Question

A particle executes SHM of amplitude A. At what distance from the mean position is its KE equal to its PE ?
A. 0.71 A
B. 0.61 A
C. 0.65 A
D. 0.8 A

Answer 1

Correct Answer - A
As, `KE = (1)/(2)m omega^(2)(A^(2)-x^(2))`
and `PE=(1)/(2)m omega^(2)x^(2)`
As, given KE = PE
`therefore" "(1)/(2)m omega^(2)(A^(2)-x^(2))=(1)/(2)m omega^(2)x^(2)or A^(2)-x^(2)=x^(2)`
or `x^(2)=A^(2)//2 rArr x = A//sqrt(2)=0.71 A`