When the displacement in SHM is one-half the amplitude A what fraction of total energy is kinetic energy and potential energy?

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Answer

$When the displacement in SHM is one-half the amplitude A what fraction of total energy is kinetic energy and potential energy?$

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Hint: In physics, SHM stands for Simple Harmonic Motion which is defined as the motion of a body between two fixed points and it vibrates from one point to another and came back to its original point and this motion repeats several times with specific time period is called Simple Harmonic Motion.

Complete step by step answer:

(b) In SHM, the total energy of a body is defined as, $E = \dfrac{1}{2}k{x_m}^2$ where $k$ is called force constant. Potential energy is given as $U = \dfrac{1}{2}k\dfrac{{{x_m}^2}}{4}$ since it’s given that amplitude is half of maximum amplitude. $U = k\dfrac{{{x_m}^2}}{8}$Now, taking the ratios of Potential energy and total energy we get,$\dfrac{U}{E} = \dfrac{2}{8}$$\therefore U = 0.25E$

Hence the fraction of potential energy to total energy is $0.25$.

(a) Now, as we know, $E = T + U$ the, we can write it as:$\dfrac{T}{E} = 1 - \dfrac{U}{E}$From part (b) we know that $\dfrac{U}{E} = \dfrac{2}{8}$ put this value in above equation we get,$\dfrac{T}{E} = 1 - 0.25$$\therefore T = 0.75E$

Hence, the fraction of Kinetic energy to total energy is $0.75$.

(c) As, we know total energy of system is $E = \dfrac{1}{4}k{x_m}^2$ and let $x$ be the displacement at which potential energy became twice of total energy and we write it as: $U = \dfrac{1}{2}k{x^2}$ Now, both energies are equal in this displacement hence, so$\dfrac{{{x_m}^2}}{4} = \dfrac{{{x^2}}}{2}$$\therefore x = \dfrac{{{x_m}}}{{\sqrt 2 }}$

Hence, the displacement at which total energy became half of potential energy is $x = \dfrac{{{x_m}}}{{\sqrt 2 }}$.

Note:Remember, Using the law of conservation of energy we know, at any point of the motion in simple harmonic motion, the total energy at a point is always equals to the sum of kinetic energy and potential energy of the body at that point and it’s written as $E = T + U$.