Question

On your first trip to Planet X you happen to take along a 180 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You're curious about the free-fall acceleration on Planet X, where ordinary tasks seem easier than on earth, but you can't find this information in your Visitor's Guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You find that the mass stretches the spring by 22.6 cm . You then pull the mass down 11.4 cm and release it. With the stopwatch you find that 11 oscillations take 19.0 s ?

Answer 1

Answer:

g_x = 3.0 m / s^2

Explanation:

Given:

- Change in length of spring x@equilibrium = 22.6 cm

- Time taken for 11 oscillations t = 19.0 s

Find:

- The value of gravitational free fall g_x at plant X:

Solution:

- We will assume a simple harmonic motion of the mass for which Time is:

                                 T  = 2*pi*sqrt(k / m )    ...... 1

- Sum of forces in vertical direction @equilibrium is zero:

                                 F_net = k*x - m*g_x = 0

                                 (k / m) = (g_x / x)    .... 2

- substitute Eq 2 into Eq 1:

                                  2*pi / T = sqrt ( g_x / x )

                                   g_x = (2*pi / T )^2 * x

- Evaluate g_x:

                                  g_x = (2*pi / (19 / 11) )^2 * 0.226

                                  g_x = 3.0 m / s^2