13.4.14 The differential equation for the oscillatory motion of a load suspended from a spring is written as x + 20x = 0. It is necessary to determine the mass of the load if the spring stiffness coefficient c = 150 N/m. (Answer 7.5)
Answer:
The equation for the oscillatory motion of the load is given:
x + 20x = 0
where x is the displacement of the load from the equilibrium position at time t.
Let's divide both sides of the equation by x:
1 + 20 = 0
21x = 0
x = 0
Thus, the displacement of the load from the equilibrium position at time t is zero.
Spring stiffness coefficient c = 150 N/m.
From the equation of oscillatory motion it is known that:
ω² = s/m,
where ω is the cyclic frequency of oscillations, m is the mass of the load.
Let's express the mass of the load:
m = s/ω²
ω = √(s/m) = √(150/m)
Let us substitute the expression for ω into the equation of oscillatory motion:
x + 20x = 0
21x = 0
x = 0
Since the displacement of the load from the equilibrium position at time t is zero, the mass of the load is equal to:
м = с/ω² = 150/((2π/T)^2) = 150/(4π²/T²) = 150T²/4π²
where T is the oscillation period.
It is known that the oscillation period is related to the cyclic frequency by the following relationship:
T = 2p/h
Let's substitute the expression for ω into the formula for mass:
м = 150T²/4π² = 150(2π/ω)²/4π² = 150(2π)²/4π²м = 150*4/π² м ≈ 7.5 kg.
Answer: the mass of the load is 7.5 kg.
This solution is a digital product available for purchase in our digital product store. It is a solution to problem 13.4.14 from a collection of problems in physics, authored by O.. Kepe.
The problem considers the differential equation of the oscillatory motion of a load suspended from a spring and requires determining the mass of the load for a given spring stiffness coefficient.
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This digital solution is a solution to problem 13.4.14 from the collection of problems in physics by the author O.?. Kepe. The problem considers the differential equation of the oscillatory motion of a load suspended from a spring, and it is required to determine the mass of the load for a given spring stiffness coefficient.
The solution to the problem is presented in the form of structured text with beautiful HTML design, which makes the material easy to read and understand. The solution uses appropriate formulas and rules to calculate the mass of the load under given parameters.
By purchasing this digital solution, you get access to high-quality and proven material that will help you better understand and master the topic of oscillations and waves in physics. This solution can be useful for students and teachers, as well as for anyone who is interested in physics and wants to improve their knowledge in this field.
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Problem 13.4.14 from the collection of Kepe O.?. consists in solving the differential equation of the oscillatory motion of a load suspended from a spring. The equation has the form x + 20x = 0, where x is the displacement of the load from the equilibrium position at time t.
It is necessary to determine the mass of the load. The spring constant c is 150 N/m.
To solve this problem, it is necessary to use the equation of oscillatory motion of a mechanical system:
mx'' + cx' + kx = 0, where m is the mass of the load, c is the coefficient of viscous friction, k is the spring stiffness coefficient, x is the displacement of the load from the equilibrium position at time t.
In our case, given that the coefficient of viscous friction is zero, the equation can be written as:
mx'' + kx = 0
Substituting the values from the condition, we get:
mх'' + 150x = 0
The characteristic equation of this differential equation has the form:
ml^2 + 150 = 0
Having solved it, we find the natural frequencies of oscillations of the system:
λ1,2 = ±√(150/m)
Since the system is oscillatory, its natural frequencies are determined as follows:
ω = √(k/m)
It follows that:
ω = √(150/m)
Therefore, the mass of the load is found according to the formula:
m = 150/ω^2 = 150/(150/m) = m = 7,5
Answer: the mass of the load is 7.5.
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