Question

Шарик массой 50 г колеблется на пружине жесткостью 500 Н/м. Амплитуда колебаний 6 см. Скорость шарика в точке максимального отклонения равна чему?

Answer 1

Ответ:0

Скорость в точке же МАКСИМАЛЬНОГО ОТКЛОНЕНИЯ

Answer 2

#### Calculation of the Velocity of the Ball To calculate the velocity of the ball at the point of maximum displacement, we can use the principles of simple harmonic motion. The formula to calculate the velocity of an object undergoing simple harmonic motion is: **v = Aω** Where: - **v** is the velocity of the object, - **A** is the amplitude of the motion, and - **ω** is the angular frequency of the motion. In this case, the amplitude of the oscillation is given as 6 cm. However, we still need to find the angular frequency of the motion. To find the angular frequency, we can use the formula: **ω = √(k/m)** Where: - **k** is the spring constant, and - **m** is the mass of the object. Given that the mass of the ball is 50 g (0.05 kg) and the spring constant is 500 N/m, we can substitute these values into the formula to find the angular frequency. **ω = √(500 N/m / 0.05 kg)** Calculating this value gives us: **ω ≈ 31.62 rad/s** Now that we have the amplitude and the angular frequency, we can calculate the velocity of the ball at the point of maximum displacement. **v = 6 cm * 31.62 rad/s** Converting the amplitude from centimeters to meters: **v ≈ 0.06 m * 31.62 rad/s** Calculating this value gives us: **v ≈ 1.8972 m/s** Therefore, the velocity of the ball at the point of maximum displacement is approximately **1.8972 m/s**.