Question

Материальная точка с массой 5 г совершает гармонические колебания с частотой 0,5 Гц. Амплитуда колебаний 3 см. Определить ускорение и силу, действующую в момент, когда смещение равно 1,5 см

Answer 1

S=0.03*sin(pi*t). s'=v=0.03*pi*cos(pi*t). s''=v'=a=-0.03*pi*pi*sin(pi*t)
s=0.015 при sin(pi*t)=1/2
a=-0.015*pi*pi м/с^2
f=m*a=-0.05*0.015*pi*pii Н

Answer 2

Given Information

- Mass of the material point: 5 g - Frequency of the harmonic oscillations: 0.5 Hz - Amplitude of the oscillations: 3 cm - Displacement at a certain moment: 1.5 cm

Calculating Acceleration

The acceleration of a material point undergoing harmonic oscillations can be calculated using the formula:

a = -ω^2x

Where: - a is the acceleration - ω is the angular frequency (2πf) - x is the displacement

In this case, the angular frequency (ω) can be calculated as follows:

ω = 2πf

Given that the frequency (f) is 0.5 Hz, we can substitute it into the formula:

ω = 2π * 0.5 = π rad/s

Now we can calculate the acceleration (a) using the given displacement (x = 1.5 cm = 0.015 m) and the angular frequency (ω):

a = -π^2 * 0.015 = -0.147 m/s^2

Therefore, the acceleration of the material point at the moment when the displacement is 1.5 cm is -0.147 m/s^2.

Calculating Force

The force acting on a material point undergoing harmonic oscillations can be calculated using the formula:

F = -mω^2x

Where: - F is the force - m is the mass - ω is the angular frequency (2πf) - x is the displacement

Given that the mass (m) is 5 g = 0.005 kg and the angular frequency (ω) is π rad/s (as calculated earlier), we can substitute these values into the formula along with the displacement (x = 1.5 cm = 0.015 m):

F = -0.005 * π^2 * 0.015 = -0.00147 N

Therefore, the force acting on the material point at the moment when the displacement is 1.5 cm is -0.00147 N.