Question

A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring

Answer 1

Answer:

a) The spring constant of the spring is $28.381\,\frac{N}{m}$, b) The angular frequency of the block is $7.038\,\frac{rad}{s}$.

Explanation:

This question is incomplete and complete version will be presented herein:

A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring (b) Find the angular frequency of the block 's vibrations.

a) Since spring is hung from the ceiling and is stretched by action of gravity on 0.573 kilogram block. According to the Hooke's Law, force experimented by the spring is directly proportional to elongation. An expression describing the phenomenon is presented and described below: (System at equilibrium - Newton's Second Law)

$m\cdot g = k\cdot \Delta x$

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$k$ - Spring constant, measured in newtons per meter.

$\Delta x$ - Spring linear deformation, measured in meters.

Now, the spring constant is cleared in this equation and outcome is computed: ($m = 0.573\,kg$, $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta x = 0.198\,m$)

$k = \frac{m\cdot g}{\Delta x}$

$k = \frac{(0.573\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.198\,m}$

$k = 28.381\,\frac{N}{m}$

The spring constant of the spring is $28.381\,\frac{N}{m}$.

b) Let suppose that mass-spring system is experimenting a simple harmonic motion, so that angular frequency is equal to:

$\omega = \sqrt{\frac{k}{m} }$

Given that $k = 28.381\,\frac{N}{m}$ and $m = 0.573\,kg$, the angular frequency, measured in radians per second, of the block is:

$\omega = \sqrt{\frac{28.381\,\frac{N}{m} }{0.573\,kg} }$

$\omega = 7.038\,\frac{rad}{s}$

The angular frequency of the block is $7.038\,\frac{rad}{s}$.