Question

A mass of 5 kg is suspended on a spring of stiffness 4000 N/m. The system is fitted with a damper with a damping ratio of 0.2. The mass is pulled down 50 mm and released. Calculate the displacement after 0.3 sec:
A.)4.07mm
B.)4.70mm
C.)7.40mm
D.)7.04mm

Answer 1

Hint: We have to find the natural frequency ${f_n}$ and angular velocity ${\omega _n}$ of the Simple harmonic motion. We also have to calculate the frequency $f$ and angular velocity $\omega $ considering damping. We have to use the damping ratio given $\delta $.Substituting this general form of the equation of an under damped simple harmonic wave we find the displacement.

Formula used:
$\omega = 2\pi f$ $\omega $ is the angular velocity $f$ and is the frequency.
${f_n} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{M}} $, here ${f_n}$ denotes the natural frequency $k$ denotes the spring constant /stiffness of the spring and $M$ denotes the mass.
$f = {f_n}\sqrt {1 - {\delta ^2}} $ Where $f$ is the frequency, ${f_n}$ denotes the natural frequency and $\delta $ denotes the damping ratio.
$\omega = {\omega _n}\sqrt {1 - {\delta ^2}} $ Here $\omega $ is the angular velocity, ${\omega _n}$ denotes the natural angular velocity and $\delta $ denotes the damping ratio.
$x = C{e^{ - \delta {\omega _n}t}}\cos \omega t$, here $x$ shows the displacement, $C$ denotes the initial /maximum displacement, $\delta $ denotes the damping ratio,$\omega $ is the angular velocity, ${\omega _n}$ denotes the natural angular velocity and $t$ shows the time.

Complete step by step answer:
In order to calculate the frequency of the wave we use the equation ${f_n} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{M}} $

It is given that $k = 4000N/m$ and $M = 5Kg$.We get
${f_n} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{4000}}{5}} $
${f_n} = \dfrac{1}{{2\pi }}\sqrt {800} $
${f_n} = 4.5Hz$

Using $\omega = 2\pi f$ we get ${\omega _n} = 9\pi $
${\omega _n} = 28.28rad/s$
The frequency and angular velocity can be found out by using $f = {f_n}\sqrt {1 - {\delta ^2}} $ and $\omega = {\omega _n}\sqrt {1 - {\delta ^2}} $

Given that $\delta = 0.2$.Substituting

$f = 4.5\sqrt {1 - {{(0.2)}^2}} $
$f = 4.41Hz$
$\omega = 28.28\sqrt {1 - {{(0.2)}^2}} $
$\omega = 27.71rad/s$

The initial / maximum displacement of this wave is ${50mm}$ downwards .Therefore $C = - 50mm$
The time for which we have to find the displacement is given $t = 0.3s$
Using all the found values in h\the general form $x = C{e^{ - \delta {\omega _n}t}}\cos \omega t$ we get,

$x = - 50{e^{ - 0.2 \times 28.28 \times 0.3}}\cos (27.71 \times 0.3)$
$x = - 50{e^{ - 1.6968}}\cos (8.313)$
$x = - 50 \times 0.183 \times - 0.443$
$x \approx 4.07mm$

Note: The angle taken here is in the units of radians. Based on the value of damping ratio and the natural angular velocity of the wave damping can be of basically three types. If ${\gamma ^2} > 4{\omega _0}^2$ it is said to be over damped. If ${\gamma ^2} = 4{\omega _0}^2$it is called critically damped and when ${\gamma ^2} < 4{\omega _0}^2$ it is under damped. (Here $\gamma $ is the damping ratio).