Mechanic Oscillations Equation Formula Latex Code and Summary

Mechanic Oscillations

Tags: #physics #mechanic #oscillations

Equation

$$m\ddot{x}=F(t)-k\dot{x}-Cx \\ F(t)=\hat{F}\cos(\omega t) \\ -m\omega^2 x=F-Cx-ik\omega x \\ \omega_0^2=C/m \\ x=\frac{F}{m(\omega_0^2-\omega^2)+ik\omega} \\ \dot{x}=\frac{F}{i\sqrt{Cm}\delta+k} \\ \delta=\frac{\omega}{\omega_0}-\frac{\omega_0}{\omega} \\ Z=F/\dot{x} \\ Q=\frac{\sqrt{Cm}}{k}$$

Latex Code

                                 m\ddot{x}=F(t)-k\dot{x}-Cx \\ 
            F(t)=\hat{F}\cos(\omega t) \\ 
            -m\omega^2 x=F-Cx-ik\omega x \\ 
            \omega_0^2=C/m \\ 
            x=\frac{F}{m(\omega_0^2-\omega^2)+ik\omega} \\ 
            \dot{x}=\frac{F}{i\sqrt{Cm}\delta+k} \\ 
            \delta=\frac{\omega}{\omega_0}-\frac{\omega_0}{\omega} \\ 
            Z=F/\dot{x} \\
            Q=\frac{\sqrt{Cm}}{k}

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Introduction

$m\ddot{x}=F(t)-k\dot{x}-Cx \\ F(t)=\hat{F}\cos(\omega t) \\ -m\omega^2 x=F-Cx-ik\omega x \\ \omega_0^2=C/m \\ x=\frac{F}{m(\omega_0^2-\omega^2)+ik\omega} \\ \dot{x}=\frac{F}{i\sqrt{Cm}\delta+k} \\ \delta=\frac{\omega}{\omega_0}-\frac{\omega_0}{\omega} \\ Z=F/\dot{x} \\ Q=\frac{\sqrt{Cm}}{k}$

Latex Code

            m\ddot{x}=F(t)-k\dot{x}-Cx \\ 
            F(t)=\hat{F}\cos(\omega t) \\ 
            -m\omega^2 x=F-Cx-ik\omega x \\ 
            \omega_0^2=C/m \\ 
            x=\frac{F}{m(\omega_0^2-\omega^2)+ik\omega} \\ 
            \dot{x}=\frac{F}{i\sqrt{Cm}\delta+k} \\ 
            \delta=\frac{\omega}{\omega_0}-\frac{\omega_0}{\omega} \\ 
            Z=F/\dot{x} \\
            Q=\frac{\sqrt{Cm}}{k}

Explanation

Latex code for the Mechanic Oscillations. I will briefly introduce the notations in this formulation.

$C$ : Construction of spring with constant
$k$ : Damping constant
$F(t)$ : Periodic force
$\dot{x}$ : Velocity
$Z=F/\dot{x}$ : Impedance of the system
$Q=\frac{\sqrt{Cm}}{k}$ : The quality of the system
Velocity resonance frequency: The frequency with minimal $|Z|$

Discussion

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Patricia Johnson

The tension is high, but I'm hopeful I'll pass this exam.

2023-03-26 00:00
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Best Wishes.

2023-04-08 00:00:00.0
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Troy Tucker

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2023-02-17 00:00
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Rose Rogers reply to Troy Tucker

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2023-02-27 00:00:00.0
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Diana George

I'm geared up to pass this exam.

2023-02-20 00:00
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Kathy Bailey reply to Diana George

Gooood Luck, Man!

2023-02-26 00:00:00.0
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