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Squeeky Toys

  1. Sep 24, 2008 #1
    What are the physics behind a squeeky toy?

    Mechanical energy from your hand decreases the volume and increases the pressure of the squeeky-toy. But what happens to the air and what about the standing waves?

    When you press the air out of it really slowly, it doesn't make any noise, why is that?
  2. jcsd
  3. Sep 28, 2008 #2
    Welcome to PF! :smile:
    Well, the air inside is forced out of the toy and through a "squeaker" component because of the pressure difference you already mentioned. I'm don't actually think there are any standing waves, as the air is forced through the squeaker without any reflection from a boundary.
    As for why it doesn't make a noise if you squeeze it slowly, the best explanation I can think of (not knowing exactly the internal workings of a squeaker) is that you'll need a certain volume of air rushing through the squeaker at any given moment to make it work. Noise is caused by something vibrating in air, and my guess is that a slow flow of air will have enough space to flow around this component of the squeaker without having to worry about pushing it out of its way en route.
  4. Sep 28, 2008 #3
    As the why when you press the toy slowly and it doesn't make a noise because, from my point of view, that you have decreased the pressure of air flowing throught the squeaker. The "squeaker" may need certain amount of pressure being passed through it in order to yield those squeaky noise. This can all be related to PV=kRT of PV=nRT.
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