What will be the amplitude and frequency of vibration

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 17K views
needhelp83
Messages
193
Reaction score
0
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??
 
Physics news on Phys.org
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.