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

When the mass is connected to the two springs individually,
`v_(1) = (1)/(2pi) sqrt((k_(1))/(m)) ….(i)`
`v_(2) = (1)/(2pi) sqrt((K_(2))/(m)) ….(ii)`

Now the block is connected with two springs considered as parallel.
Here equivalent spring constant,
`K_(eq) = k_(1) +k_(2)`
Time period of oscillation of the spring block-system is
`T = 2pi sqrt((m)/(k_(eq))) = 2pi sqrt((m)/(k_(1)+k_(2)))`
Hence frequency,
`v = (1)/(T) = (1)/(2pi) xx sqrt((k_(1)+k_(2))/(m))`
`v = (1)/(2pi) [(k_(1))/(m)+(k_(2))/(m)]^(1//2)`
From eq. (i) `(k_(1))/(m) = 4pi^(2) v_(1)^(2)` and from.
(ii), `(k_(2))/(m) = 4piv_(2)^(2)`
`rArr v = (1)/(2pi) [(4pi^(2)v_(1)^(2))/(1)+(4pi^(2)v_(2)^(2))/(1)]^(1//2)`
`= (2pi)/(2pi) [v_(1)^(2) +v_(2)^(2)]^(1//2)`
`rArr v = sqrt(v_(1)^(2)+v_(2)^(2))`