# How to apply conservation of energy to active noise-cancelling headphones?

Active noise-cancelling headphones add the opposite waveforms of the ambient sound in order to cancel out this ambient sound by destructive interference. I do not understand why adding more power in this way does not increase the total power of the played-back sound en thus its intensity (dB). I understand that if you have constructive and destructive interference at the same time, conservation of energy means a spatial redistribution of energy. But how to apply this to this application?

• Imagine a mirror but for soundwaves instead of lightwaves. And using electronics instead of... well... electrons. Jun 18, 2016 at 13:38
• "Silence Please" from Arthur C Clarke's "Tales From The White Hart" covers this topic... Jun 18, 2016 at 16:01
• @DJohnM just reread the story last week. In it, Clarke has one of his characters say "There are no side effects from the machine" but forgets to mention the total sound energy in the room doubling.
– user108787
Jun 18, 2016 at 16:56
• The headphone will cause a tiny increase in the external sound level but it's completely irrelevant. The power of sound at the eardrum is fractions of a $\mu W$, the power output of the headphone probably no more than a fraction of a $mW$. Jun 18, 2016 at 20:09
• Is there any energy loss by heat dissipation in the air inside the headphones? Jun 19, 2016 at 16:29

Active noise cancellation works by playing an inverted mirror image of the sound to erase the noise from existence and achieve complete silence (via physics of destructive interference).

This can be represented with following formula.

x - x = 0

If you notice, the waveform (x) and the inverted waveform (-x) both sum to zero,

so the total amount of energy destroyed is exactly zero.

Therefore, no energy was destroyed.

You can think of the function sin(x) as "do" and -sin(x) as "undo" (e.g. inverse functions).

The destruction of -sin(x) will undo the destruction of sin(x).