When a mass m is connected individually to two spring `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 shwon in figure., the oscillation frequecy 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

Let `k_(1)` and `k_(2)` be the spring constants of springs `S_(1)` and `s_(2)` respectively. Then
`v_(1)=(1)/(2pi)sqrt((k_(1))/(m)) and v_(2)=(1)/(2pi)sqrt((k_(2))/(m))`
If k is the spring constant of two springs `S_(1)` and `S_(2)` as connected in this problem, the `k=k_(1)+k_(2)`
If v is the effective frequency of oscillationof the mass attached in this problem, then
`v=(1)/(2pi)sqrt((k)/(m))=(1)/(2pi)sqrt((k_(1)+k_(2))/(m))=(1)/(2pi)sqrt((k_(1))/(m)+(k_(2))/(m))=(1)/(2pi)sqrt(4pi^(2)v_(1)^(2)+4pi^(2)v_(2)^(2))=sqrt(v_(1)^(2)+v_(2)^(2))`