When a mass m is connected individually to two springs `S_(1) and S_2`, the oscillation frequencies are `upsilon_(1) and upsilon_(2)`. If the same mass is attached to the two springs as shown in figure, the oscillation frequency would be

Options

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

Answer

B

Explanation

Let `k_(1) and k_(2)` be the spring constant fo springs `S_(1) and S_(2)` respectively. Then
`upsilon_(1)=(1)/(2pi)sqrt((k_(1))/(m))` . . . (i)
and `upsilon_(2)=(1)/(2pi)sqrt((k_(2))/(m))` . .. (ii)
If k is effective spring constant of two springs `S_(1) and S_(2)`. then, `k=k_(1)+k_(2)` (`because` springs are connected in parallel) if `upsilon` is the effectie frequency of oscilltion when the mass m is attached to the springs `S_(1) and S_(2)` as shown in figure, then
`upsilon=(1)/(2pi)sqrt((k)/(m))=(1)/(2pi)sqrt((k_(1)+k_(2))/(m))=(1)/(2pi)sqrt((k_(1))/(m)+(k_(2))/(m))`
`upsilon=(1)/(2pi)sqrt(4pi^(2)upsilon_(1)^(2)+4pi^(2)upsilon_(2)^(2))` (Using (i) and (ii))
`=sqrt(upsilon_(1)^(2)+upsilon_(2)^(2))`.