Question

Найти максимальную кинетическую энергию Tmax материальной точки массой m=2 г, совершающей гармонические колебания амплитудой A=4 см и частотой v=5 Гц.

Answer 1

Кинетическая энергия=mv(max)^2/2
V(max)=Aw^2
w=2П умножить на частоту

Answer 2

Calculation of Maximum Kinetic Energy

To find the maximum kinetic energy (Tmax) of a material point undergoing harmonic oscillations, we can use the formula:

Tmax = (1/2) * m * (2πv)^2 * A^2 Where: - Tmax is the maximum kinetic energy - m is the mass of the material point (given as 2 g) - v is the frequency of the oscillations (given as 5 Hz) - A is the amplitude of the oscillations (given as 4 cm)

Let's substitute the given values into the formula and calculate Tmax.

Tmax = (1/2) * 0.002 kg * (2π * 5 Hz)^2 * (0.04 m)^2

Simplifying the equation:

Tmax = 0.0002 * (2π * 5)^2 * 0.04^2 J

Tmax ≈ 0.0002 * 314.16 * 0.0016 J

Tmax ≈ 0.1007 J

Therefore, the maximum kinetic energy (Tmax) of the material point is approximately 0.1007 Joules.