Answer:
a) v = 31.67 cm / s
, b) v = -29.36 cm / s
, c) v= 29.36 cm/s, d) x = 3.46 cm
Explanation:
The angular velocity in a simple harmonic movement is
w = √ K / m
w = √ 23.2 / 0.37
w = 7,918 rad / s
a) the expression against the movement is
x = A cos (wt + Ф)
Speed is
v = dx / dt = - A w sin (wt + Ф)
The maximum speed occurs for cos = ± 1
v = A w
v = 4.0 7,918
v = 31.67 cm / s
b) as the object is released from rest
0 = -A w sin (0+ Фi)
sin Ф = 0
Ф = 0
The equation is
x = 4.0 cos (7,918 t)
v = -4.0 7,918 sin (7,918 t)
v = - 31.67 sin (7.918t)
Let's look for the time for a displacement of x = 1.5 cm, remember that the angles must be in radians
7,918 t = cos⁻¹ 1.5 / 4.0
t = 1,186 / 7,918
t = 0.1498 s
We look for speed
v = -31.67 sin (7,918 0.1498)
v = -29.36 cm / s
c) if the object passes the equilibrium equilibrium position again at this point the velocity has the same module, but the opposite sign
v = 29.36 cm / s
d) let's look for the time for the condition v = v_max / 2
31.67 / 2 = 31.67 sin ( 7,918 t)
7.918t = sin⁻¹ 0.5
t = 0.5236 / 7.918
t = 0.06613
With this time let's look for displacement
x = 4.0 cos (7,918 0.06613)
x = 3.46 cm
