Answer
a. $6.5 \times 10^2 N/m$
b. Amplitude is $2.1cm$, frequency is $2.6Hz$
Work Step by Step
a. Find the spring constant by dividing the magnitude of applied force by the displacement.
$$k=\frac{F}{x}=\frac{mg}{x}=\frac{(2.4kg)(9.80m/s^2)}{0.036 m}=6.5 \times 10^2 N/m$$
b. Find the oscillation frequency from k and the total mass.
$$f=\frac{1}{2 \pi}\sqrt{\frac{k}{m}}=\frac{1}{2 \pi}\sqrt{\frac{653 N/m }{2.4kg}}\approx 2.6Hz$$
The amplitude is the distance that the fish was pulled away from the new equilibrium, or A=2.1 cm.
