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Simple Harmonic Motion question

2022 · 24 Jun · Shift 2 · Q50
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  5. /2022 · 24 Jun · Shift 2 · Q50

Simple Harmonic Motion question

2022 · 24 Jun · Shift 2 · Q50

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
Two massless springs with spring constants 2 k and 9 k, carry 50 g and 100 g 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 :
  1. A
    1 : 2
  2. B
    3 : 2
  3. C
    3 : 1
  4. D
    2 : 3
View written solutionFree

Correct answer: B

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

  2. Here there are two vertical spring-mass systems. Gravity only shifts the equilibrium position; it does not affect the angular frequency.

  3. For the first system:

    • spring constant k1=2kk_1=2kk1​=2k
    • mass m1=50 g=0.05 kgm_1=50\,\text{g}=0.05\,\text{kg}m1​=50g=0.05kg

    So, ω1=2k0.05\omega_1=\sqrt{\frac{2k}{0.05}}ω1​=0.052k​​

  4. For the second system:

    • spring constant k2=9kk_2=9kk2​=9k
    • mass m2=100 g=0.1 kgm_2=100\,\text{g}=0.1\,\text{kg}m2​=100g=0.1kg

    So, ω2=9k0.1\omega_2=\sqrt{\frac{9k}{0.1}}ω2​=0.19k​​

  5. Since maximum velocities are equal, ω1A1=ω2A2\omega_1 A_1=\omega_2 A_2ω1​A1​=ω2​A2​ Therefore, A1A2=ω2ω1\frac{A_1}{A_2}=\frac{\omega_2}{\omega_1}A2​A1​​=ω1​ω2​​

  6. Now compute the ratio:

    =\sqrt{\frac{9k}{0.1}\cdot\frac{0.05}{2k}}$$ Cancel $k$: $$\frac{A_1}{A_2}=\sqrt{\frac{9\times 0.05}{0.1\times 2}}$$ $$=\sqrt{\frac{0.45}{0.2}}=\sqrt{2.25}=1.5$$ Hence, $$A_1:A_2=3:2$$

  7. So the correct option is: B (3:2)\boxed{\text{B }(3:2)}B (3:2)​

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