When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displaced from it

Question

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displaced from its equilibrium position and undergoes simple harmonic oscillations. What is the period of the oscillations

in progress 0
Neala 4 years 2021-08-21T04:20:07+00:00 1 Answers 15 views 0

Answers ( )

  1. Edana
    0
    2021-08-21T04:21:41+00:00
    Reply

    Answer:

    The period is T = 0.700 \ s

    Explanation:

    From the question we are told that  

        The mass is m = 0.350 \ kg

         The extension of the spring is  x = 12.0 \ cm = 0.12 \ m

           

    The spring constant for this is mathematically represented as

           k = \frac{F}{x}

    Where F is the force on the spring which is mathematically evaluated as

           F = mg = 0.350 * 9.8

           F =3.43 \ N

    So  

        k = \frac{3.43 }{ 0.12}

        k = 28.583 \ N/m

    The period of oscillation is mathematically evaluated as

          T = 2 \pi \sqrt{\frac{m}{k} }

    substituting values

         T = 2 * 3.142* \sqrt{\frac{0.35 }{28.583} }

         T = 0.700 \ s

       

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )