In the arrangement shown in the diagram, pulleys are small and springs are ideal. k 1 = k 2 = k 3 = k 4 = 10 N m - 1 are force constants of the springs and mass m = 10 kg . If the time period of small

Question

In the arrangement shown in the diagram, pulleys are small and springs are ideal. $k_{1} = k_{2} = k_{3} = k_{4} = 10 N m^{- 1}$ are force constants of the springs and mass $m = 10 kg$ . If the time period of small vertical oscillations of the block of mass $m$ is given by $2 π x seconds$ , then find the value of $x$ .

TestHub · Accepted Answer

T = 2 π m K eq where K eq = 1 4 1 k 1 + 1 k 2 + 1 k 3 + 1 k 4 = 1 0 1 6 T = 2 π m K eq = 2 π x So, x = 1 0 1 0 / 1 6 = 4 , x = 4

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