A particle is performing linear SHM, along x-axis when the displacement becomes, `(A)/(sqrt2)`, (A is the amplitude), the relation that holds between kinetic energy and potential energy is ……..

Options

A. `KE=(PE)/(2)`
B. `KE=sqrt(2)PE`
C. `KE=PE`
D. `KE=(PE)/(sqrt2)`

Answer

C

Explanation

`KE=(1)/(2)momega^(2)(A^(2)-x^(2))," "PE=(1)/(2)momega^(2)x^(2)`
At `x=(A)/(sqrt2)`
`KE=(1)/(2)m omega^(2)(A^(2)-(A^(2))/(2))=(1)/(4) m omega^(2)A^(2)`
`PE=(1)/(2)momega^(2)(A^(2))/(2)=(1)/(4)momega^(2)A^(2)`
`therefore KE=PE`.