L evitation of thin pieces of metal using a standing ultrasonic wave

1. Levitation of thin pieces of metal using a standing ultrasonic wave Suzuki Naomichi Tanaka Takahiro

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

4. “What is a standing wave ? ” 　①Wave 　②Remains in a constant position ③Most vibrates・・・anti-node Doesn’t vibrates・・・ node ④Anti-node ~ Node ・・・ ¼ wavelength Node ~ Node ・・・ ½wavelength

5. The purpose of our experiment To find out where objects will levitate if their thickness changes !!

6. Hypothesis Position of pieces is proportional to the thickness of it.

7. ①mg　＋ P↓×S =P↑×S ②Ⅴ=Sd ③ρVｇ= mg Ⅴ: Volume of aluminum foil d : Thickness of aluminum foil ρ: Density of air ρdg+P↓＝P ↑

8. 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 )

9. （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!!

10. Our expectation Thickness of aluminum foil doubles Weight will double the distance from the node will also double ?

11. The way of our experiment Levitating pieces of aluminum foil with different thickness (11μm, 26μm) Observe the position where they float

12. We compared them with the node position we calculated We investigated their error

13. Our result

14. In the case of 11μm

15. In the case of 26μm Hypothesis

16. Consideration of our experiment

17. ・Floating position ⇒node？or anti-node？・In the case of 11μm ≒ in the case of 26μm

19. Not related to the thickness of the foil !!

20. Our future subjects

21. ・ Increasing the sample size of our experiment・Thinking about why there is no difference between the 26 micrometer foil andthe 11 micrometer foil

22. Thank you for listening!