What will be the amplitude and frequency of vibration

AI Thread Summary
The spring constant of the fisherman’s scale is calculated to be 1295 N/m based on the weight of the fish. When the fish is pulled down an additional 2.5 cm and released, it vibrates with a period of 0.336 seconds. This results in a frequency of approximately 2.976 Hz. The amplitude of the vibration is confirmed to be 2.5 cm, which represents the initial displacement from the equilibrium position. The discussion clarifies the relationship between displacement, amplitude, and the behavior of oscillators.
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A fisherman’s scale stretches 2.8 cm when a 3.7 kg fish hangs from it.
a. What is the spring constant, and
b. What will be the amplitude and frequency of vibration of the fish is pulled down 2.5 cm more and released so that it vibrates up and down?
a)
F=ma
F=(3.7 kg)( 9.8 N/kg)=36.26 N

F = -kx
36.26 N=-k(- 0.028 m)
k=1295 N/m

b)
k=1295 N/m
m=3.7 kg
T = 2pi sqrt(m/k)= 2p sqrt(3.7 kg/1295 N/m)= 0.336 s

F = 1/T=1/(0.336 s)=2.976 Hz

The amplitude would be 2.5 cm more.

Would this amplitude be correct say for the problem??
 
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I think you have it. I'm not sure what you are asking. When an oscillator is displaced and released from rest, the amplitude is the intitial displacement from equilibrium.
 

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