## Question

###### Springs

A fisherman's scale stretches 3.5 cm when a 2.5 kg fish hangs from it.

(a) What is the spring constant?

N/m

(b) What will be the amplitude and frequency of vibration if the fish is pulled down 3.0 cm more and released so that it vibrates up and down?

amplitude

cm

frequency

Hz

(a) What is the spring constant?

N/m

(b) What will be the amplitude and frequency of vibration if the fish is pulled down 3.0 cm more and released so that it vibrates up and down?

amplitude

cm

frequency

Hz

## Answers

k = 2.5/0.035 = 714

T=2(pi)square root(m/k)

T= 2(3.14)square root(2.5/714)

T = .371 sec

f=1/T= 1/.371 = 2.69 Hz

a.

mg = kx

spring const = k = 2.5 x 9.8 / 0.035 = 700 N/m

b.

Amplitude = 0.03 meter

w = (k/m)^0.5 = 16.733 rad/s

Frequecny = 2.663 Hz

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