In the figure given below. a block of mass M = 490   g placed on a frictionless table is connected with two springs having same spring constant ( K = 2   N   m - 1 ). If the block is horizontally displaced through X m then the number of complete oscillations it will make in 14 π seconds will be ______.

Step-by-step solution

Both the springs are connected in parallel, therefore K e q = K + K = 2 K Now, time period of spring-block system is given by, T = 2 π m K e q = 2 π m 2 K Given here, m = 490   gm = 0 . 49   kg and K = 2   N   m - 1 So, T = 2 π 0 . 49 2 × 2 = 2 π 49 400 = 2 π 7 20 = 7 π 10 Now, number of oscillation in 14 π is N = time T = 14 π 7 π 10 = 20