A block is attached to a horizontal spring and it slides back and forth in simple harmonic motion on a frictionless horizontal s
1 answer:
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
= 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)
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)
A = A₀
w = w₀/√2
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