The cylinder is displaced 0.17 m downward from its equilibrium position and is released at time t = 0. Determine the displacement y and the velocity v when t = 3.1 s. The displacement and velocity are positive if downward, negative if upward. What is the magnitude of the maximum acceleration?

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Answer:

a

   [tex]d =0.161 \ m[/tex]

b

  [tex]v = - 0.054 \ m/s[/tex]

c

  [tex]a = 6.12 \ m/s^2[/tex]

Explanation:

From the question we are told that

      The maximum  displacement is  A =  0.17  m  

      The  time considered is  [tex]t = 3.1 \ s[/tex]

     The spring constant is  [tex]k = 137 \ N \cdot m[/tex]

      The mass is  [tex]m = 3.8 \ kg[/tex]

Generally given that the motion which the cylinder is undergoing is a simple harmonic motion , then the displacement is mathematically represented as

             [tex]d = A cos (w t )[/tex]

Where [tex]w[/tex] is the angular frequency which is mathematically evaluated as

        [tex]w = \sqrt{\frac{k}{m} }[/tex]

substituting values

       [tex]w = \sqrt{\frac{137}{ 3.8} }[/tex]

        [tex]w =6[/tex]

So the displacement is at  t

      [tex]d = 0.17 cos (6 * 3.1 )[/tex]

       [tex]d =0.161 \ m[/tex]

Generally the velocity of a  SHM(simple harmonic motion) is mathematically represented as

         [tex]v = - Asin (wt)[/tex]

substituting values

         [tex]v = - 0.17 sin ( 6 * 3.1 )[/tex]

          [tex]v = - 0.054 \ m/s[/tex]

Generally the maximum acceleration is  mathematically represented as

         [tex]a = w^2 * A[/tex]

substituting values

         [tex]a_{max} = 6^2 * (0.17)[/tex]

substituting values

         [tex]a = 6^2 * (0.17)[/tex]

        [tex]a = 6.12 \ m/s^2[/tex]

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