What is the ratio of KE at the mean position to the PE when the displacement is half of the amplitude?

Answer

Text Solution

`(4)/(1)``(2)/(3)``(4)/(3)``(1)/(2)`

Answer : A

Solution : `K.E.=(1)/(2)m omega^(2)(a^(2)-y^(2))` and `P.E.=(1)/(2)m omega^(2)y^(2)` <br> At mean position, `y=0:.K.E.=(1)/(2)m omega^(2)a^(2)` <br> At, `y=(a)/(2),P.E.=(1)/(2)m omega^(2)(a^(2))/(4)=(1)/(8)m omega^(2)a^(2)` <br> `:.` Ratio `=((1//2)m omega^(2)a^(2))/((1//8)m omega^(2)a^(2))=("K.E. at mean position")/("PE at "a//2)=(4)/(1)`

Text Solution

`(4)/(1)``(2)/(3)``(4)/(3)``(1)/(2)`.

Answer : A

Solution : Kinetic energy `E_(K)=(1)/(2)m omega^(2)(r^(2)-y^(2))` <br> At mean position `y=0,:. E_(K)=(1)/(2)m omega^(2)r^(2)` <br> Potential energy `E_(P)=(1)/(2)m omega^(2)y^(2)` <br> At ` y=( r )/(2),E_(P)=(1)/(2)m omega^(2)(r^(2))/(4)`. <br> `:." "(E_(K))/(E_(P))=(4)/(1)` <br> Correct choice is (a).