Question

Груз неподвижно висевший на пружине, растягивал её на 25мм. Затем оттянули вниз и отпустили. Определите период возникших гармонических колебаний.

Answer 1

Kx=mg
k/m=g/x
T=2pi*корень(m/k)=2pi*корень(x/g) = 2*3,14*корень(0,025/10) сек =
0,314 сек ~ 0,3 сек

Answer 2

Calculation of the Period of Harmonic Oscillations

To determine the period of the harmonic oscillations that occur when a load is suspended from a spring, we need to consider the properties of the spring and the mass of the load.

Given that the load was initially stretched the spring by 25mm and then released, we can assume that the system will undergo simple harmonic motion. Simple harmonic motion occurs when the restoring force acting on an object is directly proportional to its displacement from its equilibrium position.

The period of the harmonic oscillations can be calculated using the formula:

T = 2π√(m/k)

Where: - T is the period of the oscillations, - π is a mathematical constant approximately equal to 3.14159, - m is the mass of the load, and - k is the spring constant.

Unfortunately, the mass of the load and the spring constant are not provided in the question. Therefore, without this information, it is not possible to calculate the period of the oscillations accurately.

If you have the values for the mass of the load and the spring constant, please provide them, and I will be happy to help you calculate the period of the oscillations.