Find the variation of the displacement with time, x(t), of a...

Find the variation of the displacement with time, $x(t),$ of a damped single-degree-of-freedom system with $\zeta=0.1$ for the following initial conditions: a. $x(t=0)=x_{0}=0.2 \mathrm{~m}, \dot{x}_{0}=0$ b. $x(t=0)=x_{0}=-0.2 \mathrm{~m}, \dot{x}_{0}=0$ c. $x_{0}=0, \dot{x}_{0}=0.2 \mathrm{~m} / \mathrm{s}$ d. Plot the variations of displacement found in parts (a), (b), and (c) on the same graph in the range of 0 to $5 \mathrm{~s}$.

Find the variation of the displacement with time, $x(t),$ of a damped single-degree-of-freedom system with $\zeta=0.1$ for the following initial conditions:
a. $x(t=0)=x_{0}=0.2 \mathrm{~m}, \dot{x}_{0}=0$
b. $x(t=0)=x_{0}=-0.2 \mathrm{~m}, \dot{x}_{0}=0$
c. $x_{0}=0, \dot{x}_{0}=0.2 \mathrm{~m} / \mathrm{s}$
d. Plot the variations of displacement found in parts (a), (b), and
(c) on the same graph in the range of 0 to $5 \mathrm{~s}$.

Key Concepts

Initial_Conditions

Initial conditions—specifying the initial displacement and velocity—are essential for uniquely solving the differential equation of a damped oscillator. They determine the amplitude, phase, and overall evolution of the system's response over time. By applying appropriate initial conditions, one can accurately predict how the system will behave under different starting scenarios.

Natural_and_Damped_Frequency

The natural frequency defines the oscillation frequency of the system without damping, while the introduction of damping modifies the frequency to a lower value known as the damped natural frequency. This concept is key because it explains how the oscillatory behavior is affected by damping, altering both the amplitude decay and the period of motion.

Damping_Ratio

The damping ratio, typically denoted by ?, quantifies the damping level in relation to the critical damping level. It determines the nature of the system’s response; for example, when ? is less than 1, the system is underdamped and exhibits oscillatory decay. This parameter is crucial for predicting how quickly and in what manner the oscillations decrease over time.

Damped_Harmonic_Oscillator

A damped harmonic oscillator is a system where a restoring force proportional to displacement and a damping force proportional to velocity work together. This results in a motion that decays over time, a characteristic seen in many mechanical and electrical systems where energy is dissipated. Understanding this concept is essential to analyze systems where vibrations diminish due to energy loss.

Second_Order_Differential_Equation

The motion of a damped oscillator is described by a second-order linear differential equation, conventionally written as m*x'' + c*x' + k*x = 0. This equation encompasses the mass (m), damping coefficient (c), and stiffness (k) of the system, and forms the foundation for analyzing dynamic responses in mechanical vibrations and other similar systems.

Transcript

00:03 Okay, so starting off with part a here, we have x not is equal to 0 .2 meters, x .org, and so c1 is equivalent to 0 .2 times 31 .6 to 28 times 2 times 2 plus the square root of 3, divided by 2 times 31 .6 to 228 times the square root of 3, and that gives you 0 .21 .2 .8, times the square root of 3, and that gives you 0 .2 .2 .1.

00:35 155.

00:37 For c2, we get negative 0 .2 times 31 .6228 times 2 minus the square root of 3 divided by 2 times 31 .6228 times the square root of 3, and that gives you negative 0 .01547.

01:04 And so x of t is equals to 0 .255 .5 .5.

01:10 Times e to the negative 2 plus a square root of 3 times 31 .6228 t's minus 0 .01547 times e to the negative 2 minus the square root of 3 times 31 .6228 times t.

01:40 So you simplify this.

01:43 That gives you 0 .2155 times e to the negative 8 times e to the negative 8 times t.

01:50 8 .4 -749 t minus 0 .01547 east to the negative 118 .0163 t's times m.

02:06 For part b, x not is equal to negative 0 .2 ms, x.

02:16 Dot not is equal to 0 .2, and c1 is equal to negative 0 .2 times 31 .6 -228.

02:27 Times 2 plus the square of 3.

02:34 And you can already see that this will be 2 times 31 .6228 times the square of 3.

02:43 This is the exact same as the first part of part a.

02:46 And so it's just the negative version of 0 .255.

02:51 And then for c2, it's positive 0 .2...