A and B are two points on the path of a particle executing SHM in a straight line, at which its velocity is zero. Starting from a certain point X `(AXltBX)` the particle crosses this pint again at successive intervals of 2 seconds and 4 seconds with a speed of `5m//s`
Q. The ratio `(AX)/(BX)` is

Options

A. `(2sqrt(2)-1)/(2sqrt(2)+1)`
B. `(sqrt(3)-1)/(sqrt(3)+1)`
C. `(3sqrt(2)-2)/(3sqrt(2)+2)`
D. `(1)/(3)`

Answer

A

Explanation

The given SHM is represented in the figure.
Time period `T=2+4=6sec`
`OX=x_(0)`
and `OP=A`
`cos((theta)/(2))=(x_(0))/(A)`
`theta=2pi//6xx2=2pi//3`
so `cos((theta)/(2))=cos((pi)/(3))=(1)/(2)=(x_(0))/(A)`
or `x_(0)=A//2`
Now `v=omegasqrt(A^(2)-x^(2))`
or `5^(2)=omega^(2)A^(2)-omega^(2)x_(0)^(2)=3//4omega^(2)A^(2)`
or `omegaA=(10)/(sqrt(3))`
`impliesA=(10)/(sqrt(3)omega)=(10)/(sqrt(3))(6)/(2pi)=(10sqrt(3))/(pi)`
`(AX)/(BX)=(A-x_(0))/(A+x_(0))=(A//2)/(3A//2)=(1)/(3)`