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

Consider the diagram, two springs can be considered as parallel.
Hence, `k_(eq)="Equivalent spring constant"`
`k_(1)+k_(2)`
Time period of oscillatin of the spring block-system
`T=2pisqrt((m)/k_(eq))=2pisqrt((m)/(k_(1)+k_(2))`
`rArrv=(1)/(T)=(1)/(2pi)xxsqrt((k_(1)+k_(2))/(m))`
Equivalent oscillation frequency.
When the mass is connected to the two springs individually

`v_(1)=(1)/(2pi)sqrt(k_(1)/(m))`
`v_(2)=(1)/(2pi)sqrt(k_(2)/(m))`
From Eqs. (i),(ii) and (iii),
`v=(1)/(2pi)=[k_(1)/(m)+k_(2)/(m)]^(1//2)` [from Eq. (i)]
`(1)/(2pi)[(4pi^(2)v_(1)^(2))/(1)+(4pi^(2)v_(2)^(2))/(1)]`
`[:'"fromEq."(ii)k_(1)/(m)=4pi^(2)v_(1)^(2)"and fromEq."(iii),(k_(2))/(m)=4pi^(2)v_(2)^(2)]`
`=(2pi)/(2pi)[v_(1)^(2)+v_(2)^(2)]^(1//2)rArrv=sqrt(v_(1)^(2)+v_(2)^(2))]`
Not Do not confuse with parallel and series combinations of springs.