I don't understand what happens in reality (outside of wave theories). If I clap my hands I invest energy in the nearby air molecules, which move and transfer their energy to nearby molecules which move...and so on. Why then, when I clap louder instead of investing more energy in the same particles making them move faster I simply involve more of them? Or am I wrong? If I'm not, then what if I clap louder in a vacuum with limited air in it?


2 Answers 2


This is a very good question. I'm going to give you a more conceptual answer rather than the quick answer because I find this explanation helps my own students understand this better.

First, consider yourself standing in a gymnasium with a thousand people in it. Not a lot of room is there? Naturally, you'd want some personal space, so you push at the people near you, they have nowhere to go so they bump into the people near them and so on. Since everyone is pretty much shoulder to shoulder, the concussion wave you just produced travels only as fast as each person can stumble into the next one. But that didn't help you, everyone was still standing and you're still cramped. So you give everyone near you a giant shove. Now the people fall over, but you should notice that the wave still travels outwards at about the same speed. The reason is that everyone was still shoulder to shoulder and even though they're now falling over (giving you room to breathe), it takes the same amount of time for them to stumble and fall into the person right next to them.

That's the basic concept of the answer, but let's now talk about the air molecules when you clap. The idea that imparting more force from a stronger clap is correct. But consider in slo-mo what is happening when you clap. As your hands come down, they push air molecules out of the way, effectively imparting a net direction that each molecule moves in. This, in turn, may or may not make those initial molecules speed up, but that's not exactly relevant. After a very short distance, those molecules will hit others at arbitrary angles and impart that faster speed in many directions. The important thing to note is that your hand has moved those molecules into an area that already contained other molecules. Now each one is effectively in the gymnasium situation, all trying to push down their neighbours. To get around this, some of the excess molecules migrate to less crowded areas further away. But now those areas are super crowded and the cycle continues with the overall pressure wave traveling away from you.

So you see, you are absolutely correct, you do impart more energy and more velocity on the individual molecules. However, sound waves are not actually the molecules moving. The speed of sound depends on how long it takes the molecules to start feeling overcrowded and decide to move to a different place.

If you want the technical physics, the speed of an individual air molecule affects how fast it vibrates between regions of compression and rarefaction, which increases or decreases the pitch of the sound, but the molecule doesn't actually travel far from its original location. When you clap harder, most of the extra energy is spent moving more molecules. Slower hands move less molecules because the instantaneous velocity of each particle is much higher than that of your hand, so particles travelling away from the hand aren't moved by the clap. As the hand goes faster it catches up with the slower particles or the ones moving at angles to it, thus affecting more particles.
When the extra energy is used to speed up the molecules, It increases the pitch of the sound you'll hear, but the speed at which nearby molecules realize the increased pressure and start to flow away from it depends greatly on the overall density, temperature, etc of the air.

I probably should note that due to the increase in pressure and temperature very near to your hands, the speed of sound does technically increase right there. But as soon as the sound wave begins to move away from that area, it almost immediately returns to normal.

  • $\begingroup$ Thank you very much for the detailed and multi faceted answer! I think I almost got it(XD). I see now how more molecules are involved if the initial molecules have recieved enough energy to escape far enough to be replaced by others as energy recipients. So I want to understand better what happens in the other scenario where the same particle have recieved excessive energy. You say that it changes the pitch/molecular vibration but the speed of realization of pressure is constant. I'm not completley sure I understand why... In the gymnasium example (or better yet- billiard) if... $\endgroup$
    – ababa
    May 10, 2013 at 0:01
  • $\begingroup$ ...if the initial element was hit harder/faster the following bodies would increase their speed... I think... no? $\endgroup$
    – ababa
    May 10, 2013 at 0:02
  • $\begingroup$ they would if every collision is in line with the direction of motion, but remember that the particles can hit each other at odd angles. This is why very close to your hand, the speed of sound is increased, but further away, any net velocity increase is lost over the multitude of non-head-on collisions that have occurred. However, the initial molecules, having traveled fast, were able to bounce back and forth near your hand. When the pressure wave moves off, it leaves a low pressure area in its place, the high energy molecules rush into it faster, then back because of a new low pressure.... $\endgroup$
    – Jim
    May 10, 2013 at 17:21
  • $\begingroup$ space is created where they just left, and this repeats. Each time the pressure zone shifts, this is communicated outwards as a new pressure wave. The odd angle collisions limit the speed of far away particles and so the speed at which these waves travel, but the number of new pressure zones per second must remain more or less constant. Essentially, the frequency of vibrations has increased and the frequency of a pressure wave is its pitch. Thus, the pitch increases, but the speed of the particles doesn't. $\endgroup$
    – Jim
    May 10, 2013 at 17:28
  • $\begingroup$ Billiard ball, 2 speeds of slow and fast: if fast moving one strikes slow moving one head-on (180 degree angle), then slow moving one will speed up to be fast. If fast moving one strikes slow moving one at odd angle (say 100 degrees) then fast one is effectively reflected and slow one may stop or slow down or move in other direction but still slowly. Due to the large number of particles and collisions in air, law of large numbers says that even a short distance away from the source, the fast particles will definitely have an odd-angle collision and not transfer a high velocity outwards. $\endgroup$
    – Jim
    May 10, 2013 at 17:33

Well, I think that the extra energy is used to increase the rate of vibration (frequency) of the particles instead of increasing the velocity of the wave because the particles of the wave don't actually travel from the source to the hearer. Once the frequency is increased then the wavelength will definitely decrease and the velocity will remain the same. ($\because v = \lambda \nu$)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.