Question

К нерастянутой пружине, верхний конец которой закреплен , подвесили и без толчка отпустили тело массой m=1кг. Жесткость пружины K=98Н/М. Пренебрегая ее массой найти: А)Закон движения тела y(t), где y-его смещение из начального положения; Б) Максимальное и минимальное натяжение пружины в процессе движения очень срочно

Answer 1

g=9.8 м/с2

Пренебрегая ее массой, значит инерцию груза не учитывать

сила тяжести растягивает пружину F=mg

максимальное растяжение/амплитуда Y = F/K =mg/K=1*9.8/98 =0.1 м

период колебаний Т=2п√m/k

циклическая частота w=2п/Т =√k/m=√98/1=9.9

А)Закон движения тела y(t) = Y *sin(wt)=0.1*sin(9.9t)

Б) Максимальное натежение при sin=1 ; y(max) = 0.1

    минимальное натяжение   при sin=0  ; y(min) = 0

Answer 2

Problem Analysis

We are given a spring with an unstretched length and an upper end fixed. A mass of 1 kg is suspended from the lower end of the spring and released without any initial displacement. We need to find the equation of motion for the body and determine the maximum and minimum tension in the spring during its motion.

Solution

To solve this problem, we can use Newton's second law of motion and Hooke's law for springs.

A) Equation of Motion

The equation of motion for the body can be derived using Newton's second law, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the force acting on the body is the tension in the spring.

Let's assume that the displacement of the body from its equilibrium position is denoted by y. The force exerted by the spring is given by Hooke's law, which states that the force exerted by a spring is proportional to its displacement. Mathematically, this can be expressed as:

F = -K * y

where F is the force exerted by the spring and K is the spring constant.

Since the force acting on the body is equal to the mass of the body multiplied by its acceleration, we can write:

-K * y = m * a

where m is the mass of the body and a is its acceleration.

Since acceleration is the second derivative of displacement with respect to time, we can rewrite the equation as:

-K * y = m * d^2y/dt^2

Simplifying the equation, we get:

d^2y/dt^2 + (K/m) * y = 0

This is a second-order linear homogeneous differential equation with constant coefficients. The general solution to this equation is given by:

y(t) = A * cos(ωt) + B * sin(ωt)

where A and B are constants to be determined and ω = sqrt(K/m) is the angular frequency of the oscillation.

To find the values of A and B, we need to consider the initial conditions. In this case, the body is released from rest, which means its initial velocity is zero and its initial displacement is also zero. Therefore, we have:

y(0) = 0 (initial displacement) dy/dt(0) = 0 (initial velocity)

Substituting these initial conditions into the equation of motion, we can solve for A and B.

B) Maximum and Minimum Tension in the Spring

The tension in the spring is maximum when the displacement of the body is maximum, and it is minimum when the displacement of the body is minimum. From the equation of motion, we can see that the maximum displacement occurs when cos(ωt) = 1 and the minimum displacement occurs when cos(ωt) = -1.

Therefore, the maximum tension in the spring is given by:

T_max = K * (A + m * g / K)

where g is the acceleration due to gravity.

Similarly, the minimum tension in the spring is given by:

T_min = K * (A - m * g / K)

To find the values of A and B, we can use the initial conditions as mentioned earlier.

Conclusion

In conclusion, the equation of motion for the body is given by y(t) = A * cos(ωt) + B * sin(ωt), where A and B are constants determined by the initial conditions. The maximum tension in the spring is T_max = K * (A + m * g / K), and the minimum tension in the spring is T_min = K * (A - m * g / K).

Please note that the above solution assumes that the spring obeys Hooke's law and that there are no external forces acting on the body other than gravity.