Finding the amplitude

1. Nov 6, 2009

tjohn101

1. The problem statement, all variables and given/known data
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

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

3. The attempt at a solution
Shown above.

2. Nov 6, 2009

kuruman

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?

3. Nov 6, 2009

tjohn101

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

4. Nov 6, 2009

matt3D

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:
$$f=\frac{1}{T}$$

5. Nov 6, 2009

tjohn101

Uhh.. for amplitude?

6. Nov 6, 2009

kuruman

With the fish on the spring the spring stretches by 4.1 cm and the fish is at rest. When the fish is at rest, it is at the equilibrium position. When the fish is pulled down by 2.0 cm it is 2.0 cm below the equilibrium position. If the fish is let go, what is the amplitude?

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