91Ó°ÊÓ

The amplitudeof the forced oscillation as a function of the driving angular frequency, driving force amplitude${\mathrm{F}}_{\max }$and the damping constantbis given by

$\mathit{A}\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{m}\mathbf{a}\mathbf{x}}}{\sqrt{{\left(k-m{\mathrm{Ó\neg }}_d^2\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}{\mathbf{Ó\neg }}_{\mathbf{d}}^{\mathbf{2}}}}$ …â¶Ä¦..(¾±)

Where ${\mathrm{F}}_{\max }$is the maximum value of driving force, k is the force constant of restoring force, m is the mass, ${\mathrm{Ó\neg }}_d$is the driving angular frequency, and b is the damping constant

Here given that when the driving angular frequency ${\mathrm{Ó\neg }}_d=\sqrt{\frac{k}{m}}$, then the damping constant has a value $b_1$, the amplitude is $A_1$.

Therefore, from equation (i),

$$\begin{gathered} {\mathrm{A}}_1=\frac{{\mathrm{F}}_{max}}{\sqrt{{\left(\mathrm{k}−\mathrm{m}{\left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\right)}^2\right)}^2+{\mathrm{b}}_1^2\left(\sqrt{{\left.\frac{\mathrm{k}}{\mathrm{m}}\right)}^2}\right.}} \\ =\frac{{\mathrm{F}}_{max}}{\sqrt{{\left(\mathrm{k}−\mathrm{m}×\frac{\mathrm{k}}{\mathrm{m}}\right)}^2+{\mathrm{b}}_1^2×\frac{\mathrm{k}}{\mathrm{m}}}} \\ =\frac{{\mathrm{F}}_{max}}{\sqrt{(\mathrm{k}−\mathrm{k}{)}^2+{\mathrm{b}}_1^2×\frac{\mathrm{k}}{\mathrm{m}}}} \\ =\frac{{\mathrm{F}}_{max}}{\sqrt{{\mathrm{b}}_1^2×\frac{\mathrm{k}}{\mathrm{m}}}} \end{gathered}$$

Now for the same driving frequency ${\mathrm{Ó\neg }}_d=\sqrt{\frac{k}{m}}$and the same driving force amplitude ${\mathrm{F}}_{\max }$, the amplitude will be if the damping constant is

(a) $\mathrm{b}=3{\mathrm{b}}_1$, from equation

role="math" localid="1668148464916" $$\begin{gathered} A=\frac{F_{max}}{\sqrt{{\left(3b_1\right)}^2×\frac{k}{m}}} \\ =\frac{1}{3}\frac{F_{max}}{\sqrt{b_1^2×\frac{k}{m}}} \\ =\frac{1}{3}{A}_1 \end{gathered}$$

Hence the amplitude for the driving frequency at$3{\mathrm{b}}_1$ is$\mathrm{A}=\frac{1}{3}{\mathrm{A}}_1$

(b)$\mathrm{b}=\frac{{\mathrm{b}}_1}{2}$ from equation

$$\begin{gathered} \mathrm{A}=\frac{{\mathrm{F}}_{max}}{\sqrt{{\left({\displaystyle \frac{{\mathrm{b}}_1}{2}}\right)}^2×\frac{\mathrm{k}}{\mathrm{m}}}} \\ =\frac{2{\mathrm{F}}_{max}}{\sqrt{{\mathrm{b}}_1^2×\frac{\mathrm{k}}{\mathrm{m}}}} \\ =2{\mathrm{A}}_1 \end{gathered}$$

Hence, the amplitude for the driving frequency at $\frac{b_1}{2}$is $\mathrm{A}=2{\mathrm{A}}_1$