For a particle executing simple harmonic motion, the kinetic energy of the particle at a distance of 4 cm from the mean position is 13 rd of the maximum kinetic energy. The amplitude of the motion is

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Correct answer: A. (a) Kinetic energy of the particle at 4 cm = 13 maximum kinetic energy ∴ 12K(A^2-x^2)= 13 12 KA^2 ⇒A^2-x^2= A^23 ⇒x^2= 2A^23 ⇒ (4 )^2= 2 A^23 ⇒A^2= 3 162 Amplitude, A= 24=2 6 cm

$For a particle executing simple harmonic motion, the kinetic energy of the particle at a distance of 4 cm from the mean position is \frac{1}{3} rd of the maximum kinetic energy. The amplitude of the motion is$

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