Chapter 5, Problem 5.45P

(a)

To determine

Find the rate at which $F\left(t\right)$ does work and shot that $\langle P\rangle =m\beta {\omega }^2{A}^2$

Answer to Problem 5.45P

The rate at which $F\left(t\right)$ does the work is $P\left(t\right)=F_{0}A\omega {\cos }^2\omega t\sin \delta -F_{0}A\omega \cos \omega t\sin \omega t\cos \delta$

And average power is $\langle P\rangle =m\beta {\omega }^2{A}^2$.

Explanation of Solution

Write the mathematical equation of force driven to the oscillator

    $F\left(t\right)=F_0\cos \omega t$        (I)

Here $F_0$ is the amplitude of the force and $t$ is the time.

The equation for displacement of damped oscillation is given by

    $x\left(t\right)=A\cos \left(\omega t-\delta \right)$        (II)

Here $A$ is the amplitude.

Write the equation for amplitude of an oscillator driven by a sinusoidal force with variable frequency

  $A^2=\frac{f_0{}^2}{{\left({\omega }_0{}^2-{\omega }^2\right)}^2+4{\beta }^2{\omega }^2}$

Differentiate (II)

  $\begin{array}{c}\frac{d}{dt}x\left(t\right)=\frac{d}{dt}\left(A\cos \left(\omega t-\delta \right)\right)\\ v\left(t\right)=-A\omega \sin \left(\omega t-\delta \right)\end{array}$

The rate at which force does work is equal to the product of force and velocity.

Then,

    $\begin{array}{c}P\left(t\right)=F\left(t\right)v\left(t\right)\\ =F_0\cos \omega t\left(-A\omega \sin \left(\omega t-\delta \right)\right)\\ =-F_{0}A\omega \cos \omega t\sin \left(\omega t-\delta \right)\end{array}$

Since $\sin \left(a-b\right)=\sin a\cos b-\cos a\sin b$

    $\begin{array}{c}P\left(t\right)=-F_{0}A\omega \cos \omega t\left(\sin \omega t\cos \delta -\cos \omega t\sin \delta \right)\\ =F_{0}A\omega {\cos }^2\omega t\sin \delta -F_{0}A\omega \cos \omega t\sin \omega t\cos \delta \end{array}$

The average rate of work done is given by

    $\begin{array}{c}\langle P\rangle =\frac{1}{\tau }{\displaystyle {\int }_0^{\tau }P\left(t\right)}dt\\ =\frac{1}{\tau }{\displaystyle {\int }_0^{\tau }\left(F_{0}A\omega {\cos }^2\omega t\sin \delta -F_{0}A\omega \cos \omega t\sin \omega t\cos \delta \right)}dt\\ =\frac{1}{\tau }\left[F_{0}A\omega \sin \delta {\displaystyle {\int }_0^{\tau }{\cos }^2\omega t-F_{0}A\omega \cos \delta {\displaystyle {\int }_0^{\tau }\cos \omega t\sin \omega t}}\right]\end{array}$

For one complete cycle, ${\displaystyle {\int }_0^T\sin \omega t\cos \omega tdt=0}$ and ${\displaystyle {\int }_0^T{\cos }^2\omega t=\frac{\tau }{2}}$

    $\begin{array}{c}\langle P\rangle =\left[F_{0}A\omega \sin \delta \left(\frac{1}{\tau }\right)\left(0\right)-F_{0}A\omega \cos \delta \left(\frac{1}{\tau }\right)\left(\frac{\tau }{2}\right)\right]\\ =\frac{1}{2}{F}_{0}A\omega \sin \delta \end{array}$

But $\tan \delta =\frac{2\beta \omega }{{\omega }_0{}^2-{\omega }^2}$

So,

    $\sin \delta =\frac{2\beta \omega }{\sqrt{{\left({\omega }_0{}^2-{\omega }^2\right)}^2+4{\beta }^2{\omega }^2}}$

If number of cycles completed by the oscillator is $m$,

    $\begin{array}{c}F\left(t\right)=mf\left(t\right)\\ F_0\cos \omega t=m{f}_0\cos \omega t\\ F_0=m{f}_0\end{array}$

Then,

    $\begin{array}{c}\langle P\rangle =\frac{1}{2}m{f}_{0}A\omega \left(\frac{2\beta \omega }{\sqrt{{\left({\omega }_0{}^2-{\omega }^2\right)}^2+4{\beta }^2{\omega }^2}}\right)\\ =m{\omega }^2\beta A\frac{f_0}{\sqrt{{\left({\omega }_0{}^2-{\omega }^2\right)}^2+4{\beta }^2{\omega }^2}}\\ =m{\omega }^2\beta A^2\end{array}$

Conclusion:

The rate at which $F\left(t\right)$ does the work is $P\left(t\right)=F_{0}A\omega {\cos }^2\omega t\sin \delta -F_{0}A\omega \cos \omega t\sin \omega t\cos \delta$

And average power is $\langle P\rangle =m\beta {\omega }^2{A}^2$.

(b)

To determine

Verify that the average rate of work done over any number of complete cycles is equal to the average rate at which energy is lost to the resistive force.

Answer to Problem 5.45P

The average rate of work done over any number of complete cycles is equal to the average rate at which energy is lost to the resistive force.

Explanation of Solution

Average rate at which energy lost to the resistive force is,

  $P\text{'}=\left(\text{resistive}\text{force}\right)\left(\text{velocity}\right)$

Resistive force is proportional to the velocity

    $\begin{array}{c}\text{resistive}\text{force}=bv\\ =2\beta mv\end{array}$

Substitute $2\beta mv$ for $\text{resistive}\text{force}$ and $dx/dt$ for $\text{velocity}$

    $\begin{array}{c}P\text{'}=2\beta m\left(\frac{dx}{dt}\right)\left(\frac{dx}{dt}\right)\\ =2\beta m{\left(\frac{dx}{dt}\right)}^2\\ =2\beta m{\left(\frac{d}{dt}\left(A\cos \left(\omega t-\delta \right)\right)\right)}^2\\ =2m{A}^2{\omega }^2\beta {\sin }^2\left(\omega t-\delta \right)\end{array}$

Average rate at which energy lost to the resistive force is,

    $\begin{array}{c}\langle P\text{'}\rangle =\frac{1}{\tau }{\displaystyle {\int }_0^{\tau }p\text{'}dt}\\ =\frac{1}{\tau }{\displaystyle {\int }_0^{\tau }2m{A}^2{\omega }^2\beta \cdot {\sin }^2\left(\omega t-\delta \right)dt}\\ =2m{A}^2{\omega }^2\beta \frac{1}{\tau }{\displaystyle {\int }_0^{\tau }{\sin }^2\left(\omega t-\delta \right)}dt\end{array}$

The value of integration ${\displaystyle {\int }_0^{\tau }{\sin }^2\left(\omega t-\delta \right)dt=\frac{\tau }{2}}$

Then,

    $\begin{array}{c}\langle P\text{'}\rangle =2m{A}^2{\omega }^2\beta \frac{1}{\tau }\left(\frac{\tau }{2}\right)\\ =2m{A}^2{\omega }^2\beta \\ =\langle P\rangle \end{array}$

Conclusion:

The average rate of work done over any number of complete cycles is equal to the average rate at which energy is lost to the resistive force.

(c)

To determine

Show that $\langle P\rangle$ is maximum when $\omega ={\omega }_0$

Answer to Problem 5.45P

$\langle P\rangle$ is maximum when $\omega ={\omega }_0$ or resonance of power occur at $\omega ={\omega }_0$

Explanation of Solution

The average rate at which the force does the work is

    $\begin{array}{c}\langle P\rangle =m\beta {\omega }^2{A}^2\\ =m\beta {\omega }^2{A}^2\end{array}$

Substitute for $A$

    $\langle P\rangle =m\beta {\omega }^2\frac{f_0{}^2}{{\left({\omega }_0{}^2-{\omega }^2\right)}^2+4{\beta }^2{\omega }^2}$

For maximum value of $\langle P\rangle$, the value in the denominator should be as minimum as $\beta$ and $\omega$, and should be non-zero.

  $\begin{array}{c}{\omega }_0{}^2-{\omega }^2=0\\ {\omega }_0=\omega \end{array}$

Conclusion:

Therefore, $\langle P\rangle$ is maximum when $\omega ={\omega }_0$ or resonance of power occur at $\omega ={\omega }_0$