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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2019
  • Medium
  • MCQ

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 harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is \(K_2\) then

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Answer & Solution

Correct Answer

(A) \({K_2} = 2{K_1}\)

Step-by-step Solution

Detailed explanation

Maximum kinetic energy at lowest point \(B\) is given by

\(\mathrm{K}=\mathrm{mg} \ell(1-\cos \theta)\)

where \(\theta=\) angular amp.

\(\mathrm{K}_{1}=\mathrm{mg} \ell(1-\cos \theta)\)

\(\mathrm{K}_{2}=\mathrm{mg}(2 \ell)(1-\cos \theta)\)

\(\mathrm{K}_{2}=2 \mathrm{K}_{1}\)

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