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

Levitation of thin pieces of metal

using a standing ultrasonic wave

Suzuki NaomichiTanaka Takahiro

It is a sound over 20,000 H ｚ Short wavelength

Big amplitude.

First, “what is an ultrasonic wave? ”

　① Wave

　② Remains in a constant position

　③ Most vibrates ・・・ anti-node

　　 Doesn’t vibrates ・・・ node　 ④Anti-node ~ Node ・・・ ¼ wavelength　　 Node ~ Node ・・・ ½ 　 wavelength

“What is a standing wave ? ”

The purpose of our experiment

To find out where objects will levitate if their thickness

changes !!

HypothesisPosition of pieces is proportional

to the thickness of it.

①mg　＋ P↓×S =P↑×S ②Ⅴ=S d

③ρV ｇ = mgⅤ: Volume of aluminum foild : Thickness of aluminum

foilρ: Density of air

ρdg+P ↓ ＝P ↑

Using the formula for acoustic

velocity

V=fλ 　 V=331.6+0.6t 　 　 V : Acoustic velocity (m/s) 　　　　 　 f 　 : Pitch (Hz)　 λ: Wavelength (m) 　　　 t : Temperature (℃ or )

　（ m ）　　　　　＝ 8.5(mm)

anti-node ~ node 　　 1/4λ≒2.13(mm)

node ~ node 1/2λ≒4.25(mm) 　　

We can calculate the theoretical

levitation position from this value!!

Weight will double

Thickness of aluminum foil

doubles

the distance from the node will also double ?

Our expectation

The way of our experiment

Levitating pieces of aluminum foil with different thickness (11μm, 26μm)

Observe the position where they float

We compared them with the node position we calculated

We investigated their error

Our result

1.5 2 2.5 3 3.5 4 4.5 5 5.50

1

2

3

4

5

6

7

8

9

10

(mm) 　　　　　　　　　　　　　　　　　　　　　　　　　 The length of one side of aluminum foil 　　　　　　　

In the case of 11μm

The length of one side of aluminum foil(mm) 2.00 3.00 4.00 5.00The different from a calculated value⊿ ｘ(mm) 2.2063 1.8508 1.1793 1.9204

Standard deviation 0.1313 0.4164 0.1183 0.7172

1.5 2 2.5 3 3.5 4 4.5 5 5.50

1

2

3

4

5

6

7

8

9

10

(mm) 　　　　　　　　　　　　　 The length of one side of aluminum foil 　　　　　　

In the case of 26μm

Hypothesis

The length of one side of aluminum foil(mm) 2.00 3.00 4.00 5.00The different from a calculated value⊿ｘ(mm) 1.9769 1.7183 1.6225 2.3250

Standard deviation 0.4167 0.4629 0.4104 0.6050

Consideration of our experiment

・ Floating position ⇒node ？ or anti-node ？

・ In the case of 11μm ≒ in the case of 26μm 　

　Not related to the

thickness of the foil !!

Our future subjects

・ Increasing the sample size of our experiment

・ Thinking about why there is no difference between the 26 micrometer foil

andthe 11 micrometer foil

Thank you for listening!