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. 0.72 s
B. 0.64 s
C. 0.48 s
D. 0.36 s

Answer

C

Explanation

Under the influence of one force `F_(1) = m omega_(1)^(2)y` and under the action force, `F_(2) = momega(2)^(2)y`
Under the action of both the forces `F = F_(1) + F_(2)`
`rArr momega^(2)y = momega_(1)^(2)y + momega_(2)^(2)y`
`rArr omega^(2) = omega_(1)^(2) + omega_(2)^(2) rArr ((2pi)/T)^(2) = ((2pi)/T_(1))^(2) + ((2pi)/T_(2))^(2)`
`rArr T = sqrt((T_(1)^(2)T_(2)^(2))/(T_(1)^(2) + T_(2)^(2))) = sqrt((4/5)^(2) (3/5)^(2))/((4/5)^(2) + (3/5)^(2)) = 0.48 s`