A fisherman\'s scale stretches 4.0cm when a 2.3kg fish hangs from it. A) What is

Question

A fisherman's scale stretches 4.0cm when a 2.3kg fish hangs from it.

A) What is the spring stiffness constant?

Express your answer to two significant figures and include the appropriate units.

B) What will be the amplitude of vibration if the fish is pulled down 2.5cm more and released so that it vibrates up and down?

Express your answer to two significant figures and include the appropriate units.

C) What will be the frequency of vibration if the fish is pulled down 2.5cm more and released so that it vibrates up and down?

Express your answer to two significant figures and include the appropriate units.

Please list all steps involve in reaching the answer.

Explanation / Answer

A)

Mfish = 2.3kg
change in spring lenth = x = 4cm
we know that F force applied by a spring when stretched by x distance = -kx
so for equilibrium
Ffish = Fspring
Mfish * g = kx
2.3 * 9.8 = k * 2.3 *10-2

k = 9800 kg/s2  ------------ Ans

B)

If the fish is pulled down 2.5cm more than the more pulling will only become the amplitude as when the fish hangs freely is the equilibrium position and the fish will oscillate with an amplitude of 2.5cm around the previous equilibrium position

so Amplitude = 2.5 cm

C)

we frequency = (1/2pi) * sqrt(k/m) = (1/2pi) sqrt (9800/2.3) = 10.38 Hz
frequency = 10.38 Hz = 10.38 s-1

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