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What will be the amplitude and frequency of vibration

  • Thread starter needhelp83
  • Start date
199
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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??
 

OlderDan

Science Advisor
Homework Helper
3,021
1
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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