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Sagot :
Answer:
T₀ = 2π [tex]\sqrt{\frac{m}{k} }[/tex] T = [tex]\sqrt{\frac{5}{6} }[/tex] T₀
Explanation:
When the block is oscillating it forms a simple harmonic motion, which in the case of a spring and a mass has an angular velocity
w = [tex]\sqrt{k/m}[/tex]
To apply this formula to our case, let's look for the equivalent constant of the springs.
Let's start with the springs in parallels.
* the three springs in the upper part, when stretched, lengthen the same distance, therefore the total force is
F_total = F₁ + F₂ + F₃
the springs fulfill Hooke's law and indicate that the spring constant is the same for all three,
F_total = - k x - k x - kx = -3k x
therefore the equivalent constant for the combination of the springs at the top is
k₁ = 3 k
* the two springs at the bottom
following the same reasoning the force at the bottom is
F_total2 = - 2 k x
the equivalent constant at the bottom is
k₂ = 2 k
now let's work the two springs are equivalent that are in series
the top spring is stretched by an amount x₁ and the bottom spring is stretched x₂
x₂ = x -x₁
x₂ + x₁ = x
if we consider that the springs have no masses we can use Hooke's law
[tex]-\frac{F_{1} }{k_{1} } - \frac{F_{2}}{k_{2} } = \frac{F}{k_{eq} }[/tex]
therefore the equivalent constant is the series combination is
[tex]\frac{1}{k_{eq} } = \frac{1}{k_{1} } + \frac{1}{k_{2} }[/tex]
we substitute the values
\frac{1}{k_{eq} } = \frac{1}{3k } + \frac{1}{2k }
\frac{1}{k_{eq} } = \frac{5}{6k} }
k_eq = [tex]\frac{6k}{5}[/tex]
therefore the angular velocity is
w = [tex]\sqrt{\frac{6k}{5m} }[/tex]
angular velocity, frequency, and period are related
w = 2π f = 2π / T
T = 2π / w
T = 2π [tex]\sqrt{\frac{5m}{6k} }[/tex]
T₀ = 2π [tex]\sqrt{\frac{m}{k} }[/tex]
T = [tex]\sqrt{\frac{5}{6} }[/tex] T₀
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