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

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

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L evitation of thin pieces of metal using a standing ultrasonic wave. Suzuki Naomichi Tanaka Takahiro . F irst, “what is an ultrasonic wave? ”. It is a sound over 20,000 H z S hort wavelength B ig amplitude. “What is a standing wave ? ”.  ① W ave  ② Remains in a constant position - PowerPoint PPT Presentation

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

Page 1: 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

Page 2: L evitation of thin pieces of metal using a standing ultrasonic wave
Page 3: L evitation of thin pieces of metal using a standing ultrasonic wave

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

Big amplitude.

First, “what is an ultrasonic wave? ”

Page 4: L evitation of thin pieces of metal using a standing 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 ? ”

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The purpose of our experiment

To find out where objects will levitate if their thickness

changes !!

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HypothesisPosition of pieces is proportional

to the thickness of it.

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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 ↑

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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 )

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 ( 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!!

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Weight will double

Thickness of aluminum foil

doubles

the distance from the node will also double ?

Our expectation

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The way of our experiment

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

Observe the position where they float

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We compared them with the node position we calculated

We investigated their error

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Our result

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(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

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1.5 2 2.5 3 3.5 4 4.5 5 5.50

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(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

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Consideration of our experiment

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・ Floating position ⇒node ? or anti-node ?

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

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 Not related to the

thickness of the foil !!

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Our future subjects

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・ Increasing the sample size of our experiment

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

andthe 11 micrometer foil

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Thank you for listening!