The mass of cargo attached to the spring was increased 9 times. How will the period and oscillation frequency of this spring pendulum change?

m1 = 9 * m2.
T1 / T2 -?
The period of natural oscillations of the spring pendulum T is determined by the formula: T = 2 * п * √ (m / k), where п is the number п, m is the mass of the load, k is the stiffness of the spring.
T1 = 2 * п * √ (m1 / k) = 2 * п * √ (9 * m2 / k) = 3 * 2 * п * √ (m2 / k).
T2 = 2 * п * √ (m2 / k).
T1 / T2 = 3 * 2 * п * √ (m2 / k): 2 * п * √ (m2 / k) = 3.
Answer: the period of natural oscillations of the spring pendulum, when the mass of the cargo is reduced by 9 times, decreases by 3 times: T1 / T2 = 3.

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