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A spring of stiffness constant `k` and natural length `l` is cut into two parts of length `3l//4 and l//4`, respectively, and an arrangement is made as shown in figure . If the mass is slightly displaced , find the time period of oscillation. |
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Answer» The stiffness of a spring is inversely proportional to its length . Therefore the stiffness of each part is `k_(1) = (4)/(3) k` and `k_(2) = 4 k` Time period `T = 2 pi sqrt((m)/(k_(1) + k_(2)))` `= 2 pi sqrt ((3m)/(16 k)) = (pi)/(2) sqrt((3 m)/(k))` |
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