Finding the amplitude

  • Thread starter tjohn101
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  • #1
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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.
 

Answers and Replies

  • #2
kuruman
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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
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It is already 4.1 cm away, right? Wouldn't adding 2 cm just make it 6.1 cm away?
 
  • #4
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You can use the period of oscillation:
[tex]T=2\pi\sqrt{\frac{m}{k}}[/tex]
then put what you found for the period in this equation:
[tex]f=\frac{1}{T}[/tex]
 
  • #5
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Uhh.. for amplitude?
 
  • #6
kuruman
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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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