The difference in sound level between two values expressed in decibels (dB) can be calculated using the formula:
Difference in decibels (dB) = 10 * log10(I1/I2),
where I1 and I2 are the sound intensities corresponding to the respective decibel values.
To determine how many times louder a sound is when comparing 60 dB to 20 dB, we can use the above formula. However, it's important to note that decibels are logarithmic units, and the relationship between decibels and perceived loudness is not linear.
Let's assume that the reference intensity for 0 dB is I0 = 1 * 10^-12 watts per square meter (W/m^2). With this reference value, we can calculate the sound intensities for 60 dB and 20 dB.
For 60 dB: I1 = I0 * 10^(60/10)
For 20 dB: I2 = I0 * 10^(20/10)
Now we can calculate the difference in decibels:
Difference in decibels (dB) = 10 * log10(I1/I2) = 10 * log10((I0 * 10^(60/10)) / (I0 * 10^(20/10))) = 10 * log10(10^6) = 10 * 6 = 60 dB
Therefore, 60 dB is 60 dB louder than 20 dB.
Regarding the second part of your question, the sound intensity is defined by the amplitude of the sound wave. The amplitude represents the maximum displacement of particles in the medium (such as air molecules) caused by the wave. The greater the amplitude, the higher the sound intensity. In other words, the amplitude of the wave determines how "loud" the sound is perceived.