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

Explanation

If y is th edisplacement of particle from the mean position, then
P.E. of particle , `U=(1)/(2)ky^(2)`
and K.E. of particle, `K=(1)/(2)k (A^(2)-y^(2))`
where A is the amplitude of oscillation and k is a force constant.
As per question, `K=(U)/(2), so`
`(1)/(2)k(A^(2)-y^(2))=(1)/(2)((1)/(2)ky^(2))`
or `A^(2)-y^(2)=(y^(2))/(2) or 2A^(2)-2y^(2)=y^(2)`
or `y=ssqrt((2)/(3))A`
` :.` Distnace of location from extreme position
`=A-sqrt((2)/(3))A=A(sqrt(3)-sqrt(2))//sqrt(3)`
`=A(1.732-1514)//1.732=0.83A`
`=0.183` times the amplitude
(ii) `v_(max)=Aomega` and speed at displacement from mean position is
`v=omegasqrt(A^(2)-y^(2))`
As per question, `v=v_(max)//2`
As `omega sqrt(A^(2)-y^(2))=Aomega//2 or A^(2)-y^(2)=A^(2)//4`
or `y^(2)=A^(2)-A^(2)=3A^(2)//4`
or ` y =sqrt(3)A//2`
`:.` Distance from the extreme position is
`=A-y=A-(sqrt(3)A)/(2)=A[(2-sqrt(3))/(2)]`
`=A[(2-1.732)/(2)]=0.134A`
`=0.134` times the amplitude.