(a) the weight of the fish is:
[tex]W=mg=(3.5 kg)(9.81 m/s^2)=34.3 N[/tex]
and this is the force that stretches the spring by [tex]x=3.7 cm=0.037 m[/tex]. So, we can use Hook's law to find the constant of the spring:
[tex]k= \frac{F}{x}= \frac{34.3 N}{0.037 m}=927.0 N/m [/tex]
(b) The fish is pulled down by 2.8 cm = 0.028 m more, so now the total stretch of the spring is
[tex]x'=3.7 cm+2.8 cm=6.5 cm[/tex]
But this is also the amplitude of the new oscillation, because this is the maximum extension the spring can get, so A=6.5 cm.
The angular frequency of oscillation is given by:
[tex]\omega= \sqrt{ \frac{k}{m} }= \sqrt{ \frac{927.0 N/m}{3.5 kg} }=16.3 Hz [/tex]
and so the frequency is given by
[tex]f= \frac{\omega}{2 \pi} = \frac{16.3 Hz}{2 \pi}=2.6 Hz [/tex]
