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Simple Harmonic Motion question

2019 · 11 Jan · Shift 2 · Q49
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  5. /2019 · 11 Jan · Shift 2 · Q49

Simple Harmonic Motion question

2019 · 11 Jan · Shift 2 · Q49

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2. Then :
  1. A
    K2{K_2}K2​=K12{{{K_1}} \over 2}2K1​​
  2. B
    K2 = 2K1
  3. C
    K2 = K1
  4. D
    K2 = K14{{{K_1}} \over 4}4K1​​
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