The ratio of kinetic energy to the potential energy of a particle executing SHM at a distance equal to half its amplitude, the distance being measured from its equilibrium position is

Question

Accepted Answer

Correct answer: C. (c) The kinetic energy of the particle executing SHM at a distance x from its equilibrium position is K= 12m ^2(A^2-x^2) and the potential energy is U= 12m ^2x^2 where A is the amplitude, is the angular frequency and M is the mass of the body. At x= A2 ⇒K= 12m ^2(A^2- ( A2 )^2) and U= 12m ^2 ( A2 )^2 Their corresponding ratio is KU= 12m ^2(A^2- ( A2 )^2) 12m ^2 ( A2 )^2= (A^2- A^24) A^24 = 34A^2 A^24= 31

$The ratio of kinetic energy to the potential energy of a particle executing SHM at a distance equal to half its amplitude, the distance being measured from its equilibrium position is$

Solution

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