Respuesta :
Answer:
A = A₀ ,  w = w₀/√2
Explanation:
This is a problem that we must solve with Newton's second law. They indicate that at the end of the initial movement where the speed is zero, add a mass to the block, we assume that it has the same mass, therefore the total mass is m_total = 2 m. Let's write Newton's second law at this point
          [tex]F_{e}[/tex] = m_total a
the elastic force is
          F_{e} = - k x
acceleration is
          a = d²x / dt²
we substitute
          - k x = m_total  d²x / dt²
           d²x / dt² + (k / m_total) x = 0
we substitute
           d²x / dt² + (k /2m) x = 0
the solution to this differential equation is
          x = A cos (wt + Ф)
where
         w = √ (k / 2m)
to find the constant Ф we use the velocity
          v = dx / dt = - Aw sin (wt + Ф)
         Â
At the most extreme point and when the new movement begins (t = 0) they indicate that v = 0
          0 = - A w sin Ф
for this expression to be zero the sine must be zero therefore Ф = 0
when replacing
         x = A cos (wt)
         w = 1 /√2  √ (k / m)
if we want to relate to the initial movement (before placing the block)
         w₀ = √ (k / m)
         w = w₀ /√ 2
The amplitude of the movement is the distance from the equilibrium point to where the movement begins, in this case it is the same as in the initial movement
         A = A₀
the subscript is used to refer to the oscillations before placing the second block
we substitute to have the final equation
         x = A₀ cos (w₀ t /√2)
          Â
         A = A₀
         w = w₀/√2
