L evitation of objects using a standing ultrasonic wave

1. T.Chikazawa H.Nakayama R.Makii Levitation of objects using a standing ultrasonic wave

2. First, “what is an ultrasonic wave? ” It is a sound over 20,000 Hz Short wavelength Big amplitude.

3. ①Wave 　②Remains in a constant position ③Most vibration・・・anti-node No vibration・・・ node ④Anti-node ~ Node ・・・ ¼ wavelength Node ~ Node ・・・ ½wavelength “What is a standing wave ? ”

4. Using the formula for acoustic velocity V=fλV=331.6+0.6t V : Acoustic velocity (m/s) f: Pitch (Hz) λ: Wavelength (m)t: Temperature (℃ or )

5. （m）　　　　　＝8.5(mm)anti-node ~ node1/4λ ≒2.13(mm)node ~ node 1/2λ ≒4.25(mm)We can calculate the theoretical levitation position from this value!!

6. We were interested in an experiment about acoustic levitation done by last year’s students. When we used a ultrasonic wave, we found that the object levitate more easily. We studied the relationship between the sound wave and object. 【Study motive】

7. At the beginning…… We used a couple of objects. We also use ultrasonic wave because it is highly directional.

8. 【experiment】 ＜Purpose＞ ・We find whether or not there was only other resistance force causing the objects to float. ⊿x１

9. 【Expectation】 Acting forces are only 　　　　　　　　　　　　　　　　　‘gravity’ . ⊿xwould have a constant value.

10. 【Experimental method】 Ⅰ　・We levitated object 1. ・ We levitated object 2 , and measured the distance between object 1 and 2.(⊿X1) ⊿x１

11. Ⅱ・We levitated object 3 ,and measured the distance between object 2 and 3.(⊿X２) 　　・We also measured the distance between object 1and 2 again.(⊿X３) ⊿x２ ⊿x３

12. 【Result】 Value of each

13. We showed to graph the average value. ⊿x have a similar value between 4.1mm and 4.5mm.

14. 【Consideration】 We found that ⊿X is nearly λ/2(≒４．２５mm) in any case . ↓ There is no other resistance force

15. 【Future】 We want to increase the sample size of our experiment to make sure that our result are reproducible!

16. THANKFORLISTENING

17. ⊿X≒λ/2(=4.25mm) object