K=21mω2x2 ⇒Kmax=21mω2A2 A =Lθ ω=Lg ⇒K=21m⋅Lg⋅L2θ2 =21mgLθ2 ∴K2K1=2LL=21⇒K2=2K1 ∴ K2K1=2 LL=21⇒K2=2 K1
A pendulum is executing simple harmonic motion and its maximum kinetic energy is K_1. If the length of the pendulum is doubled and it performs simple…
JEE Main 2019 — Physics Waves & Oscillations
2019mcqmedium
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
Official previous-year question
Held on 11 Jan 2019 · Verified 6 Jul 2026.
Options
- A
K2=2K1
- B
K2=2K1
- C
K2=4K1
- D
K2=K
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