A fisherman’s scale stretches 2.8 cm when a 3.7 kg fish hangs from it.(adsbygoogle = window.adsbygoogle || []).push({});

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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# What will be the amplitude and frequency of vibration

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