From this, https://electronics.stackexchange.com/questions/42904/square-wave-sine-wave-is-more-audible I now understand that a Square soundwave will be perceived louder than that of a Sine sound wave when played at the same frequency (for example 500 Hz). However, how about a sawtooth sound wave; which order from highest to lowest sound intensity is it placed?
I know that the Sine wave is only composed of the fundamental frequency and hence produces the lowest pitch - which is less sensitive to human ears. Square sound waves are composed of infinite odd harmonics, and sawtooth waves are composed of both odd and even harmonics. From experimenting, I have found that square sound waves have greatest sound intensity (dB) than sawtooth, which has greater sound intensity than sine waves. What is the explanation behind this system of order?
// Extra info. I conducted an experiment where I put a speaker and a soundmeter (Decibel 10th App) at a distance away (say 10 cm) from each other, then played 1000 Hz Sine wave and measured the intensity of it. Similarly, I did this for 1000 Hz Square, and 1000 Hz Sawtooth. What I found was that the sine had an intensity of 94dB,square had 95dB and sawtooth had 89 dB. I want to investigate the effects of the waveforms on the intensity of sound.*Oh, and the amplitude was set at a constant.On the soundmeter app, the update frequency set to 10 Hz, and calibration set to 0.2 dB. In a quiet room (during the experiment), the sound intensity was measured to be 52dB