13.4.5 For the oscillatory motion of a mass t = 0.5 kg suspended from a spring, the differential equation has the form y + 60y = 0. It is necessary to determine the spring stiffness coefficient. (Answer 30)
To solve this problem, it is necessary to use the formula for the differential equation of oscillatory motion:
m u'' + k u = 0,
where m is the mass of the load, k is the spring stiffness coefficient.
Substituting known values into this formula, we get:
0.5 u'' + k u = 0.
To further solve this equation, it is necessary to find a general solution to an equation of the form:
у = A cos(ωt + φ),
where A is the amplitude of oscillations, ω is the circular frequency, φ is the initial phase.
Differentiating this function twice, we get:
у'' = -A ω^2 cos(ωt + φ).
Substituting the found values into the original differential equation, we get:
-0.5 A ω^2 cos(ωt + φ) + k A cos(ωt + φ) = 0.
This equation is valid for any t, therefore, the cosine can be eliminated:
-0.5 A ω^2 + k A = 0.
Expressing the spring stiffness coefficient from this equation, we obtain:
k = 0.5 ω^2.
Substituting the frequency value ω = 2πf = 2π/T = 2π√(k/m), we obtain:
k = (2π/T)^2 m = (2π/1)^2 0.5 = 4π^2 × 0.5 = 2π^2.
Thus, the spring stiffness coefficient is:
k = 2π^2 ≈ 19,739.
Answer: 19.739 (nearest integer is 20).
So, having solved this problem, we found that the spring stiffness coefficient is equal to 20 in conventional units.
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This digital product is a solution to problem 13.4.5 from Kepe O.’s collection on physics. The problem is to determine the spring stiffness coefficient for the oscillatory motion of a load weighing 0.5 kg suspended from this spring, provided that the differential equation describing this motion has the form y + 60y = 0.
To solve the problem, it is necessary to use the formula for the differential equation of oscillatory motion and find a general solution to an equation of the form y = A cos(ωt + φ), where A is the amplitude of oscillations, ω is the circular frequency, φ is the initial phase. By substituting the found values into the original differential equation, you can obtain a formula for determining the spring stiffness coefficient.
This product is presented in the form of a detailed description using formulas and logical conclusions, which will make it easy to understand and solve this problem. The design is made in accordance with the requirements for high-quality HTML code layout, which ensures ease of use.
The solution to problem 13.4.5 from the collection of Kepe O. is an excellent choice for students and teachers who are studying physics and want to deepen their knowledge in this area. In addition, this product can be useful to anyone interested in physical phenomena and their solutions.
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The product is the solution to problem 13.4.5 from the collection of Kepe O.?.
This problem presents a differential equation for the oscillatory motion of a load weighing 0.5 kg suspended from a spring, which is written as y + 60y = 0, where y is a function of time that describes the displacement of the load from the equilibrium position.
To solve the problem, it is necessary to determine the spring stiffness coefficient.
To do this, you can use the formula that describes the oscillatory motion of a load suspended on a spring with stiffness k:
my'' + ky = 0,
where m is the mass of the load, y is a function of time, describing the displacement of the load from the equilibrium position, y'' is the second derivative of the function y with respect to time.
By comparing this formula with the equation from the problem, we can derive the relationship between the spring stiffness coefficient and the mass of the load:
k = m*w^2,
where w is the oscillation frequency.
The problem gives an equation of oscillatory motion of the form y + 60y = 0. Compared with the general formula, one can see that the oscillation frequency is sqrt(60), and the mass of the load is 0.5 kg. Substituting these values into the formula for the spring stiffness coefficient, we obtain:
k = 0.5*(sqrt(60))^2 = 30.
Thus, the spring constant is 30, which is the answer to the problem.
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