Respuesta :
Answer:
x = 0.100 cos (3.65 t)
Explanation:
We must solve Newton's second law for this case
F = m a
the spring force is given by Hooke's law
F = -kx
and the definition of acceleration is
a =[tex]\frac{d^2x}{dt^2}[/tex]
let's substitute
- kx = m \frac{d^2x}{dt^2}
\frac{d^2x}{dt^2} + [tex]\frac{k}{m}[/tex] x = 0
this equation has as a solution
x = A cos (wt + Ф)
where
w = [tex]\sqrt{\frac{k}{m} }[/tex]
let's calculate
w = [tex]\sqrt{\frac{20}{1.5} }[/tex]
w = 3.65 rad / s
to find the constant fi let's use the condition t = 0 x = 10.0 cm
x = A cos (0 +Фi)
0.100 = A
therefore Ф = 0
the complete solution is
x = 0.100 cos (3.65 t)
this solution is in meters
