91Ó°ÊÓ

The process of recurring changes of any quantity or measure around its equilibrium value in time is defined as oscillation. A periodic change of a matter between two values or about its central value is also known as oscillation.

1- Newton's second law : The net force $F$on a body with mass $m$is related to the body's acceleration $a$by:

$F=ma$

2- The maximum transverse speed and acceleration of a particle in simple hormonic motion are found in terms of the angular speed $w$and amplitude $A$as follows:

$$\begin{gathered} v_{max}=wA \\ {a}_{max}=w^{2}A \end{gathered}$$

3- The angular frequency $w$of a wave is related to the frequency f by:

$w=2\mathrm{Ï€}\mathrm{f}$

• The mass of the disk: $m=(0.10\mathrm{g})\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)=0.10×{10}^{-3}\mathrm{kg}$
• The frequency of the disk is: $f=\left(1.0{\mathrm{MHz}}^{\mathrm{z}}\right)\left(\frac{{10}^6{\mathrm{Hz}}_{\mathrm{z}}}{1{\mathrm{MHz}}_{\mathrm{z}}}\right)=1.0×{10}^6\mathrm{Hz}$
• The maximum restoring force applied to the disk is:$F_{\max }=40,000\mathrm{N}$

(a) The maximum restoring force applied to the disk is related to the maximum acceleration of the disk by Newton's second law from Equation (1):

• $F_{max=}m{a}_{max}$

Substitute for $a_{max}$from Equation (3):

• $F_{max}=m{w}^{2}A$
  

Substitute for $w$from Equation (4):

• role="math" localid="1649945464662" $$\begin{gathered} F_{max}=m{(2\mathrm{Ï€}f)}^{2}A \\ =4{\mathrm{Ï€}}^{2}m{f}^{2}A \end{gathered}$$

$A=\frac{F_{\max }}{4{\mathrm{Ï€}}^{2}m{f}^2}$

Now, Substitute numerical values: $A=\frac{40,000\mathrm{N}}{4{\mathrm{π}}^2\left(0.10×{10}^{-3}\mathrm{kg}\right){\left(1.0×{10}^6\mathrm{Hz}\right)}^2}$

$=\left(10.1×{10}^{-6}\mathrm{m}\right)\left(\frac{{10}^6\mathrm{μ}\mathrm{m}}{1\mathrm{m}}\right)$

$=10.1\mathrm{μ}\mathrm{m}$

The maximum speed of the disk is found from Equation (2):

• $v_{max}=wA$
  

Substitute for $w$from Equation (4):

• $v_{max}=2\mathrm{Ï€}\mathrm{f}\mathrm{A}$
  

Substitute numerical values:

$$\begin{gathered} v_{\max }=2\mathrm{π}\left(1.0×{10}^6\mathrm{Hz}\right)\left(10.1×{10}^{-6}\mathrm{m}\right) \\ =64m/s \end{gathered}$$

The result of the maximum amplitude that won't rupture the disk and maximum speed of the disk is:

$\text{(a)}A=10.1\mathrm{μ}\mathrm{m}$