A vertical spring scale can measure weights up to 225 N . The scale extends by an amount of 12.5 cm from its equilibrium position at 0 N to the 225 N mark. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.65 Hz. Ignoring the mass of the spring, what is the mass m of the fish?

The spring-mass system experiments a Simple Harmonic Motion, then the angular frequency ([tex]\omega[/tex]), in radians per second, is determined by the following equation:

[tex]\omega = 2\pi\cdot f[/tex] (1)

Where [tex]f[/tex] is the frequency, in hertz.

In addition, the mass of the fish ([tex]m[/tex]), in kilograms, is:

[tex]m = \frac{k}{\omega^{2}}[/tex] (2)

Where [tex]k[/tex] is the spring constant, in newtons per meter.

And the spring constant is determined by Hooke's Law:

[tex]k = \frac{F}{\Delta x}[/tex] (3)

Where:

[tex]F[/tex] - Elastic force, in newtons.

[tex]\Delta x[/tex] - Spring elongation, in meters.

If we know that [tex]F = 225\,N[/tex], [tex]\Delta x = 0.125\,m[/tex] and [tex]f = 2.65\,hz[/tex], then the mass of the fish is:

[tex]\omega = 2\pi\cdot f[/tex]

[tex]\omega \approx 16.650\,\frac{rad}{s}[/tex]

[tex]k = \frac{F}{\Delta x}[/tex]

[tex]k = 1800\,N[/tex]

[tex]m = \frac{k}{\omega^{2}}[/tex]

[tex]m = 6.493\,kg[/tex]

The mass of the fish is 6.493 kilograms.

hamzaahmeds hamzaahmeds · Answer 2 · 2021-11-04T11:36:05+01:00

This question involves the concepts of Hooke's Law, and the frequency of spring-mass system.

The mass of the fish is "6.5 kg".

First, we will calculate the angular speed of the system:

[tex]\omega=2\pi f[/tex]

where,

ω = angular speed = ?

f = frequency = 2.65 Hz

Therefore,

[tex]\omega = 2\pi(2.65\ Hz)\\\omega = 16.65\ rad/s\\[/tex]

Now, we will use the Hooke's Law to find out the spring constant of the spring:

[tex]K =\frac{F}{\Delta x}[/tex]

where,

K = spring constant = ?

F = force applied = maximum weight = 225 N

Δx = extension = 12.5 cm = 0.125 m

Therefore,

[tex]k=\frac{225\ N}{0.125\ m}\\\\k = 1800\ N/m\\[/tex]

Now, we will use the formula for the angular speed of a spring-mass system to find out the mass of the fish:

[tex]\omega = \sqrt{\frac{k}{m}}\\\\m = \frac{k}{\omega^2}\\\\m=\frac{1800\ N/m}{(16.65\ rad/s)^2}[/tex]

m = 6.5 kg

Learn more about Hooke's Law here:

brainly.com/question/13348278?referrer=searchResults

The attached picture illustrates Hooke's Law.