An ultrasonic transducer, of the type used in medical ultrasound imaging, is a very thin disk $\left(m=0.10g\right)$driven back and forth in $SHM$at $1.0MHz$by an electromagnetic coil.

a. The maximum restoring force that can be applied to the disk without breaking it is $40,000N$. What is the maximum oscillation amplitude that won't rupture the disk?

b. What is the disk's maximum speed at this amplitude?

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:

$w=2\mathrm{\pi }\mathrm{f}$

• The mass of the disk: $m=(0.10\mathrm{g})\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)=0.10\times {10}^{-3}\mathrm{kg}$
• The frequency of the disk is: $f=\left(1.0{\mathrm{MHz}}^{\mathrm{z}}\right)\left(\frac{{10}^6{\mathrm{Hz}}_{\mathrm{z}}}{1{\mathrm{MHz}}_{\mathrm{z}}}\right)=1.0\times {10}^6\mathrm{Hz}$
• 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" $$\begin{gathered} F_{max}=m{(2\mathrm{\pi }f)}^{2}A \\ =4{\mathrm{\pi }}^{2}m{f}^{2}A \end{gathered}$$

$A=\frac{F_{\max }}{4{\pi }^{2}m{f}^2}$

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

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

$=10.1\mu \mathrm{m}$

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

• $v_{max}=wA$
  

Substitute for $w$from Equation (4):

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

Substitute numerical values:

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

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

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