_2}{K_1}}}AB​AA​​=K1​K2​​​​

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