A body weighing 600 g is suspended from a chain of two parallel springs with stiffness coefficients
| August 18, 2021 | education
A body weighing 600 g is suspended from a chain of two parallel springs with stiffness coefficients of 500 N / m and 250 N / m. Determine the natural oscillation period of the system.
To find the oscillation period of the resulting system, we will use the formula: T = 2Π * √ (m / k) = 2Π * √ (m / (k1 + k2)).
Values of variables: m – body weight (m = 600 g = 0.6 kg); k1 is the stiffness of the first spring (k1 = 500 N / m); k2 is the stiffness of the second spring (k2 = 250 N / m).
Calculation: T = 2Π * √ (m / (k1 + k2)) = 2 * 3.14 * √ (0.6 / (500 + 250)) = 0.178 s.
Answer: The period of natural oscillations of the resulting system is 0.178 s.
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