Due to some force `F_(1)` a body oscillates with period `4//5s` and due to other force `F_(2)` it oscillates with period `3//5s`. If both the forces acts simultaneously in same direction then new period is

Options

A. `(12)/(25)s`
B. `(35)/(24)s`
C. `(25)/(12)s`
D. `(24)/(25)s`

Answer

A

Explanation

In S.H.M. `F= -m omega^(2)x`
`F_(1)= - m omega_(1)^(2)x and F_(2) = - m omega_(2)^(2)x`
and `F= F_(1)+F_(2) therefore - m omega^(2)x= - m omega_(1)^(2)x -momega_(2)^(2)x`
`therefore omega^(2)=omega_(1)^(2)+omega_(2)^(2)`
`therefore ((2pi)/(T))^(2)=((2pi)/(T_(1)))^(2)+((2pi)/(T_(2)))^(2)`
`therefore (1)/(T^(2))=(1)/(T_(1)^(2))+(1)/(T_(2)^(2))=(T_(2)^(2)+T_(1)^(2))/(T_(1)^(2)T_(2)^(2))`
`therefore T=(T_(1)T_(2))/(sqrtT_(1)^(2)+T_(2)^(2))=((4)/(5)xx(3)/(5))/(sqrt(((4)/(5))^(2)+((3)/(5))^(2)))`
`=(12)/(25sqrt((16)/(25)+(9)/(25)))`
`therefore T=(12)/(25)s`