Solution to problem 14.3.19 from the collection of Kepe O.E.

14.3.19

There is a body 1 weighing 2 kg, which moves relative to body 2 weighing 8 kg under the action of a spring. The law of motion of body 1 is given by the formula: s = 0.2 + 0.05 cos(ωt), where s is the coordinate of body 1, and ω is the angular velocity of the spring oscillations.

Body 2 can slide along horizontal guides. At time t = 2 s, body 2 begins to move from a state of rest. It is necessary to determine the speed of body 2 at this moment in time.

Answer:

Initially, we determine the angular velocity of the spring oscillations:

ω = 2π/T, where T is the oscillation period of the spring.

Since the movement of body 1 is connected with the movement of body 2, we can express the coordinate of body 1 through the coordinate of body 2:

s = x - l, where x is the coordinate of body 2, and l is the length of the stretched spring.

Differentiating this expression with respect to time, we obtain:

v = dx/dt - dl/dt = dx/dt - v₂, where v is the speed of body 1, and v₂ - speed of body 2.

Since body 1 moves under the action of a spring, its acceleration is determined by the formula:

a = -ω²s = -ω²(x - l).

Then the acceleration of body 2 will be determined by the expression:

a₂ = -a(m₁/m₂) = ω²(x - l)(m₁/m₂), where m₁ = 2 kg - body weight 1, and m₂ = 8 kg - body weight 2.

Since body 2 begins to move from a state of rest, its initial speed is 0. Then, to determine the speed of body 2 at the time t = 2 s, you can use the formula:

v₂ = ∫₀²a₂dt = (ω²m₁/m₂)∫₀²(x - l)dt = (ω²m₁/m₂)(s₀t - l₀sin(ωt)),

where are you₀ = s(t=2) = 0.35 m - coordinate of body 1 at time t = 2 s, and l₀ - length of the stretched spring in a given state.

Substituting known values, we get:

v₂ = (2π/T)²(2 kg)/(8 kg)(0.35 m - l₀sin(4π

Solution tasks 14.3.19

There is a body 1 weighing 2 kg, which moves relative to body 2 weighing 8 kg under the action of a spring. The law of motion of body 1 is given by the formula: s = 0.2 + 0.05 cos(ωt), where s is the coordinate of body 1, and ω is the angular velocity of the spring oscillations.

Body 2 can slide along horizontal guides. At time t = 2 s, body 2 begins to move from a state of rest. It is necessary to determine the speed of body 2 at this moment in time.

Answer:

Initially, we determine the angular velocity of the spring oscillations:

ω = 2π/T, where T is the oscillation period of the spring.

Since the movement of body 1 is connected with the movement of body 2, we can express the coordinate of body 1 through the coordinate of body 2:

s = x - l, where x is the coordinate of body 2, and l is the length of the stretched spring.

Differentiating this expression with respect to time, we obtain:

v = dx/dt - dl/dt = dx/dt - v₂, where v is the speed of body 1, and v₂ - speed of body 2.

Since body 1 moves under the action of a spring, its acceleration is determined by the formula:

a = -ω²s = -ω²(x - l).

Then the acceleration of body 2 will be determined by the expression:

a₂ = -a(m₁/m₂) = ω²(x - l)(m₁/m₂), where m₁ = 2 kg - body weight 1, and m₂ = 8 kg - body weight 2.

Since body 2 begins to move from a state of rest, its initial speed is 0. Then, to determine the speed of body 2 at the time t = 2 s, you can use the formula:

v₂ = ∫₀²a₂dt = (ω²m₁/m₂)∫₀²(x - l)dt = (ω²m₁/m₂)(s₀t - l₀sin(ωt)),

where are you₀ = s(t=2) = 0.35 m - coordinate of body 1 at time t = 2 s, and l₀ - length of the stretched spring in a given state.

Substituting known values, we get:

v₂ = (2π/T)²(2 kg)/(8 kg)(0.35 m - l₀

Solution to problem 14.3.19 from the collection of Kepe O..

that digital product is the solution to problem 14.3.19 from the collection of Kepe O.. in physics. If you are a student or schoolchild studying physics, then this solution will be useful for you in the learning process.

This problem considers the motion of two bodies connected by a spring. It is necessary to determine the speed of one of the bodies at a certain moment in time. The solution to the problem is presented in the form of detailed step-by-step instructions that will allow you to understand how the answer was obtained and how to apply this technique in solving similar problems.

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By purchasing this digital product, you receive a useful tool for studying physics that will help you better understand the material and successfully complete tasks.

This product is a solution to problem 14.3.19 from the collection of Kepe O.?. in physics. The problem considers the motion of two bodies connected by a spring, and it is necessary to determine the speed of one of the bodies at a certain point in time. The solution is presented in the form of detailed instructions with a step-by-step solution algorithm.

According to the conditions of the problem, body 1 with a mass of 2 kg moves relative to body 2 with a mass of 8 kg under the action of a spring. The law of motion of body 1 is given by the formula s = 0.2 + 0.05 cos(ωt), where s is the coordinate of body 1, and ω is the angular velocity of the spring oscillations. Body 2 can slide along horizontal guides.

To solve the problem, it is necessary to determine the angular velocity of oscillations of the spring and express the coordinate of body 1 through the coordinate of body 2. Then you need to differentiate this expression with respect to time to obtain the speed of body 1. The acceleration of body 1 is determined by the formula a = -ω^2s, and the acceleration of body 2 - expression a2 = -a(m1/m2).

Since body 2 begins to move from a state of rest, its initial speed is 0. To determine the speed of body 2 at time t = 2 s, you can use the formula v2 = ∫0^2a2dt. Substituting the known values, we get the answer: v2 = 0.

This product is presented in html format, which makes it easy to read and study the material. It will be useful for students and schoolchildren studying physics, as it contains a detailed solution to the problem with step-by-step instructions.

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Solution to problem 14.3.19 from the collection of Kepe O.?. consists in determining the speed of body 2 weighing 8 kg at time t = 2 s, if it begins to move from a state of rest and, under the action of a spring, moves relative to body 1 weighing 2 kg according to the law s = 0.2 + 0.05 cos ?t, where s is the displacement of body 1 relative to the equilibrium position, t is time in seconds, ? - angular frequency of spring oscillations in radians per second.

To solve the problem it is necessary to use the laws of dynamics and the law of conservation of momentum. First, the speed of body 1 at time t = 2 s is determined using the formula for speed during harmonic oscillations: v = -Asin(ωt), where A is the amplitude of oscillations, ω is the angular frequency of oscillations of the spring. Then, using the law of conservation of momentum, the speed of body 2 is determined.

In this problem, the angular frequency of oscillation of the spring is unknown, so it must be determined from the oscillation equation s = 0.2 + 0.05 cos ?t. For this equation it is necessary to reduce it to the form s = A cos(ωt + φ), where A is the amplitude of oscillations, ω is the angular frequency of oscillations of the spring, φ is the initial phase of oscillations. After reducing the equation to this form, we get:

s = 0,25 cos (?t - 1,107)

Comparing this equation with s = A cos(ωt + φ), we find that A = 0.25, φ = -1.107 rad. Then the angular frequency of oscillations of the spring is equal to ω = ?, where ? = ωt + φ. We substitute the values ​​t = 2 s and ω = ?/t - φ/t and find the angular frequency of the spring oscillations:

ω = 1.107/2 + arccos(0.2/0.25)/2 ≈ 0.785 rad/s

Next, using the formula for speed during harmonic vibrations, we determine the speed of body 1 at time t = 2 s:

v1 = -Asin(ωt) = -0.25sin(0.785*2) ≈ -0.306 m/s

Finally, using the law of conservation of momentum, we find the speed of body 2 at time t = 2 s:

m1v1 + m2v2 = 0

v2 = -m1v1 / m2 = 0.306 * 2 / 8 = 0.0765 m/s

So, the speed of body 2 at time t = 2 s, if it began to move from a state of rest, is equal to 0.0765 m/s.

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