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In the forced harmonic oscillator the velocity of the oscillator is given as -
$$V=Ap\cos(pt-\varphi)$$ where ,
$p$= the driving angular frequency,
$A$= amplitude of the forced harmonic oscillator.

Thus "$Ap$" together creates what we call the velocity amplitude and it is given as
$$V_0=\frac{f_0p}{\sqrt[2]{(\omega^2-p^2)^2+4b^2p^2}}$$ where ,
$\omega$=the natural frequency of the oscillator,
$b$=damping factor,
$V_0$= the velocity amplitude

Now, the book I am following again mentions something called as the resonant frequency and it is given as - $$V=V_0\cos(pt-\varphi)$$ Its is given that $v$ is maximum i.e resonant velocity occurs ,when $p=\omega$ in $V_0$ but it is not said the $\cos(pt-\varphi)$ should also be 1 in order for $v$ to be maximum. This is one of my doubts

The confusion I am facing is, there is not concrete definition of velocity amplitude and velocity resonance in my book and in the internet too ,what I assume velocity amplitude is that it is the maximum value of velocity that is given to the oscillator by the driving periodic force and the velocity resonance is the maximum velocity of the oscillator but I wanted to be confirmed that I am right or is there something else to these two guys.

So basically I want a proper definition about velocity resonance and velocity amplitude and also why $\cos(pt-\varphi)$ is not considered to be 1 for $v$ to become maximum only $p=\omega$ (in $V_0$) is considered for $V$ to become maximum.

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    $\begingroup$ Please do not post formulae as screenshots, but use MathJax instead. $\endgroup$ Sep 14 at 10:31
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Velocity resonance is the condition when your system has the highest velocity amplitude and velocity amplitude is the highest velocity your system can achieve during the oscillation.

In a forced harmonic oscillator, in the steady state, the solution for displacement is of the form $x = A\cos(\omega t - \psi)$ where $\omega$ is the driving frequency and A is the oscillation amplitude. $\psi$ and A have the following form:

$A = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + \omega^2\gamma^2}}$

$\psi = \tan^{-1}(\frac{\omega\gamma}{\omega_0^2 - \omega^2})$

I have used a slightly different notation. $\omega_0 = \sqrt{\frac{k}{m}}$ is the natural frequency of the oscillator, $\gamma = 2b$ (as per your notation) is the damping rate and $F_0 = mf_0$ (again as per your notation) is the driving force.

Now if you were to differentiate A with $\omega$ and equate it to zero, you would get the value of $\omega$ which maximizes A. This is known as amplitude resonance and happens at $\omega = (\omega_0^2 - \frac{\gamma^2}{2})^{1/2}$

Next, from simple differentiation we have $v = -A\omega\sin(\omega t - \psi)$. the velocity amplitude is $A\omega$. For a velocity resonance, we must choose an $\omega$ such that $A\omega$ is maximized. The same method as before can be followed to maximize $V_0 = A\omega$ and it turns out that it is maximized for $\omega = \omega_0$. You would also find that $\psi = \pi/2$. This means that velocity resonance occurs when the driving force drives the system at its natural frequency and has a $pi/2$ phase difference with the oscillations.

The reason why the sine term (or the cosine term as in your case) is not maximized is that the velocity changes with time. Even when the velocity amplitude is maximum, there will be a time ($t=\frac{\psi}{\omega}$) when velocity is $0$. The amplitude simply denotes the maximum velocity that can be achieved. If you were to drive the system at a frequency different than $\omega_0$, the highest velocity your system can achieve during the oscillation will be less than what it can achieve if driven at $\omega_0$

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  • $\begingroup$ Everything is understandable but the explantion of why cos term is not considered during maximizing the velocity is not quit understandable,thank you for your efforts sir.... $\endgroup$
    – PATRICK
    Sep 15 at 13:06
  • $\begingroup$ The cos term is time-dependent. If you maximize it you are finding a "particular" time when it's maximum. Cosine value is anyway between -1 and 1 so it is implicit that multiplying term will be the maximum value. $\endgroup$
    – Neel
    Sep 16 at 10:43
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this is your differential equation

$${\frac {d^{2}}{d{t}^{2}}}x \left( t \right) +{p}^{2}x \left( t \right) +2\,b{\frac {d}{dt}}x \left( t \right) ={\frac {f_{{0}}}{m}} $$

transformed to Laplace domain with the initial conditions $~x(0)=0~,\dot x(0)=0~$ you obtain

$$x(s)={\frac {f_{{0}}}{m\,s \left( {s}^{2}+{p}^{2}+2\,b\,s \right) }}$$

the velocity $v=\dot x\,\Rightarrow\,v(s)=s\,x(s)$ thus

$$v(s)={\frac {f_{{0}}}{m\, \left( {s}^{2}+{p}^{2}+2\,b\,s \right) }}$$

substitute $s\mapsto i\,\omega$ and obtain the amplitude $$|v(i\,\omega)|={\frac {f_{{0}}}{m\sqrt { \left( {\omega}^{2}-{p}^{2} \right) ^{2}+4\, {b}^{2}{\omega}^{2}}}} $$

you get resonance at

$$\left( {\omega}^{2}-{p}^{2} \right) ^{2}+4\,{b}^{2}{\omega}^{2}=0\\ \Rightarrow\\ \omega_r=\sqrt {{p}^{2}-2\,{b}^{2}\,\pm\,2\sqrt {-{p}^{2}{b}^{2}+{b}^{4}}}$$

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  • $\begingroup$ I didn't quit get you sir... $\endgroup$
    – PATRICK
    Sep 15 at 13:04
  • $\begingroup$ I don’t think that your equation fir the magnitude ist correct $\endgroup$
    – Eli
    Sep 15 at 13:42
  • $\begingroup$ It is correct sir ,I have not written it myself it is from a book. $\endgroup$
    – PATRICK
    Sep 15 at 14:30

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