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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 |
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Answer» `v_(1)+ v_(2)`  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))`. |
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