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A fisherman’s scale stretches 3.05 cm when a 2.5-kg fish hangs from it. What will be the frequency of vibration if the fish is pulled down 13.60 cm more and released so that it vibrates up and down?

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k=mg/x=2.5•9.8/0.0305 =803.3 N/m
the natural frequency is
ω =sqrt(k/m) = sqrt(803.3/2.5)=17.9 rad/s

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