What should be the displacement of a particle, executing SHM, from its position of equilibrium so that the velocity of the particle is half of its maximum velocity?

Explanation

If the angular frequency of a particle executing SHM is `omega` and the amplitude of motion is A, then maximum velocity of the particle = `omegaA`
For a displacement x, velocity of the particle,
`v=omegasqrt(A^(2)-x^(2))`
According to the question,
`omegasqrt(A^(2)-x^(2))=1/2omegaA`
or, `A/2=sqrt(A^(2)-x^(2))or,A^(2)/4=A^(2)-x^(2)`
or, `x^(2)=A^(2)-A^(2)/4or,x^(2)=3/4A^(2)`
or, `x=pmsqrt3/2A`, this is the required displacement.