Solution to problem 15.2.4 from the collection of Kepe O.E.

15.2.4. Let us consider a material point M with a mass m = 0.5 kg, which is attached to a flexible thread of length OM = 2 m. It oscillates in the vertical plane in accordance with the equation = (?/6)sin 2 ?t. It is necessary to determine the kinetic energy of a material point at the lowest point of oscillation.

To solve the problem, we use the formula for the kinetic energy of a material point: K = (mv^2)/2, where m is the mass, v is the speed of the point.

At the lowest point of oscillation, the speed of the point will be maximum, the amplitude of oscillations will reach its maximum value. The maximum speed of a point can be found by taking the first derivative of the oscillation equation: v_max = (d?/dt)max = (π?/3)√(g/ℓ), where g is the acceleration of free fall, ℓ is the length of the thread.

Substituting known values, we obtain v_max = π√(5g/6) ≈ 3.07 m/s.

Now we can find the kinetic energy of a material point at the lowest point of oscillation: K = (mv_max^2)/2 = (0.5*3.07^2)/2 ≈ 10.8 J. Answer: 10.8.

Solution to problem 15.2.4 from the collection of Kepe O.?.

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Solution to problem 15.2.4 from the collection of Kepe O.?. consists in determining the kinetic energy of a material point at the lowest point of its oscillations. To do this, it is necessary to use the formula for the kinetic energy of a material point: K = (mv^2)/2, where m is the mass of the point, v is its speed.

First, let's find the speed of the material point at the lowest point of oscillation. To do this, we use the equation of motion of a harmonic oscillator: x = Asin(ωt + φ), where x is the coordinate of the point, A is the amplitude of oscillations, ω is the angular frequency, t is time, φ is the initial phase. In this problem, the angular frequency is equal to ω = (√(g/OM)), where g is the acceleration of free fall, and the initial phase is φ = π/2, since at the bottom point of oscillation the speed of the point is maximum and the displacement relative to the equilibrium position is minimum. Then x = Asin(√(g/OM)t + π/2). Let's differentiate this expression with respect to time to find the speed: v = dx/dt = A(√(g/OM))*cos(√(g/OM)*t + π/2).

The amplitude and time values ​​are not given in the condition, but they are not required to find the speed at the bottom point. At the bottom point of oscillations cos(π/2) = 0, therefore the speed of the material point at the bottom point of oscillations will be equal to v = A*(√(g/OM))*0 = 0.

Since the speed at the bottom point of oscillation is zero, the kinetic energy of the material point at the bottom point of oscillation will be equal to zero: K = (mv^2)/2 = 0.

Answer: the kinetic energy of a material point in its lower position is 0.

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