Question

On your first trip to Planet X you happen to take along a 290 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You're curious about the acceleration due to gravity 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 23.0 cm . You then pull the mass down 11.7 cm and release it. With the stopwatch you find that 9.00 oscillations take 17.7 s.

Can you now satisfy your curiosity?
what is the new g?

Answer 1

The new value of g
`= 1.19 m/s^2`

Step-by-step explanation:

Mass =290 g

Spring size = 40 cm

The size of spring after streches = 23 cm

Pulling the mass down at = 11.7 cm

The stopwatch shows the number of oscillations in 17.7 s = 9 osscillations.

The oscillation frequency is independent to g.

`So, w = \sqrt{(k)/(m)} \\T = (17.7)/(9) = 1.966 \\W = (2\pi)/(T) = 3.197 \\w = \sqrt{(k)/(m)} \\k = w^(2)m \\k = (3.197)^(2) * 0.29 = 2.96 N/M \\`

The amount of spring stretch due to the hanging mass, x, is dependent on g.

`Now, \ mg = kx \\g = (kx)/(m) \\g = (2.96 * 0.117)/(0.29) = 1.19 m/s^(2)`

Therefore, the new value of g
`= 1.19 m/s^2`