What will be the amplitude and frequency of vibration

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SUMMARY

The discussion focuses on calculating the spring constant and the amplitude and frequency of vibration for a fish hanging from a fisherman’s scale. The spring constant was determined to be 1295 N/m using the formula F = -kx, where F is the force exerted by the fish. The period of vibration was calculated as 0.336 seconds, leading to a frequency of 2.976 Hz. The amplitude of the vibration is confirmed to be 2.5 cm, which is the additional displacement from the equilibrium position.

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  • Basic knowledge of Newton's second law of motion (F=ma)
  • Familiarity with oscillatory motion and its parameters
  • Ability to perform calculations involving square roots and trigonometric functions
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  • Study the principles of oscillatory motion and harmonic oscillators
  • Learn about the relationship between mass, spring constant, and frequency in oscillations
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Physics students, mechanical engineers, and anyone interested in the dynamics of oscillatory systems will benefit from this discussion.

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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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