Finding upper and lower bound superposition frequencies of ultrasound pulses

[Mugen112]

Homework Statement

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

a) How many complete cycles are in each pulse?

b) What is the lower bound of the range of frequencies must be superimposed to create each pulse?

c)What is the upper bound of the range of frequencies must be superimposed to create each pulse?

Homework Equations

ΔX=ΔtV
f=1/T

The Attempt at a Solution

a) Δt= 12mm/1500 = 8 * 10^-6
T = 1/f
T = 1/10^6 = 10^-6

So... (8 * 10^-6)/(10^-6) = 8 full cycles in each pulse

b) and c) I have no idea... All I have is..

Δf = 1/Δt = 125,000 Hz.. How do I find the upper/lower bound?

 

[Redbelly98]

It looks like the range of frequencies will cover a 125,000 Hz range, centered on 1.000 Mhz. From that you can figure out the lowest and highest frequencies in the range.

 

[nlightner]

To find the lower and upper bound frequencies for superposition, we need to consider the concept of bandwidth. Bandwidth refers to the range of frequencies that can be transmitted or received by a system. In this case, the bandwidth of the ultrasound pulses is determined by the spatial length of each pulse and the speed of sound in water. Using the equation ΔX=ΔtV, we can calculate the bandwidth of the pulses as 12 mm/8 * 10^-6 s = 1.5 * 10^9 Hz or 1.5 GHz. This means that the lower bound frequency for superposition would be 1.000 MHz - 1.5 GHz = 999.999 MHz and the upper bound frequency would be 1.000 MHz + 1.5 GHz = 1.501 GHz. Therefore, the range of frequencies that must be superimposed to create each pulse is 999.999 MHz to 1.501 GHz. It is important to note that this is an ideal scenario and in reality, there may be some overlap or variation in the actual superposition frequencies due to factors such as the properties of the transmitting and receiving devices.