JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 12)
For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal ?
x = $${A \over 2}$$
x = $$\pm$$ A
x = $$\pm$$ $${A \over {\sqrt 2 }}$$
x = 0
Giải thích
KE = PE
$${1 \over 2}k({A^2} - {X^2}) = {1 \over 2}K{X^2}$$
$$ \Rightarrow $$ $${A^2} - {X^2} = {X^2}$$
$$ \Rightarrow $$ $$2{X^2} = {A^2}$$
$$ \Rightarrow $$ $${X^2} = {{{A^2}} \over {\sqrt 2 }}$$
$$ \Rightarrow $$ $$X = \pm {A \over {\sqrt 2 }}$$
$${1 \over 2}k({A^2} - {X^2}) = {1 \over 2}K{X^2}$$
$$ \Rightarrow $$ $${A^2} - {X^2} = {X^2}$$
$$ \Rightarrow $$ $$2{X^2} = {A^2}$$
$$ \Rightarrow $$ $${X^2} = {{{A^2}} \over {\sqrt 2 }}$$
$$ \Rightarrow $$ $$X = \pm {A \over {\sqrt 2 }}$$