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 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
Held on 11 Jan 2019 · Verified 6 Jul 2026.
K2=2K1
K2=2K1
K2=4K1
K2=K
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