Question

5. (II) A fisherman’s scale stretches 3.6 cm when a 2.4-kg fish hangs from it. (a) What is the spring stiffness constant and (b) what will be the amplitude and frequency of oscillation if the fish is pulled down 2.1 cm more and released so that it oscillates up and down?

Answer 1

Answer:

(1)  K = 6.5 N/m

(2) amplitude =2.1 cm

Explanation:

extension (x) = 3.6 cm

mass (m) = 2.4 kg

acceleration due to gravity (g) = 9.8 m/s^{2}

(1).  we can get the spring stiffness constant from the formula F = Kx

  where

• F = force = m x g = 2.4 x 9.8 = 23.52 N
• x = extension = 3.6 cm
• K = spring stiffness constant
• substituting all required values into the formula we have

        23.52 = 3.6K

       K = 6.5 N/m

(2).  what will be the amplitude and frequency of oscillation if the fish is pulled down 2.1 cm more and released so that it oscillates up and down?

• Amplitude is the maximum extent of a vibration or an oscillation measured from its position of equilibrium. In this case, the position of equilibrium of the scale was when the fish was hanging with the scale stretched by 3.6 cm and that is where the amplitude will be measured from. with the fish pulled down 2.1 cm and released to oscillate, it would not go higher than the initial length to which it was pulled hence the initial length of 2.1 cm would be its amplitude.