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

Options

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

Answer

B

Explanation

Since `v_(1) = (1)/(2pi) sqrt((k_(1))/(m))`
and `v_(2) = (1)/(2pi) sqrt((k_(2))/(m))`
Here the spring are connected in parallel
So equivalent spring constant `k = k_(1) + k_(2)`
The frequency (v) of the given oscillating system
`v = (1)/(2pi) sqrt((k)/(m)) = (1)/(2pi) sqrt((k_(1) k_(2))/(m))`
` = (1)/(2pi) sqrt((k_(1)/(m) + (k_(2))/(m))`
or ` v = (1)/(2pi) sqrt(4pi^(2)v_(1)^(2) + 4pi^(2)v_(2)^(2))`
` = sqrt(v_(1)^(2) + v_(2)^(2))`