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A block of mass m is hung from the ceiling by the system of massless springs consisting of two layers. The upper layer consists of 3 strings in paralle, and the lower layer consists of 2 strings in parallel. The horizontal bar between the two layers has negligible mass. The force constants of all springs are k. Calculate the period of the vertical oscillations of the block.

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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