Question

1. If I=35,000 time the threshold of hearing, what is the difference in db?

2. Mathematically prove that the threshold of pain is 130db. (Mathematically prove means to show an equation and solution)

3. If one speaker is quieter than another, what is the reduction in db?

a. If it has only 1/3 the intensity of the other?

b. If it has only 1/6 the intensity of the other?

4. Mathematically prove that a 10db increase is 10 times the power, and only 3.16 times the pressure/voltage. (Show an equation and solution)

5. Describe the difference between A weighting and C weighting (be thorough but concise).

Answer 1

1)

Difference in dB means sound Intensity level

so L_I = 10 log₁₀ (I / I₀) here, I₀ = Threshold of hearing

given, I = 35000 I₀

L_I = 10 log₁₀ (35000 I₀ / I₀)

= 10 log₁₀ (3.5 * 10⁴) = 45.44 db

2)

as we all are aware that threshold of pain is 10¹³ times the threshold of hearing

so here I = 10¹³ I₀

so, Threshold of pain

L_I-pain = 10 log₁₀ (I / I₀) here, I₀ = Threshold of hearing

L_I = 10 log₁₀ (10¹³ I₀ / I₀)

= 10 log₁₀ ( 10¹³) = 130 db proved

3)  

a)

let louder has intensity of I₀ so quieter has I₀/3 .

so, reduction in db = 10 log₁₀ ((I₀/3)/I₀) = - 4.77 db = 4.77 db

b)

let louder has intensity of I₀ so quieter has I₀/6 .

so, reduction in db = 10 log₁₀ ((I₀/6) / I₀) = -7.78 db = 7.78 db

4)

as we know that power is directly proportional to the square of the pressure

so, here power P = 3.16²

also, Power P = 10 log ( I/I₀)

Increase in the sound intensity level L_I = 10 * log₁₀ (3.16²)

= 10 * 2 * log₁₀ (3.16) = 10 db

5)

The A-weightings and C-weightings are different sensitivity scales used for measuring noise activity of different frequency levels.

A weighting sound level is able to discriminates the low frequencies as in the case of our ear where as C weighting can not discriminate the low frequencies sound level and mainly used for monitoring sources such as engine , explosions etc.