-2
\documentclass[10pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{newtxtext,newtxmath} 
\usepackage[vmargin=3cm, left=4cm, right=2cm]{geometry}
\usepackage{graphicx}
\usepackage{amsmath,amssymb}
\begin{document}
    \section*{Forced Oscillator}
    \subsection*{Introduction}
    In damped harmonic oscillator we see that due to air friction damping occurs and after some time the oscillation dies out.So if we want that our oscillation dosen't dies out we need to provide external force to it so as to maintain the oscillation against the damping due to air.
    \section*{Theory}
    In damped harmonic oscillator the amplitude of the oscillation goes on decreasing as the energy is dissipated in overcoming the resistive force.If the oscillations are to be maintained, energy must be supplied to the system to make up for the losses.If this is done by applying an external driving force which is time independent,the oscillations are called the force oscillations.The equation of motion of such system is written as
    \begin{equation*} m\ddot{x}+2m\mu\dot{x}+kx=f(t) \end{equation*}
    \begin{equation} \ddot{x}+2\mu\dot{x}+\omega^{2}x=\frac{f(t)}{m} \end{equation}
    If the driving force is sinusoidal, we have
    \begin{equation} f(t)=f_{0}\sin\omega t \end{equation}
    where \begin{math} \nu=\frac{\omega}{2\pi} \end{math} is the frequency of the applied force. \\
     Then the equation becomes \\
     \begin{equation} \ddot{x}+2\mu \dot{x}+\omega^{2}x=\dfrac{f_{0}\sin\omega t}{m} \end{equation}
    This is a linear inhomogenous differential equation and can be solved in different ways. The simplest method is to assume a solution\\
    \begin{equation} x=A\sin(\omega t-\theta) \end{equation}
    and determine the values of constants A and $\theta$.\\
    Then,substituing solution(5) in equation(2) we get\\
    \begin{equation*} -A\omega^{2}\sin(\omega t- \theta)+2\mu A \omega\cos{\omega t- \theta}+\omega_{0}^2 A \sin(\omega t-\theta)=\frac{f_{0}\sin\omega t}{m} \end{equation*}
    \begin{equation*} = \frac{f_{0}\sin((\omega t-\theta)+\theta)}{m} \end{equation*} 
    \begin{equation*}  = \frac{f_{0}\sin(\omega t-\theta)}{m}\cos{\theta}+\frac{f_{0}\cos(\omega t-\theta)}{m}\sin{\theta}  \end{equation*}
    Comparing the coefficent of $\sin$($\omega$ t-$\theta$) and $\cos${$\omega$ t -$\theta$} on the two sides of the equation, we get\\
    \begin{equation} -A\omega^{2}+\omega_{0}^2 A=\frac{f_{0}\cos\theta}{m} \end{equation}
    \begin{equation} 2\mu A\omega=\frac{f_{0}\sin{\theta}{m} \end{equation} 
    From equations (6) and (7) we get expressions for phase angle $\theta$ and amplitude A:\\
    \begin{equation} \tan{\theta} = \frac{2\mu\omega}{(\omega^{2}-\omega_{0}^2)} \end{equation}
    and \begin{equation} A = \frac{f_0/m}{\sqrt{(\omega^{2}-\omega_{0}^2)^2+4\mu^{2}\omega^{2}}} \end{equation}
    With these constants,solution(to write equation no. here) is
    \begin{equation} x=\frac{f_0/m}{\sqrt{(\omega^{2}-\omega_{0}^2)^2+4\mu^{2}\omega^{2}}} \sin{(\omega t-\theta)} \end{equation}
    This solution of damped forced oscillator shows that the particle performs forced oscillationshaving amplitude and phase dependent on damping and forcing frequency $\omega$.\\
    Equation (12) gives that part of the solution which has the same periodicity as that of applied force. This corresponds to the inhomogenous or particular solution. The other solution which corresponds to the homogenous equation is obtained in the previous article by solving \begin{equation} \ddot{x}+2\mu\dot{x}+\omega^{2}x=0 \end{equation}. But, this leads to oscillations that are damped. This is many times called the transient solution. The particular solution does not die out since oscillations are sustained by the external oscillating force. This solution is, therfore called the steady state solution. In forced oscillations we are mainly intrested in the oscillations given by equation (12). From equation (10) it is clear that if angular frequency $\omega$ of the applied force is increased from zero to natural frequency $\omega_{0}$ of the system, the phase angle between applied force and displacement increases from zero to 90$^{\circ}$. If $\omega$$>$$\omega_{0}$, $\tan$$\theta$ is negative. Hence, $\theta$ takes value greater than 90$^{\circ}$ and approaches 180$^{\circ}$ as $\omega$ becomes very large as compared to $\omega_{0}$. In other words, when the frequency of the applied force is very large, the displacement and the applied force are in opposition.\\
    graph space\\
    The variation in phase difference $\theta$ with frequency of the applied force is also influenced by the value of $\mu$. This change is shown in figure above. If there exists no damping, angle $\theta$ changes abruptly from 0 to 180$^{\circ}$ when frequency $\omega$ of the applied force becomes equal to he natural frequency of the system.\\
    \subsection*{Displacement (Amplitude) Resonance}\\ \\
    Amplitude A of the forced oscillations varies with the frequency of the applied force. There exists some frequency called the resonance frequency at which the amplitude becomes maximum. At this stage maximum transfer of energy occurs from the driving force to the driven force to the driven system. This is called resonance. The frequency t the time of resonance, i.e., the resonant frequency will obviously correspond to the maximum value of amplitude A. Thus, A is maximum when \frac{dA}{d$\omega$}=0 and \frac{d^2A}{d$\omega$^2}$<$0. This condition leads to\\
    \begin{equation*} 2(\omega^{2}-\omega_{0}^2)=4\mu^2 \end{equation*}
    \begin{equation*} \omega^2=\omega_{0}^2-2\mu^2 \end{equation*}
    with the condition \begin{equation*} \omega_0^2>2\mu^2 \end{equation*}.
    Hence, the resonant frequency is\\
    \begin{equation} \omega_r = \sqrt{\omega^2-2\mu^2} \end{equation}
    The maximum value of the amplitude is obtained by putting $\omega$=$\omega_r$ in equation 11. Thus
    \begin{equation*} A_m_a_x = \frac{f_0/m}{\sqrt{4\mu^4+4\mu^2(\omega_0^2-2\mu^2)}} \end{equation*}
    \begin{equation} \frac{f_0/m}{2\mu\sqrt{\omega_0^2-\mu^2}} \end{equation}
    This shows that the value of $A_m_a_x$ depends upon $\mu$, the damping factor. The variation of A with $\omega$ for various values of $\mu$ as given by equation (11) is represented by the curves of figure below. Thus, for undamped oscillations ($\mu$=0) when forcing frequency $\omega$=$omega_0$, the natural frequency, the amplitude of the oscillations becomes very large, theoretically infinte.\\
    Graph area\\
    However, due to damping, which is always present in any naturally oscillation system, the amplitude becomes large but finite.
    \subsection{Sharpness of Resonance}
    We have seen that the displacement resonance takes place at a particular frequency. The resonance curves drawn in figure above show that the smaller the value of the damping factor, the greater is amplitude $A_m_a_x$ or$(A\omega)_m_a_x$ and steeper is the resonance curve. Thus, the steepness or sharpness of the resonance depends entirely upon the magnitude of the damping factor. We can define a measure for the sharpness as follows:\\
    Cosnider a resonance curve for the displacement resonance at figure (). The amplitude is $A_m_a_x$ at frequency $\omega_r$ and becomes \frac{A_m_a_x}{2} at frequency\\
    \begin{equation} \omega_h = \omega_r \pm \Delta \end{equation}
    where quantity\\
    \begin{equation} \Delta =|\omega_h-\omega_r| \end{equation}
    is called the half-width of the resonance curve. It can be used to measure the sharpness of resonance. Thus, if $\Delta$ is small, the resonance is sharp. This can be seen from the following considerations. When amplitude is half of $A_m_a_x$
    \begin{equation*} \frac{A_m_a_x}{2}= A\end{equation*}  ( at \begin{equation*} \omega_h=\omega_r\pm\Delta)   \end{equation*}
    or from equation () and ()
    \begin{equation*} \frac{f_0/m}{2\sqrt{4\mu^2\omega_0^2-4\mu^4}}  = \frac{f_0/m}{\sqrt{(\omega^{2}-\omega_{h}^2)^2+4\mu^{2}\omega_h^{2}}} \end{equation*}
    Hence, solving thid equation, we find that the half maximum value of A occurs at a frequency given by \\
    \begin{equation*} \omega_h^2=\omega_r^2\pm2\sqrt{3}\mu\sqrt{(\omega_0^2-\mu^2)}\end{equation*}
    or for small values of $\mu$, ($\omega$_r$\approx$$\omega_0$)\\
    \begin{equation*} \omega_{h}=\omega_r\simeq\pm\sqrt{3}\mu\end{equation*}
    where we have neglected the term of the order of $\mu^2$.\\
    Thus\\
    \begin{equation} |\omega_{h}-\omega_r|=\Delta=\mu\sqrt{3} \end{equation}.
    This shows that for small $\mu$, $\omega_{h}$ is close to $\omega_r$, $\Delta$ is small and the resonance is sharp.\\
    Graph area\\
    The sharpness of resonance is also measured in terms of another quantity called the Q-factor of the system. It is defined as 2$\pi$ times the ratio of the average energy stored in the system to the energy dissipated per cycle by the applied force.\\
    Hence\\
    \begin{equation*} Q=\dfrac{2\pi<\frac{1}{2}m\dot{x}^2+\frac{1}{2}m\omega_0^2x^2>}{\tau<2m\mu\dot{x}^2>} \end{equation*}
    for a damped simple harmonic oscillator.\\
    Here, we have used the notation$<$$>$ to denote time average of the quantity.\\
    The average value of $\dot{x^2}$ is\\
    \begin{equation*} <\dot{x}>=<A^2\omega^2\cos^2(\omega t- \theta)> \end{equation*}
    \begin{equation*} =\frac{1}{2}A^2\omega^2 \end{equation*}
    Similarly, \begin{equation*} <x^2>=\frac{A^2}{2} \end{equation*}
    where we have used the fact that the time average of $\sin^2${($\omega$ t- $\theta$)} and $\cos^2${($\omega$ t- $\theta$)} over a cycle is \frac{1}{2}. Hence, the Q-factor is given by\\
    \begin{equation} Q=\frac{\omega^2+\omega_{0}^2}{4\mu\omega} \end{equation}
    where we used $\tau$=\frac{2$\pi$}{$\omega$}.
    Near resonance \begin{equation*} \omega \approx \omega_{0} \end{equation*} \begin{equation*} Q=\frac{\omega_{0}}{2\pi} \end{equation*}
    This shows that the value of Q is greater for smaller values of $\mu$ and corresponds to greater sharpness of the resonance. Hence, we conclude that the gretaer the value of Q, The greater is the sharpness of resonance.







\end{document}

I know the question i asked makes no sense and the tag i made, but i am unable to figure out the probelm as its just showing the hbox error and pdf is not being created. i know the \ command is the problem but please help me out in this anyone. by correcting my code...actually we cant or i dont know can send the tex file.please help me

9
  • 3
    I downvoted because a wall of code isn't helpful. This post has no question, just code. May 18 '18 at 7:09
  • 4
    Welcome to TeX.SX! Please reduce your code to a minimal working example, starting with \documentclass and ending with \end{document}.
    – Bobyandbob
    May 18 '18 at 7:10
  • 1
    Don't use \\ for ending paragraphs, but a blank line. Check the manual of amsmath for multiline displays.
    – egreg
    May 18 '18 at 7:14
  • 2
    the underfull hbox warning (it is not an error) comes from things like becomes \\ \begin{equation} Just remove all those \\ they are all wrong. May 18 '18 at 8:02
  • 2
    The first actual error is in the equation that starts with 2\mu A\omega (unclosed brace), but really, there are so many things wrong with this code that it is not reasonable to ask members here to fix it. May 18 '18 at 8:32
2

I have been bored and did some cleanup of your code:

  • Removed all the \\ before your equation environment which lead to the hbox errors as mentioned by @DavidCarlisle.
  • Removed duplicate package includes and fixed some formatting and typos.
  • Added the xfrac package for the sfrac command. Added the hyperref and cleveref package for cross-references.
  • Combined equation environments.
  • Added two cross-referencing examples. As your equation numbers seem somehow off, I did not apply this to the whole document.

The code probably is still not perfect, but does not produce the warnings and errors raised by your original code.

\documentclass[10pt,a4paper]{article}

\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{newtxtext,newtxmath} 
\usepackage[vmargin=3cm, left=4cm, right=2cm]{geometry}

% Better fractions inside fractions.
\usepackage{xfrac}

% For crossreferences.
\usepackage[hidelinks]{hyperref}
\usepackage{cleveref}

\begin{document}
    \section*{Forced Oscillator}

    \subsection*{Introduction}

    In damped harmonic oscillator we see that due to air friction damping occurs and after some time the oscillation dies out. So if we want that our oscillation dosen't dies out we need to provide external force to it so as to maintain the oscillation against the damping due to air.

    \section*{Theory}

    In damped harmonic oscillator the amplitude of the oscillation goes on decreasing as the energy is dissipated in overcoming the resistive force. If the oscillations are to be maintained, energy must be supplied to the system to make up for the losses. If this is done by applying an external driving force which is time independent, the oscillations are called the force oscillations. The equation of motion of such system is written as
    \begin{align}
        m \ddot{x} + 2m \mu \dot{x} + kx &= f(t) \nonumber \\
       \ddot{x} + 2 \mu \dot{x} + \omega^{2} x = \frac{f(t)}{m}.
    \end{align}

    If the driving force is sinusoidal, we have
    \begin{equation}
        \label{eq:eq2}
        f(t) = f_{0} \sin\omega t
    \end{equation}
    where $ \nu = \frac{\omega}{2\pi}$  is the frequency of the applied force.

    Then the equation becomes
    \begin{equation}
        \ddot{x} + 2 \mu \dot{x} + \omega^{2} x = \frac{f_{0} \sin\omega t}{m}.
    \end{equation}

    This is a linear inhomogenous differential equation and can be solved in different ways. The simplest method is to assume a solution
    \begin{equation}
        x = A \sin(\omega t-\theta)
    \end{equation}
    and determine the values of constants A and $\theta$.

    Then, substituting solution(5) in \cref{eq:eq2} we get
    \begin{align*}
        -A \omega^{2} \sin(\omega t- \theta) + 2 \mu A \omega \cos(\omega t- \theta) + \omega_{0}^2 A \sin(\omega t-\theta) &= \frac{f_{0} \sin\omega t}{m} \\
        &= \frac{f_{0} \sin((\omega t-\theta)+\theta)}{m} \\
        &= \frac{f_{0} \sin(\omega t-\theta)}{m} \cos{\theta} + \frac{f_{0} \cos(\omega t-\theta)}{m} \sin\theta.
    \end{align*}

    Comparing the coefficent of $\sin(\omega t-\theta)$ and $\cos(\omega t -\theta)$ on the two sides of the equation, we get
    \begin{align}
        -A \omega^{2} + \omega_{0}^2 A &= \frac{f_{0} \cos\theta}{m} \\
        \label{eq:eq6}
        2 \mu A \omega &= \frac{f_{0} \sin\theta}{m}
    \end{align}

    From equations \labelcref{eq:eq6} and (7) we get expressions for phase angle $\theta$ and amplitude A:
    \begin{equation}
        \tan\theta = \frac{2\mu\omega}{(\omega^{2} - \omega_{0}^2)}
    \end{equation}
    and
    \begin{equation}
        A = \frac{f_0/m}{\sqrt{(\omega^{2}-\omega_{0}^2)^2+4\mu^{2}\omega^{2}}}
    \end{equation}

    With these constants, solution(to write equation no. here) is
    \begin{equation}
        x = \frac{f_0/m}{\sqrt{(\omega^{2}-\omega_{0}^2)^2+4\mu^{2}\omega^{2}}} \sin{(\omega t-\theta)}
    \end{equation}

    This solution of damped forced oscillator shows that the particle performs forced oscillationshaving amplitude and phase dependent on damping and forcing frequency $\omega$.

    Equation (12) gives that part of the solution which has the same periodicity as that of applied force. This corresponds to the inhomogenous or particular solution. The other solution which corresponds to the homogenous equation is obtained in the previous article by solving
    \begin{equation}
        \ddot{x} + 2 \mu \dot{x} + \omega^{2} x = 0.
    \end{equation}

    But, this leads to oscillations that are damped. This is many times called the transient solution. The particular solution does not die out since oscillations are sustained by the external oscillating force. This solution is, therfore called the steady state solution. In forced oscillations we are mainly intrested in the oscillations given by equation (12). From equation (10) it is clear that if angular frequency $\omega$ of the applied force is increased from zero to natural frequency $\omega_{0}$ of the system, the phase angle between applied force and displacement increases from zero to 90$^{\circ}$. If $\omega > \omega_{0}$, $\tan\theta$ is negative. Hence, $\theta$ takes value greater than 90$^{\circ}$ and approaches 180$^{\circ}$ as $\omega$ becomes very large as compared to $\omega_{0}$. In other words, when the frequency of the applied force is very large, the displacement and the applied force are in opposition.

    graph space

    The variation in phase difference $\theta$ with frequency of the applied force is also influenced by the value of $\mu$. This change is shown in figure above. If there exists no damping, angle $\theta$ changes abruptly from 0 to 180$^{\circ}$ when frequency $\omega$ of the applied force becomes equal to he natural frequency of the system.

    \subsection*{Displacement (Amplitude) Resonance}

    Amplitude $A$ of the forced oscillations varies with the frequency of the applied force. There exists some frequency called the resonance frequency at which the amplitude becomes maximum. At this stage maximum transfer of energy occurs from the driving force to the driven force to the driven system. This is called resonance. The frequency $t$ the time of resonance, i.\,e., the resonant frequency will obviously correspond to the maximum value of amplitude $A$. Thus, $A$ is maximum when $\frac{dA}{d\omega}=0$ and $\frac{d^2A}{d\omega^2}<0$. This condition leads to
    \begin{align*}
        2 (\omega^{2} - \omega_{0}^2) &= 4 \mu^2 \\
        \omega^2 & =\omega_{0}^2 - 2 \mu^2
    \end{align*}
    with the condition 
    \begin{equation*}
        \omega_0^2 > 2 \mu^2.
    \end{equation*}

    Hence, the resonant frequency is
    \begin{equation}
        \omega_r = \sqrt{\omega^2-2\mu^2}.
    \end{equation}

    The maximum value of the amplitude is obtained by putting $\omega=\omega_r$ in equation 11. Thus
    \begin{align}
        A_{\max} &= \frac{\sfrac{f_0}{m}}{\sqrt{4 \mu^4 + 4 \mu^2 (\omega_0^2 - 2 \mu^2)}} \nonumber \\
        &= \frac{\sfrac{f_0}{m}}{2 \mu \sqrt{\omega_0^2 - \mu^2}}
    \end{align}

    This shows that the value of $A_{\max}$ depends upon $\mu$, the damping factor. The variation of $A$ with $\omega$ for various values of $\mu$ as given by equation (11) is represented by the curves of figure below. Thus, for undamped oscillations ($\mu=0$) when forcing frequency $\omega=\omega_0$, the natural frequency, the amplitude of the oscillations becomes very large, theoretically infinite.

    Graph area

    However, due to damping, which is always present in any naturally oscillation system, the amplitude becomes large but finite.

    \subsection*{Sharpness of Resonance}

    We have seen that the displacement resonance takes place at a particular frequency. The resonance curves drawn in figure above show that the smaller the value of the damping factor, the greater is amplitude $A_{\max}$ or $(A\omega)_{\max}$ and steeper is the resonance curve. Thus, the steepness or sharpness of the resonance depends entirely upon the magnitude of the damping factor. We can define a measure for the sharpness as follows:

    Consider a resonance curve for the displacement resonance at figure (). The amplitude is $A_{\max}$ at frequency $\omega_r$ and becomes $\frac{A_{\max}}{2}$ at frequency
    \begin{equation}
        \omega_h = \omega_r \pm \Delta
    \end{equation}
    where quantity
    \begin{equation}
        \Delta = |\omega_h - \omega_r|
    \end{equation}
    is called the half-width of the resonance curve. It can be used to measure the sharpness of resonance. Thus, if $\Delta$ is small, the resonance is sharp. This can be seen from the following considerations. When amplitude is half of $A_{\max}$
    \begin{equation*}
        \frac{A_{\max}}{2} = A
    \end{equation*}
    (at $\omega_h = \omega_r \pm \Delta$) or from equation () and ():
    \begin{equation*}
        \frac{\sfrac{f_0}{m}}{2 \sqrt{4 \mu^2 \omega_0^2 - 4 \mu^4}} = \frac{\sfrac{f_0}{m}}{\sqrt{(\omega^{2} - \omega_{h}^2)^2 + 4 \mu^{2} \omega_h^{2}}}
    \end{equation*}

    Hence, solving third equation, we find that the half maximum value of $A$ occurs at a frequency given by
    \begin{equation*}
        \omega_h^2 = \omega_r^2 \pm 2 \sqrt{3} \mu \sqrt{(\omega_0^2 - \mu^2)}
    \end{equation*}
    or for small values of $\mu$, ($\omega_r \approx \omega_0$)
    \begin{equation*}
        \omega_{h} = \omega_r \simeq \pm \sqrt{3} \mu
    \end{equation*}
    where we have neglected the term of the order of $\mu^2$.

    Thus
    \begin{equation}
        |\omega_{h} - \omega_r| = \Delta = \mu\sqrt{3}.
    \end{equation}

    This shows that for small $\mu$, $\omega_{h}$ is close to $\omega_r$, $\Delta$ is small and the resonance is sharp.

    Graph area

    The sharpness of resonance is also measured in terms of another quantity called the Q-factor of the system. It is defined as $2\pi$ times the ratio of the average energy stored in the system to the energy dissipated per cycle by the applied force.

    Hence
    \begin{equation*}
        Q = \dfrac{2 \pi \left\langle \frac{1}{2} m \dot{x}^2 + \frac{1}{2} m \omega_0^2 x^2 \right\rangle}{\tau \left\langle 2m \mu \dot{x}^2 \right\rangle}
    \end{equation*}
    for a damped simple harmonic oscillator.

    Here, we have used the notation $\langle \cdot \rangle$ to denote time average of the quantity.

    The average value of $\dot{x}^2$ is
    \begin{align*}
        \langle \dot{x} \rangle &= \left\langle A^2 \omega^2 \cos^2 (\omega t- \theta) \right\rangle \\
        &= \frac{1}{2} A^2 \omega^2.
    \end{align*}

    Similarly,
    \begin{equation*}
        \left\langle x^2 \right\rangle = \frac{A^2}{2}
    \end{equation*}
    where we have used the fact that the time average of $\sin^2 (\omega t- \theta)$ and $\cos^2 (\omega t- \theta)$ over a cycle is $\frac{1}{2}$. Hence, the Q-factor is given by
    \begin{equation}
        Q = \frac{\omega^2 + \omega_{0}^2}{4 \mu \omega}
    \end{equation}
    where we used $\tau = \frac{2\pi}{\omega}$.

    Near resonance $\omega \approx \omega_{0}$
    \begin{equation*}
        Q = \frac{\omega_{0}}{2\pi}.
    \end{equation*}

    This shows that the value of $Q$ is greater for smaller values of $\mu$ and corresponds to greater sharpness of the resonance. Hence, we conclude that the greater the value of $Q$, the greater is the sharpness of resonance.
\end{document}
3
  • 1
    If the equation is part of the sentence, you should not leave a blank line, as it is not a separate paragraph.
    – Johannes_B
    May 18 '18 at 10:17
  • @Johannes_B Thanks for pointing out - must have somehow missed this after search-and-replace.
    – epR8GaYuh
    May 18 '18 at 12:20
  • 4
    Your generosity know no bounds! May 18 '18 at 12:31

Not the answer you're looking for? Browse other questions tagged or ask your own question.