Question

A fisherman’s scale stretches 3.3 cm when a 2.6-kg fish hangs from it. The fish is pulled down 2.5 cm more and released so that it oscillates up and down with a simple harmonic motion (SHM). ? (a) What is the spring stiffness constant? (b) What is be the amplitude? (c) What is the period and frequency of oscillations?

Answer 1

a) The spring constant is 772.1 N/m

b) The amplitude is 0.025 m

c) The period is 0.365 s, the frequency is 2.74 Hz

Explanation:

a)

At equilibrium, the weight of the fish hanging on the spring is equal to the restoring force of the spring. Therefore, we can write:

$mg=kx$

where

(mg) is the weight of the fish

kx is the restoring force

m = 2.6 kg is the mass of the fish

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 3.3 cm = 0.033 m is the stretching of the spring

Solving for k, we find:

$k=\frac{mg}{x}=\frac{(2.6)(9.8)}{0.033}=772.1 N/m$

b)

Later, the fish is pulled down by 2.5 cm (0.025 m), and the system starts to oscillate.

The amplitude of a simple harmonic motion is equal to the maximum displacement of the system. In this case, when the fish is pulled down by 2.5 cm and then released, immediately after the releasing the fish moves upward, until it reaches the same displacement on the upper side (+2.5 cm).

This means that the amplitude of the motion corresponds also to the initial displacement of the fish when it is pulled down: therefore, the amplitude is

A = 2.5 cm = 0.025 m

c)

The period of oscillation of a mass-spring system is given by

$T=2\pi \sqrt{\frac{m}{k}}$

where

m is the mass

k is the spring constant

Here we have

m = 2.6 kg

k = 772.1 N/m

So, the period is

$T=2\pi \sqrt{\frac{2.6}{772.1}}=0.365 s$

And the frequency is given by the reciprocal of the period, therefore:

$f=\frac{1}{T}=\frac{1}{0.365}=2.74 Hz$

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