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Simple Harmonic Motion question

2003 · Shift 0 · Q159
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  5. /2003 · Shift 0 · Q159

Simple Harmonic Motion question

2003 · Shift 0 · Q159

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
Two particles AAA and BBB of equal masses are suspended from two massless springs of spring of spring constant k1{k_1}k1​ and k2{k_2}k2​, respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of AAA and BBB is
  1. A
    k1k2\sqrt {{{{k_1}} \over {{k_2}}}}k2​k1​​​
  2. B
    k2k1{{{{k_2}} \over {{k_1}}}}k1​k2​​
  3. C
    k2k1\sqrt {{{{k_2}} \over {{k_1}}}}k1​k2​​​
  4. D
    k1k2{{{{k_1}} \over {{k_2}}}}k2​k1​​
View written solutionFree

Correct answer: C

  1. For a mass–spring system executing SHM, the maximum speed is vmax⁡=ωAv_{\max}=\omega Avmax​=ωA where AAA is amplitude and ω=km.\omega=\sqrt{\frac{k}{m}}.ω=mk​​.

  2. For particle AAA: vmax⁡,A=AAk1mv_{\max,A}=A_A\sqrt{\frac{k_1}{m}}vmax,A​=AA​mk1​​​

    For particle BBB: vmax⁡,B=ABk2mv_{\max,B}=A_B\sqrt{\frac{k_2}{m}}vmax,B​=AB​mk2​​​

  3. Given that the two maximum velocities are equal, AAk1m=ABk2m.A_A\sqrt{\frac{k_1}{m}}=A_B\sqrt{\frac{k_2}{m}}.AA​mk1​​​=AB​mk2​​​.

    Since masses are equal, mmm cancels: AAk1=ABk2.A_A\sqrt{k_1}=A_B\sqrt{k_2}.AA​k1​​=AB​k2​​.

  4. Therefore, AAAB=k2k1.\frac{A_A}{A_B}=\sqrt{\frac{k_2}{k_1}}.AB​AA​​=k1​k2​​​.

  5. Comparing with the options, this matches: C: k2k1.\boxed{\text{C: } \sqrt{\frac{k_2}{k_1}}}.C: k1​k2​​​​.

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