(a) Using the equation for the amplitude$\mathit{A}$ , show that if the viscous friction is small, the amplitude is large when ${\mathit{\omega }}_{\mathbf{D}}$is approximately equal to${\mathit{\omega }}_{\mathbf{F}}$ . Using the equation involving the phase shift$\mathit{\phi }$ , show that the phase shiftis approximately${\mathbf{0}}^{\mathbf{°}}$ for very low driving frequency${\mathit{\omega }}_{\mathbf{D}}$ , approximately${\mathbf{180}}^{\mathbf{°}}$ for very high driving frequency${\mathit{\omega }}_{\mathbf{D}}$ , and${\mathbf{90}}^{\mathbf{°}}$ at resonance, consistent with your experiment.

(b) Show that with small viscous friction, the amplitude$\mathit{A}$ drops to$\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}$ of the peak amplitude when the driving angular frequency differs from resonance by this amount:

$\left|{\omega }_F-{\omega }_D\right|\mathbf{\approx }\frac{\mathbf{c}}{\mathbf{2}\mathbf{m}}{\mathit{\omega }}_{\mathbf{F}}$

(Hint: Note that near resonance${\mathit{\omega }}_{\mathbf{D}}\mathbf{\approx }{\mathit{\omega }}_{\mathbf{F}}$ , So${\mathit{\omega }}_{\mathbf{F}}\mathbf{+}{\mathit{\omega }}_{\mathbf{D}}\mathbf{\approx }\mathbf{2}{\mathit{\omega }}_{\mathbf{F}}$ .) Given these results, how does the width of the resonance peak depend on the amount of friction? What would the resonance curve look like if there were very little friction?

A periodic variable's amplitude is a measure of how much it changes in a single period (such as time or spatial period). A non-periodic signal's amplitude is its magnitude in comparison to a reference value. In a variety of ways, the amount of the differences between the variable's extreme values is utilised to define amplitude.

Write the equation for the amplitude$A$ .

$A=\frac{{{\omega }_F}^2}{\sqrt{{\left({\omega }_F^2-{\omega }_D^2\right)}^2+{\left(\frac{c}{m}{\omega }_D\right)}^2}}D$

When$c\approx 0$ then ${\left(\frac{c}{m}{\omega }_D\right)}^2\approx 0$ and when${\omega }_F\approx {\omega }_D$ then ${\left({\omega }_F^2-{\omega }_D\right)}^2\approx 0$.

By using the approximation $c\approx 0$ and ${\omega }_F\approx {\omega }_D$ the result is .

$\sqrt{{\left({\omega }_F^2-{{\omega }_D}^2\right)}^2+{\left(\frac{c}{m}{\omega }_D\right)}^2}\approx 0$

On using this result in the equation of amplitude the value of amplitude is$A\to \infty$ .

Write the phase shift equation.

$\cos \left(\varphi \right)=\frac{\left({\omega }_F^2-{\omega }_D^2\right)}{\sqrt{{\left({\omega }_F^2-{\omega }_D^2\right)}^2+{\left({\displaystyle \frac{c}{m}}{\omega }_D\right)}^2}}$

When${\omega }_D<<{\omega }_F$ then ${\left({\omega }_F^2-{\omega }_D^2\right)}^2\approx {\omega }_F^2$ and ${\omega }_F^2-{\omega }_D^2\approx {\omega }_F^2$.

By using the approximation${\omega }_D<<{\omega }_F$ the result is .

role="math" localid="1668513686939" $\sqrt{{\left({\omega }_F^2-{\omega }_D^2\right)}^2+{\left(\frac{c}{m}{\omega }_D\right)}^2}\approx \sqrt{{\omega }_F^4+{\left(\frac{c}{m}\right)}^2{\omega }_D^2\approx {\omega }_F^2}$

use these results into the equation of phase shift.

$\begin{array}{rcl}\cos \left(\varphi \right)& \approx & \frac{{\omega }_F^2}{{\omega }_{F^2}^2}\\ \cos \left(\varphi \right)& \approx & 1\\ \varphi & \approx & 0°\end{array}$

Write the phase shift equation.

$\cos \left(\varphi \right)=\frac{\left({\omega }_F^2-{\omega }_D^2\right)}{\sqrt{{\left({\omega }_F^2-{\omega }_D^2\right)}^2+{\left({\displaystyle \frac{c}{m}}{\omega }_D\right)}^2}}$

When${\omega }_D>>{\omega }_F$ then $\sqrt{{\left({\omega }_F^2-{\omega }_D^2\right)}^2+{\left(\frac{c}{m}{\omega }_D\right)}^2}\approx \sqrt{{\omega }_D^4+{\left(\frac{c}{m}{\omega }_D\right)}^2},\sqrt{{\omega }_D^4+{\left(\frac{c}{m}{\omega }_D\right)}^2}\approx {\omega }_D^2$ , and ${\omega }_F^2-{\omega }_D^2\approx -{\omega }_D^2$.

use these results into the equation of phase shift.

$\begin{array}{rcl}\cos \left(\varphi \right)& \approx & -\frac{{\omega }_D^2}{{\omega }_D^2}\\ & \approx & -1\\ \varphi & \approx & 180°\end{array}$

In the resonance case,${\omega }_F={\omega }_D$the result is ${\omega }_F^2-{\omega }_D^2=0$.

Use this into the equation of phase shift.

$\begin{array}{rcl}\cos \left(\varphi \right)& =& 0\\ \varphi & =& 90°\end{array}$

Therefore, the below results are concluded:

• For a low viscous friction coefficient and${\omega }_F\approx {\omega }_D$ , the amplitude is$A\to \infty$ .
• For${\omega }_D<<{\omega }_F$ , the phase shift is$\varphi =0°$.
• For ${\omega }_D>>{\omega }_F$, the phase shift is $\varphi =180°$.
• For${\omega }_D={\omega }_F$ , the phase shift is$\varphi =90°$ .

The amplitude is given by $A=\frac{{\omega }_F^2}{\sqrt{{\left({\omega }_F^2-{\omega }_D^2\right)}^2+{\left({\displaystyle \frac{c}{m}}{\omega }_D\right)}^2}}D$ when the amplitude drops to $\frac{1}{\sqrt{2}}$of the maximum value, solve the denominator.

$\begin{array}{rcl}{\left({\omega }_F^2-{\omega }_D^2\right)}^2& =& {\left(\frac{c}{m}{\omega }_D\right)}^2\\ \left|{\omega }_F^2-{\omega }_D^2\right|& =& \frac{c}{m}{\omega }_D\end{array}$

The left part of the equation can be written as${\omega }_F^2-{\omega }_D^2=\left({\omega }_F-{\omega }_D\right)\times \left({\omega }_F+{\omega }_D\right)$ .

Near resonance,${\omega }_F+{\omega }_D\approx 2{\omega }_F\approx 2{\omega }_D$ use this into the above equation.

$\begin{array}{rcl}\left({\omega }_F-{\omega }_D\right)\times 2{\omega }_D& =& \frac{c}{m}{\omega }_D\\ \left({\omega }_F-{\omega }_D\right)& =& \frac{c}{2m}\end{array}$

When the expression above is valid than the amplitude will be reduced by$\frac{1}{\sqrt{2}}$.

Therefore, when the expression$\left({\omega }_F-{\omega }_D\right)=\frac{c}{2m}$ is valid than the amplitude will be reduced by$\frac{1}{\sqrt{2}}$ but, on the other hand$\left|{\omega }_F^2-{\omega }_D^2\right|=\frac{c}{2m}{\omega }_D$ must also be true.