Respuesta :
Answer:
w = 16.73 rad / s, v = - 1.04 m / s
Explanation:
This is an exercise in oscillatory motion of a system of a spring and a mass, the angular velocity is
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
To calculate the spring constant we use the data that Δx = 3.5 cm = 0.035 m with a mass of m = 0.38 kg, let's write Hooke's law
F = -kx
in this case the applied force is the weight of the body
F = W
we substitute
mg = k x
k = m g / x
k = 0.38 9.8 / 0.035
k = 106.4 N / m
let's find the angular velocity
w = Ra 106.4 / 0.38
w = 16.73 rad / s
the movement of this system is described by the expression
ax = A cos (wt + Ф)
the quantity A is called the amplitude of the movement in this case A = 7.5 cm = 0.075 m
to encode the constant we use that the system is released from its position of maximum displacement (x = A)
we substitute
A = A cos (0 + Ф)
cos Ф = 1
Ф = cos-1 1
Ф = 0º
by substituting in the equation of motion
x = 0.075 cos (16.73 t)
Let's find the time it takes to reach the desired position x = 4.2 cm = 0.042 m
cos 16.73 t = 0.042 / 0.075 = 0.56
16.73 t = cos -1 0.56
remember that angles must be expressed in radians
t = 0.976 / 16.73
t = 0.0583 s
body velocity is
v = [tex]\frac{dx}{dt}[/tex]
v = -0.075 16.73 sin 16.73t
let's calculate for the time found
v = - 1.25475 sin (16.73 0.0583)
v = - 1.04 m / s
