Question

A fisherman's scale stretches 3.7 cm when a 3.5 kg fish hangs from it. (a) what is the spring constant? n/m (b) what will be the amplitude and frequency of vibration if the fish is pulled down 2.8 cm more and released so that it vibrates up and down? amplitude cm frequency hz

Answer 1

(a) the weight of the fish is:
$W=mg=(3.5 kg)(9.81 m/s^2)=34.3 N$
and this is the force that stretches the spring by $x=3.7 cm=0.037 m$. So, we can use Hook's law to find the constant of the spring:
$k= \frac{F}{x}= \frac{34.3 N}{0.037 m}=927.0 N/m$

(b) The fish is pulled down by 2.8 cm = 0.028 m more, so now the total stretch of the spring is
$x'=3.7 cm+2.8 cm=6.5 cm$
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:
$\omega= \sqrt{ \frac{k}{m} }= \sqrt{ \frac{927.0 N/m}{3.5 kg} }=16.3 Hz$
and so the frequency is given by
$f= \frac{\omega}{2 \pi} = \frac{16.3 Hz}{2 \pi}=2.6 Hz$