A particle executes simple harmonic motion of amplitude A. (i) At what distance from the mean positio is its kinetic energy equal to its potential energy? (ii) At what points is its speed half the maximum speed?

Options

A. 0.81 A
B. 0.71 A
C. 0.41 A
D. 0.91 A

Answer

B

Explanation

Potential energy of a particle executing SHM at a displacement y from the mean position is
`PE = (1)/(2)ky^(2)" "...(i)`
Final kinetic energy of the particle is
`KE = (1)/(2)k(A^(2)-y^(2))" "...(ii)`
As PE and KE are equal, so equating them, we get
`(1)/(2)ky^(2)=(1)/(2)k(A^(2)-y^(2))`
or `A^(2)-y^(2)=y^(2)`
`rArr" "2y^(2)=A^(2)`
or `y=(A)/(sqrt(2))=0.07 A = 0.71 A`