Question

In a physics lab, you attach a 0.200-kg air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring’s force constant. 12.0 6.0 - 6.0 -12.0 0.40t (s) 14.9 . When a body of unknown ma

Answer 1

Answer:

  k = 0.292 N / m

Explanation:

The glider with the spring forms a system with a simple harmonic movement, in this type of movement the angular velocity is given by

      w = √ k / m

The angular velocity is related to the frequency and is related to the period

      w = 2π f

      f = 1 / T

We replace

      2π / T = ra (k / m

      T = 2π √ (m / k)

The period is the time of a complete oscillation, in this case the glider starts from the equilibrium point and returns for the first time, this corresponds to the middle of the period

 

       ½ T = 2.60 s

       T = 2 2.60

       T= 5.20 s

We clear the equation

       T² = 4π² m / k

        k = 4π² m / T²

Let's calculate

        k = 4π² 0.200 / 5.20²

        k = 0.292 N / m