Wendler_K_152_lab10_

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Karsyn Wendler Physics 152- Lab Section: Wednesday 2:00pm Lab #10- Applications of Ultrasound Transducers November 15, 2023 INTRODUCTION My lab partner Sophia and I used a computer-controlled function generator and oscilloscope to observe and understand the properties of ultrasound transducers. We examined Lissajous curves, the inverse-square law for sound intensity, and theoretically- standing waves and interference. We also looked into the applications of ultrasounds, even getting to look at our own hand under the ultrasound probe. PART 4.1: ULTRASOUND TRANSDUCER We began the lab by setting up the transmitter-receiver pair, function generator, Velleman PCG1000 software, and the oscilloscope. We applied a 40kHz sine wave and confirmed that 4.1-1 we did see both the input signal on the transmitter and the output signal from the receiver. Thus, no adjustments were made to our set up. PART 4.2: PHASE OF A TRAVELING WAVE We tried moving the receiver closer and father away on the apparatus to observe phase changes in the received wave. This data allowed us to calculate the speed and wavelength of the sound wave. 4.2-1 We found that as we moved the receiver farther away, the output wave shifted to the right along the horizontal axis, and when we moved the receiver closer to the transmitter it shifted left. 4.2-2 After centering one of our peaks along the oscilloscope’s vertical line, we recorded the initial position of the receiver. Then we moved the peak over 5 divisions and recorded that point as the final position. This data was used to calculate the speed of sound c in m/s . d initial 46.30 cm d final 47.20 cm ∆࠵? 25 μ s c 360 m/s

࠵? = ࠵? & − ࠵? ( ∆࠵? = . 4720࠵? − .4630࠵? . 000025࠵? = ࠵?࠵?࠵?࠵?/࠵? When compared to the actual value of c=343 m/s , we find our measurement to be within 5% accuracy. 360 − 343 = 17 17 343 × 100 = ࠵?. ࠵?࠵?% 4.2-3 We followed a similar process, writing the measurements for horizontal divisions spanned by 10 peaks. We used these values to calculate the wavelength. d initial 52.00 cm d final 42.90 cm ࠵? 9.1 mm Wavelength is determined by dividing the wave-speed by frequency. Thus, we solved for wavelength using the following calculations: ∆࠵? = 50࠵?࠵? × 5 = 250࠵?࠵? = ࠵?. ࠵?࠵?࠵?࠵?࠵?࠵? ࠵? = ࠵? ࠵? = E 1 ࠵? F G ࠵? & − ࠵? ( ∆࠵? H = E 1 40000࠵?࠵? F E . 5200࠵? − .4290࠵? . 00025࠵? F = ࠵?. ࠵?࠵?࠵?࠵?࠵? PART 4.3: LISSAJOUS CURVE We manipulated the oscilloscope settings to view both the transmitted and received signals parametrically on the xy-plane. We investigated the effect of the distance between the transmitter and receiver on the resulting Lissajous curve. 4.3-1 When plotted on the xy-plane the apparatus set up creates a Lissajous curve. When the two voltages are plotted parametrically as such, time, t, becomes an implicit parameter that is not explicitly shown in the graph. 4.3-2 & 3

The blue and yellow lines on the graph (pictured above) correspond to two sine waves with exactly the same phase and exactly the opposite phase. One represents a 0 sin wave for both the x and y parameters and the other represents a 0 sin wave and 180 sin wave respectively. By observing how the Lissajous curve changed in response to the transducer being moved, we identified the phase angles as follows: Blue: ࠵? = 0 Red: ࠵? = 30 Teal: ࠵? = 60 Green: ࠵? = 90 Brown: ࠵? = 120 Purple: ࠵? = 150 Yellow: ࠵? = 180 4.3-4 We attempted to set the receiver at a position where the Lissajous curve corresponded to a zero phase angle. Then we moved it so that the curve rotated 10 times- corresponding to 10 360- degree phase shifts. Both positions are recorded. We then calculated the wavelength: d initial 37.40 cm d final 46.60 cm ࠵? 0.92 cm (9.2 mm) ࠵? = G ࠵? & − ࠵? ( ∆࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?ℎ࠵? H = E 46.60࠵?࠵? − 37.40࠵?࠵? 10 F = ࠵?. ࠵?࠵?࠵?࠵? = ࠵?. ࠵?࠵?࠵? This wavelength is almost exactly equivalent to that obtained in 4.2-3. 4.3-5 We attempted to set the phase angle to about 30 ° (we were having some trouble determining exactly where that was) by moving the receiver along the bench. We tried to measure the phase difference in time as accurately as possible, then use that to calculate the phase difference in angle. Based on the result below, we believe there may have been some human error in our measurements. ∆࠵? 3.700 cm Phase diff in angle 53.28 ° ࠵? = ࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵? ࠵? (࠵?࠵?࠵?࠵?࠵?࠵?) × 360° = 3.700 ࠵?࠵? 25.00 ࠵?࠵? × 360° = ࠵?࠵?. ࠵?࠵?°

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4.3-6 In the y(x) mode, we measured the following values for the Lissajous curve from the oscilloscope screen: A 2.575 V B 40.00 mV Asin q 1.700 V Bsin q 25.00 mV q (A) 41.31 ° q (B) 38.68 ° 4.3-7 From our calculations, I believe that the second method (4.3-6) was more accurate in measuring the phase angle as the values are much closer to 30 ° . This may have been human error though as the first method required us to eyeball measurements where as we were able to use tools on the oscilloscope for the second part. PART 4.4: INVERSE SQUARE LAW FOR THE SOUND INTENSITY The inverse square law says that in the case of sound, energy travels as pressure waves and the energy is proportional to the square of the pressure. We tested the validity of this claim using the receiver and transmitter. 4.4-1 We took incremental data points recording the position of the receiver and the voltage of the received signal on the oscilloscope, starting at few centimeters away from the transmitter and going up to approximately one meter. We plotted this data on a graph as Voltage (V on the y- axis) vs. Inverse Distance (1/m on the x-axis). The resulting graph is attached below:

4.4-2 We added a linear fit line to the data. The resulting R-value is 0.999 The resulting RMS error is 4.71 The R-value being almost a perfect 1.00 indicates that the curve is almost completely linear. Additionally, I accdinetally added a (0,0) data point that altered the curve, but we were short on time, so I could not figure out how to remove it. I think if we had removed the (0,0) data point, the linearity of the curve may have been altered. 4.4-3 Based on our graph, the data has an inverse-linear law, not an inverse square law; however, this may have been slightly different with out the (0,0) point as previously mentioned. PART 4.5: ABSORPTION OF ULTRASOUND 4.5-1 Unfortunately my lab partner and I did not make it to this part of the lab; however I was able to find a friend’s data to use for observations. This portion of the lab tests to see what materials allow ultrasound to pass through more easily. The data is attached below: Material Voltage w/o Material Voltage WITH Material Absorption Ratio Qualitative Description Plain Paper 330 mV 50 mV 0.85 Most absorptive Paper Tissue 330 mV 240 mV 0.27 Slightly absorptive Plastic Sheet 330 mV 120 mV 0.64 Kind of absorptive Cloth 330 mV 310 mV 0.06 Least absorptive Unfortunately, due to time constraints, my lab partner and I were unable to finish Parts 4.6-4.8 of the lab manual. I have tried to summarize some of what would have been covered below. PART 4.6: STANDING WAVES Standing Waves are formed when two waves of the same frequency and related phase travel in the opposite direction. We would have used a solid screen on the screen holder to create our own standing waves if we had completed this section. The distance between the node and antinode would have been used to calculate the wavelength. This then would have been compared to the wavelength found in 4.2-3. PART 4.7: INTERFERENCE USING A LLOYD’S MIRROR This section would have used a Lloyd’s mirror to do a Young’s interference (double-slit) experiment. We would have filled a table with data points informing us of the vertical position of the maxima form the surface of the table for several maxima. We would have used the formula given by the theory of double-slit diffraction to compare the measured value to a theoretical value. PART 4.8: PRINCIPLES OF ULTRASONOGRAPHY AND ECHO DETECTION This activity was to learn about practical application of ultrasound. Ultrasound waves use echoes to detect objects. A wave packet is sent out, reflected by an object, and returned as an

echo. The known speed of the wave can then be used to calculate the distance traveled by the wave, which is twice the distance between the wave source and the object. 4.8-1 This part could be done without data. We found the maximum distance of an object that can be detected for Tpacket=100ms. ࠵? bcdefg = 1 ࠵? bcdefg = 10࠵?࠵? = 100࠵?࠵? ࠵? = ࠵?∆࠵? 2 343 ࠵? ࠵? × 0.1࠵? 1 × 1 2 = ࠵?࠵?. ࠵?࠵?࠵? 4.8-2. This part could be done without data. We found the minimum distance of an object that can be detected for a wave packet of 8 periods of a square wave with a period of T=25microseconds if the same single source transducer is used as both a transmitter and receiver. ࠵? i bfj(klm = 25࠵?࠵?(8) = 200࠵?࠵? 343 ࠵? ࠵? × 200 × 10 op ࠵? 1 × 1 2 = ࠵?. ࠵?࠵?࠵?࠵?࠵? 4.8-3. This part could be done without data. I calculated the round-trip distance and object distance that corresponds to a time division of 1.00ms on the oscilloscope. Round Trip Dist: 343 ࠵? ࠵? × 1.00 × 10 oq ࠵? 1 = ࠵?. ࠵?࠵?࠵?࠵? Object Dist: 343 ࠵? ࠵? × 1.00 × 10 oq ࠵? 1 × 1 2 = ࠵?. ࠵?࠵?࠵?࠵? 4.8-4-6 all require experimental data. PART 4.9: MEDICAL ULTRASOUND The final part of this experiment involved an ultrasound machine where we were able to see, first hand, how their higher sound frequency allows for better spatial resolution and imaging. 4.9-1 . Given that the speed of sound in human soft tissue is 1540 m/s on average, what is the wavelength of sound in human soft tissue for the frequencies 3.5 MHz and 5.0 MHz? ࠵? = ࠵? ࠵? = E 1540࠵?/࠵? 3.5 × 10 p ࠵?࠵? F = ࠵?. ࠵? × ࠵?࠵? o࠵? ࠵? ࠵? = ࠵? ࠵? = E 1540࠵?/࠵? 5.0 × 10 p ࠵?࠵? F = ࠵?. ࠵? × ࠵?࠵? o࠵? ࠵?

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4.9-2 When I stuck my hand in the water with the ultrasound probe I was able to see an image of my hand on the machine. 4.9-3 We did this section of the lab as a group and we did not make it to this part.

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