A particle is executing a linear S.H.M. `v_(1) " and " v_(2)` are as speeds at dis"tan"ces `x_(1) " and " x_(2)` from the equilibrium position. What is its ampiltude of oscillation ?

Options

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

Answer

A

Explanation

`v_(1)=omega sqrt(A^(2)-x_(1)^(2)), v_(2)=omega sqrt(A^(2)-x_(2)^(2))`
`therefore (v_(1))/(v_(2))=(sqrt(A^(2)-x_(1)^(2)))/(sqrt(A^(2)-x_(2)^(2))) therefore(v_(1)^(2))/(v_(2)^(2))=(A^(2)-x_(1)^(2))/(A^(2)-x_(2)^(2))`
`therefore v_(1)^(2)(A^(2)-x_(2)^(2))=v_(2)^(2)(A^(2)-x_(1)^(2))`
`therefore v_(1)^(2)A^(2)-v_(1)^(2)x_(2)^(2)=v_(2)^(2)A^(2)-v_(2)^(2)x_(1)^(2)`
`therefore A^(2)(v_(1)^(2)-v_(2)^(2))=v_(1)^(2)x_(2)^(2)-v_(2)^(2)x_(1)^(2)`
`therefore A=sqrt((v_(1)^(2)x_(2)^(2)-v_(2)^(2)x_(1)^(2))/(V_(1)^(2)-v_(2)^(2)))`