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Simple Harmonic Motion question

2019 · 11 Jan · Shift 2 · Q61
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Simple Harmonic Motion question

2019 · 11 Jan · Shift 2 · Q61

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be :
  1. A
    32{{\sqrt 3 } \over 2}23​​ s
  2. B
    32{3 \over 2}23​ s
  3. C
    23{2 \over {\sqrt 3 }}3​2​ s
  4. D
    232\sqrt 323​ s
View written solutionFree

Correct answer: D

  1. For a simple pendulum, T=2πlgT = 2\pi\sqrt{\frac{l}{g}}T=2πgl​​ For the same pendulum, lll is constant, so T∝1gT \propto \frac{1}{\sqrt{g}}T∝g​1​

  2. Acceleration due to gravity on a planet is g=GMR2g = \frac{GM}{R^2}g=R2GM​

  3. Given for the new planet:

    • Mass: Mp=3MEM_p = 3M_EMp​=3ME​
    • Diameter: Dp=3DED_p = 3D_EDp​=3DE​

    Since radius is proportional to diameter, Rp=3RER_p = 3R_ERp​=3RE​

  4. Therefore, gp=G(3ME)(3RE)2=3GME9RE2=13GMERE2=gE3g_p = \frac{G(3M_E)}{(3R_E)^2} = \frac{3GM_E}{9R_E^2} = \frac{1}{3}\frac{GM_E}{R_E^2} = \frac{g_E}{3}gp​=(3RE​)2G(3ME​)​=9RE2​3GME​​=31​RE2​GME​​=3gE​​

  5. Now use the relation between periods: TpTE=gEgp=gEgE/3=3\frac{T_p}{T_E} = \sqrt{\frac{g_E}{g_p}} = \sqrt{\frac{g_E}{g_E/3}} = \sqrt{3}TE​Tp​​=gp​gE​​​=gE​/3gE​​​=3​

  6. Given TE=2 sT_E = 2\,\text{s}TE​=2s, Tp=23 sT_p = 2\sqrt{3}\,\text{s}Tp​=23​s

  7. Comparing with options:

    • A: 32\frac{\sqrt{3}}{2}23​​ s
    • B: 32\frac{3}{2}23​ s
    • C: 23\frac{2}{\sqrt{3}}3​2​ s
    • D: 232\sqrt{3}23​ s

    Hence, the correct option is D.

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