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

The block is considered with two springs considered as parallel. Hence equivalent spring constant `k= k_(1) +k_(2)` where `k_(1)" and "k_(2)` are spring constant of spring `S_(1)" and "S_(2)` respectively.

Time period of oscillation of the spring block system is
`T= 2pi sqrt((m)/(k))= 2pi sqrt((m)/((k_1)+(k_2)))`
where k = equivalent spring constant
Frequency of the system,
`v= (1)/(2pi) sqrt((k_(1)+k_(2))/(m))" ""........"(1)`
Frequency of oscillation of a body of mass m with spring `S_(1)` is
`v_(1)= (1)/(2pi) sqrt((k_1)/(m))" ""........."(2)`
and frequency of oscillation of this body with sprin `S_(2)` is
`v_(2)= (1)/(2pi) sqrt((k_2)/(m))" ""........."(3)`
From equation (2),
` (k_1)/(m)= 4pi^(2) v_(1)" ""........."(4)`
and from equation (3),
` (k_2)/(m)= 4pi^(2) v_(2)" ""........."(5)`
From equation (1),
` v= (1)/(2pi) [(k_1)/(m)+(k_2)/(m)]^(1/2)`
`v= (1)/(2pi) [4pi^(2) v_(1)^(2) + 4pi^(2)v^(2)]^(1/2)" "[therefore ` From equations (4) and (5)]
`therefore v= (2pi)/(2pi) [v_(1)^(2) +v_(2)^(2)]^(1/2)`
`therefore v= sqrt(v_(1)^(2) + v_(2)^(2))`.