Question

(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?

Short answer

a. Using equation of amplitude,the results are below:

• 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 $\varphi =90°$.

(b) When the driving angular frequency differs from resonance by the amount$\left|{\omega }_F^2-{\omega }_D^2\right|=\frac{c}{m}{\omega }_D$ , the amplitude will be reduced by $\frac{1}{\sqrt{2}}$.

Step-by-step solution

Definition of Amplitude

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.

Analysing the equation for 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$A\to \infty$ .

Analysing phase shift when ωD<<ωF

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.

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

Analysing phase shift when ωD>>ωF

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.

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

Analysing phase shift when  ωF=ωD

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$\varphi =90°$ .

Analysing theexpression for amplitude when it drops 

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.

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

Calculation of resonance peak

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.