Question

Ultrasound pulses with a frequency of $1.000MHz$are transmitted into water, where the speed of sound is $1500m/s$. The spatial length of each pulse is localid="1650889451408" $12$localid="1650889457691" $mm$.

a. How many complete cycles are contained in one pulse?

b. What range of frequencies must be superimposed to create each pulse?

Short answer

.a) n=8

b) $0.938\mathrm{MHz}\le f\le 1.063MHz$

Step-by-step solution

part (a) step 1: Given information

a) The number of complete cycles in one pulse can be obtained by dividing the spatial length of the pulse by the wavelength of the ultrasound pulse, where the wavelength can be calculated as follows

$\lambda =\frac{v}{f}=\frac{1500\mathrm{m}/\mathrm{s}}{1\times {10}^6\mathrm{Hz}}=1.5\times {10}^{-3}\mathrm{m}$

Hence, the number of complete cycles in one pulse is

$n=\frac{\Delta x}{\lambda }=\frac{12\times {10}^{-3}\mathrm{m}}{1.5\times {10}^{-3}\mathrm{m}}=8$

part (b) step 2: Given information

.b) The ultrasound pulse is a wave packet that satisfies the equation

$\Delta f\Delta t\approx 1$

first, we need to find the pulse duration$\left(\Delta t\right)$, and that can be done by finding the period of the wave and multiply it by the number of complete cycles in one pulse. The period is

$T=\frac{1}{f}=\frac{1}{1\times {10}^6\mathrm{Hz}}=1\times {10}^{-6}\mathrm{s}$

thus, the duration of the pulse (the wave packet ) is

$\Delta t=n\times T=8\times 1\times {10}^{-6}\mathrm{s}=8\times {10}^{-6}\mathrm{s}$

now $\Delta f$can be calculated as

$\Delta f=\frac{1}{\Delta t}=\frac{1}{8\times {10}^{-6}\mathrm{s}}=1.25\times {10}^5\mathrm{Hz}$

Finally, the range of frequencies that must be superimposed to create the given pulse is

$$\begin{gathered} \left(1\mathrm{MHz}-\frac{\Delta f}{2}\right)\le f\le \left(1\mathrm{MHz}+\frac{\Delta f}{2}\right) \\ \left(1\times {10}^6\mathrm{Hz}-\frac{1.25\times {10}^5\mathrm{Hz}}{2}\right)\le f\le \left(1\times {10}^6\mathrm{Hz}+\frac{1.25\times {10}^5\mathrm{Hz}}{2}\right) \\ 9.38\times {10}^5\mathrm{Hz}\le f\le 10.63\times {10}^5\mathrm{Hz} \\ 0.938\mathrm{MHz}\le f\le 1.063\mathrm{MHz} \end{gathered}$$