JEE Mains · Physics · STD 11 - 13. oscillations
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
- A \({K_2} = 2{K_1}\)
- B \({K_2} = \frac{{{K_1}}}{2}\)
- C \({K_2} = \frac{{{K_1}}}{4}\)
- D \({K_2} = {K_1}\)
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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