A particle executing harmonic motion is having velocities `v_1` and `v_2` at distances is `x_1` and `x_2` from the equilibrium position. The amplitude of the motion is

Options

A. `sqrt((v_(1)^(2)x_(2)^(2)-v_(2)^(2)x_(1)^(2))/(v_(1)^(2)+v_(2)^(2)))`
B. `sqrt((v_(1)^(2)x_(2)^(2)-v_(2)^(2)x_(2)^(2))/(v_(1)^(2)+v_(2)^(2)))`
C. `sqrt((v_(1)^(2)x_(2)^(2)-v_(2)^(2)x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))`
D. `sqrt((v_(1)^(2)x_(2)^(2)-v_(2)^(2)x_(1)^(2))/(v_(1)^(2)+v_(2)^(2)))`

Answer

B

Explanation

As `v=omegasqrt(A^(2)-x^(2))`
`therefore v_(1)=omegasqrt(A^(2)-x_(1)^(2))orv_(1)^(2)=omega^(2)(A^(2)-x_(1)^(2))" "...(i)`
and `v_(2)=omegasqrt(A^(2)-x_(2)^(2))orv_(2)^(2)=omega^(2)(A^(2)-x_(2)^(2))" "...(ii)`
Dividing (i) by (ii), we get `(v_(1)^(2))/(v_(2)^(2))=((A^(2)-x_(1)^(2)))/((A^(2)-x_(2)^(2)))`
or `v_(1)^(2)A^(2)-v_(1)^(2)x_(2)^(2)=v_(2)^(2)A^(2)-v_(2)^(2)x_(1)^(2)`
or `A^(2)(v_(1)^(2)-v_(2)^(2))=v_(1)^(2)x_(2)^(2)-v_(2)^(2)x_(1)^(2)orA=sqrt((v_(1)^(2)x_(2)^(2)-v_(2)^(2)x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))`