Where does the energy which would other wise be transferred to sound energy go to when a bell is struck in a vacuum? I'm guessing that the bell heats up a little but I'm not sure, can anyone explain this to me?
2007-01-11 10:24:42 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
It is exactly that.
Say you ring a bell in a vacuum, it will vibrate. Because no solid is perfect (even in theory, it is more efficient to have some defaults), there will be a damping effect, by which the vibration of the whole structure will be absorbed by the atoms as heat.
So, your bell will heat up just a bit and then lose that energy through radation, according for instance to the stefen-boltzmann equation: P/A = sigma*T^4
That means that the power radiated depends on the 4th power of the temperature.
Hope it helps :-)
2007-01-11 10:34:17 · answer #1 · answered by Vincent L 3 · 2⤊ 0⤋
That sound energy is the energy of the bell compressing air. Since there's no air, the bell will "ring" for longer giving away less energy. The energy will eventually be given up to heat in the bell or in whatever's holding it.
2007-01-11 10:33:52 · answer #2 · answered by Nicknamr 3 · 2⤊ 0⤋
The bell vibrates just like it does in the atmosphere, it just doesn't slow down as fast because it isn't leaking energy to the air, just to whatever is holding it.
2007-01-11 10:31:56 · answer #3 · answered by Anonymous · 2⤊ 0⤋