Question

груз,подвешенный на пружине,вызвал её удлинение на 4см.Найдите период колебаний пружины вместе с грузом

Answer 1

при равновесии груза сила тяжести равна силе упругости mg=kx, отсюда жесткость пружины k=mg/x (1). период колебаний груза на пружине равен T=2pi/w=2pi(m/k)^1/2 (2), т.к. циклическая частота w=(k/m)^1/2. подставляем (1) в (2) и получаем T=2pi(x/g)^1/2

Answer 2

Calculation of the Period of Oscillation

To find the period of oscillation of a spring with a suspended mass, we can use the formula:

T = 2π√(m/k)

where: - T is the period of oscillation, - m is the mass of the object suspended on the spring, and - k is the spring constant.

In this case, the spring elongation caused by the suspended mass is given as 4 cm. However, we need additional information to calculate the period of oscillation. Specifically, we need the mass of the object and the spring constant.

Please provide the mass of the object and the spring constant so that we can calculate the period of oscillation for you.