Question

A proud deep-sea fisherman hangs a fish of mass 61.0 kg from an ideal spring having negligible mass. The fish stretches the spring a distance of 0.119 m

What is the force constant of the spring?

What is the period of oscillation of the fish if it is pulled down and released?

Answer 1

part 1

For equilibrium of the fish,

spring force = weight of fish

where k is the force constant

= 61.0 x 9.81/0.119

= 5030 N/m

part 2

period of oscillation is given by

$T= 2 \pi \sqrt{m/k}$

  $= 2 \pi \sqrt{61.0/5030}$

= 0.692 s