Chapter 14, Problem 105P

(a)

To determine

The average power delivered by a driving force.

Explanation of Solution

Formula used:

Write the expression for average power delivered in driving an oscillator.

  $\begin{array}{c}P=\overrightarrow{F}\cdot \overrightarrow{v}\\ =Fv\cos \theta \end{array}$   ........ (1)

Here, $P$ is the average power, $F$ is the driving force of oscillator, $v$ is the driving velocity and $\theta$ is the angle between $F$ and $v$ .

Calculation:

Write the expression for the force as a function of time.

  $F=F_0\cos \omega t$   ........ (2)

Here, $F$ is the driving force of oscillator, $F_0$ is the maximum force, $\omega$ is the angular frequency and $t$ is the time.

Write the expression for the position of the oscillator.

  $x=A\cos \left(\omega t-\delta \right)$   ........ (3)

Here, $x$ is the instantaneous position, $A$ is the maximum amplitude of oscillator, $\omega$ is the angular frequency, $t$ is the time taken and $\delta$ is the phase difference.

Differentiate the above equation.

  $v=-\left(A\omega \sin \left(\omega t-\delta \right)\right)$   ......... (4)

Substitute $F_0\cos \omega t$ for $F$ and $-\left(A\omega \sin \left(\omega t-\delta \right)\right)$ for $v$ in equation (1).

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

Conclusion:

Thus,the average power delivered by a driving force is $-A\omega F_0\cos \omega t\sin \left(\omega t-\delta \right)$ .

(b)

To determine

The average power delivered by a driving force.

Explanation of Solution

Given:

Formula used:

Write the expression for average power delivered in driving an oscillator.

  $\begin{array}{c}P=\overrightarrow{F}\cdot \overrightarrow{v}\\ =Fv\cos \theta \end{array}$   ........ (1)

Here, $P$ is the average power, $F$ is the driving force of oscillator, $v$ is the driving velocity and $\theta$ is the angle between $F$ and $v$ .

Calculation:

Write the expression for the force as a function of time.

  $F=F_0\cos \omega t$   ........ (2)

Here, $F$ is the driving force of oscillator, $F_0$ is the maximum force, $\omega$ is the angular frequency and $t$ is the time.

Write the expression for the position of the oscillator.

  $x=A\cos \left(\omega t-\delta \right)$   ........ (3)

Here, $x$ is the instantaneous position, $A$ is the maximum amplitude of oscillator, $\omega$ is the angular frequency, $t$ is the time taken and $\delta$ is the phase difference.

Differentiate the above equation.

  $v=-\left(A\omega \sin \left(\omega t-\delta \right)\right)$   ......... (4)

Substitute $F_0\cos \omega t$ for $F$ and $-\left(A\omega \sin \left(\omega t-\delta \right)\right)$ for $v$ in equation (1).

  $\begin{array}{c}P=\left(F_0\cos \omega t\right)\left(-\left(A\omega \sin \left(\omega t-\delta \right)\right)\right)\\ =-A\omega F_0\cos \omega t\sin \left(\omega t-\delta \right)\end{array}$   ......... (5)

Write the expression for $\sin \left({\theta }_1-{\theta }_2\right)$ expansion.

  $\sin \left({\theta }_1-{\theta }_2\right)=\sin {\theta }_1\cos {\theta }_2-\cos {\theta }_1\sin {\theta }_2$   ......... (6) Substitute $\omega t$ for ${\theta }_1$ and $\delta$ for ${\theta }_2$ in equation (6).

  $\sin \left(\omega t-\delta \right)=\sin \omega t\cos \delta -\cos \omega t\sin \delta$   ......... (7)

Substitute above value in equation (5).

  $\begin{array}{c}P=\left(F_0\cos \omega t\right)\left(-\left(A\omega \left(\sin \omega t\cos \delta -\cos \omega t\sin \delta \right)\right)\right)\\ =-A\omega F_0{\cos }^2\omega t\sin \delta -A\omega F_0\cos \delta \cos \omega t\sin \omega t\end{array}$

Conclusion:

Thus, the average power delivered by a driving forceis $-A\omega F_0{\cos }^2\omega t\sin \delta -A\omega F_0\cos \delta \cos \omega t\sin \omega t$ .

(c)

To determine

Average value of $\left(-A\omega F_0{\cos }^2\omega t\sin \delta -A\omega F_0\cos \delta \cos \omega t\sin \omega t\right)$ is zero, average power is $P=\frac{1}{2}\left(A\omega F_0\sin \delta \right)$ .

Explanation of Solution

Given:

Formula used:

Calculate the average value of $\langle \sin \theta \cos \theta \rangle$ .

  $\begin{array}{c}\langle \sin \theta \cos \theta \rangle =\frac{1}{2\pi }\left({\displaystyle \underset{0}{\overset{2\pi }{\int }}\sin \theta \cos \theta d\theta }\right)\\ =\frac{1}{2\pi }\left(\frac{1}{2}{{\sin }^2\theta |}_0^{2\pi }\right)\\ =0\end{array}$

Calculate the average value of $\langle {\cos }^2\theta \rangle$ .

  $\begin{array}{c}\langle {\cos }^2\theta \rangle =\frac{1}{2}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\cos }^2\theta }d\theta \\ =\frac{1}{2}\left[\frac{1}{2}{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(1+\cos 2\theta \right)d\theta }\right]\\ =\frac{1}{2\pi }\left(\pi +0\right)\\ =\frac{1}{2}\end{array}$

  $$

Calculation:

Write the expression for average power.

  $P_{av}=\left(-A\omega F_0\langle {\cos }^2\omega t\rangle \sin \delta -A\omega F_0\cos \delta \langle \cos \omega t\sin \omega t\rangle \right)$ . ….. (1)

Substitute $0$ for $\langle \sin \omega t\cos \omega t\rangle$ and $\frac{1}{2}$ for $\langle {\cos }^2\theta \rangle$ in equation (1).

  $\begin{array}{c}{P}_{av}=\left(-\frac{1}{2}A\omega F_0\sin \delta -A\omega F_0\cos \delta \left(0\right)\right)\\ =\frac{1}{2}\left(A\omega F_0\sin \delta \right)\end{array}$

Conclusion:

Thus, average power is $\frac{1}{2}\left(A\omega F_0\sin \delta \right)$ .

(d)

To determine

Use triangle to show that $\begin{array}{c}\sin \delta =\frac{b\omega }{\sqrt{m^2{\left({\omega }_0{}^2-{\omega }^2\right)}^2+b^2{\omega }^2}}\\ =\frac{b\omega A}{F_0}\end{array}$

Explanation of Solution

Introduction:

Draw the triangle to calculate the value using triangle law.

  

Use the above triangle to find $\sin \delta$ .

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

Conclusion:

Thus, the value of $\sin \delta$ is $\frac{b\omega }{\sqrt{m^2{\left({\omega }_0{}^2-{\omega }^2\right)}^2+b^2{\omega }^2}}$ .

(e)

To determine

Average power is $P_{av}=\frac{1}{2}\left(\frac{b{\omega }^2{F}_0}{m^2{\left({\omega }_0^2-{\omega }^2\right)}^2+b^2{\omega }^2}\right)$ .

Explanation of Solution

Given:

  $\sin \delta =\frac{b\omega A}{F_0}$

Formula used:

Write the expression for $\sin \delta$

  $\sin \delta =\frac{b\omega A}{F_0}$   ........ (1)

Rearrange above equation for the value of $\omega$ .

  $\omega =\frac{F_0}{bA}\left(\sin \delta \right)$   ......... (2)

Write the expression for average power.

  $P=\frac{1}{2}\left(A\omega F_0\sin \delta \right)$   ........ (3)

Calculation:

Substitute $\frac{F_0}{bA}\left(\sin \delta \right)$ for $\omega$ in equation (3).

  $P=\frac{1}{2b}\left(F_0{}^2{\sin }^2\delta \right)$   .......... (4)

Substitute $\frac{b\omega }{\sqrt{m^2{\left({\omega }_0{}^2-{\omega }^2\right)}^2+b^2{\omega }^2}}$ for $\left(\sin \delta \right)$ in equation (4).

  $P_{av}=\frac{1}{2}\left(\frac{b{\omega }^2{F}_0}{m^2{\left({\omega }_0^2-{\omega }^2\right)}^2+b^2{\omega }^2}\right)$ .

Conclusion:

Thus, proved thatAverage power is $\frac{1}{2}\left(\frac{b{\omega }^2{F}_0}{m^2{\left({\omega }_0^2-{\omega }^2\right)}^2+b^2{\omega }^2}\right)$ .