Question

Ultrasound pulses with a frequency of 1.000 MHz are transmitted into water, where the speed of sound is 1500 m /s. The spatial length of each pulse is 12 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) Number of cycle complete in one pulse is 8

.b) Range of frequencies 0.938 MHz$\le$ f $\le$1.063 MHz

Step-by-step solution

Given Information Frequency of Ultrasound pulse 1.000 MHz, Speed of sound is  1500 m/s and length of each pulse is 12 mm

.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

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

Number of cycle complete in one pulse

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

b) The ultrasound pulse is a wave packet that satisfies the equationΔfΔt≈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

$\left(1\mathrm{MHz}-\frac{\Delta f}{2}\right)\le f\le \left(1\mathrm{MHz}+\frac{\Delta f}{2}\right)$

$\left(1\times {10}^6\mathrm{Hz}-\frac{1.25\times {10}^5\mathrm{Hz}}{2}\right)\le f\le \left(1\times {10}^6\mathrm{Hz}+\frac{1.25\times {10}^5\mathrm{Hz}}{2}\right)$

$9.38\times {10}^5\mathrm{Hz}\le f\le 10.63\times {10}^5\mathrm{Hz}$

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