# Finding the amplitude

## Homework Statement

A fisherman's scale stretches 4.1 cm when a 2.7 kg fish hangs from it.

(a) What is the spring stiffness constant?
645.3658537 N/m
(b) What will be the amplitude and frequency of vibration if the fish is pulled down 2.0 cm more and released so that it vibrates up and down?
amplitude ? cm
frequency 2.460601261 Hz

## Homework Equations

I thought it was as simple as 2.0+4.1=6.1 cm. I was wrong.

## The Attempt at a Solution

Shown above.

kuruman
Homework Helper
Gold Member
The amplitude is the maximum displacement from the equilibrium position. How far from the equilibrium position is the fish when it is pulled down 2.0 cm?

It is already 4.1 cm away, right? Wouldn't adding 2 cm just make it 6.1 cm away?

You can use the period of oscillation:
$$T=2\pi\sqrt{\frac{m}{k}}$$
then put what you found for the period in this equation:
$$f=\frac{1}{T}$$

Uhh.. for amplitude?

kuruman