ExamBro
ExamBro
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

  1. A \({K_2} = 2{K_1}\)
  2. B \({K_2} = \frac{{{K_1}}}{2}\)
  3. C \({K_2} = \frac{{{K_1}}}{4}\)
  4. D \({K_2} = {K_1}\)
Verified Solution

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. \(\mathrm{K}_{1}=\mathrm{mg} \ell(1-\cos \theta)\) \(\mathrm{K}_{2}=\mathrm{mg}(2 \ell)(1-\cos \theta)\) \(\mathrm{K}_{2}=2 \mathrm{K}_{1}\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app