Physics Homework Help?

Question

A fisherman's scale stretches 3.5 cm when a 2.2 kg fish hangs from it.

 

Part A

What is the spring stiffness constant?

 

Part B 

What will be the amplitude of vibration if the fish is pulled down 2.8 cm more and released so that it vibrates up and down?

Part C

What will be the frequency of vibration if the fish is pulled down 2.8 cm more and released so that it vibrates up and down?

Answer 1

Part A:

A fisherman's scale stretches 3.5 cm when a 2.2 kg fish hangs from it.

Here  kg and .

Find the spring stiffness constant.

Hooks law : .

 is the restoring elastic force exerted by the spring and  is  the displacement from the equilibrium position.

Substitute  and g = 9.8 in .

.

Substitute corresponding values in .

 N/m.

Spring stiffness constant is  N/m.

Answer 2

Part B:

When the fish is hung from the scale, it defines the equilibrium position at 3.5 cm.

When the fish is pulled down an additional distance, the initial amplitude of oscillation is then established to be 2.8 cm.

Thus, amplitude of vibration is 2.8 cm.

Answer 3

Part C:

A fishermans scale stretches 3.5 cm when a 2.2 kg fish hangs from it.

Find the frequency of vibration if the fish is pulled down 2.8 cm more and released so that it vibrates up and down.

Thus, the motion is simple harmonic motion.

Frequency of  object in simple harmonic motion is  .

From part A, N/m and kg.

Substitute corresponding values in .

Thus, the frequency of vibration is 2.66 Hz.