hw11b - HW#11b Note: numbers used in solution steps are...

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HW#11b Note: numbers used in solution steps are different from your WebAssign values. Page 1 of 5 Problem 1. A person wears a hearing aid, which makes a sounds louder by 30 dB. This means the sound intensity is increased times. This means the amplitude of the sound-wave is increased times. Solution. The number of decibels of difference between sound levels I 2 and I 1 is given by 2 1 10log I I β ⎛⎞ = ⎜⎟ ⎝⎠ . We can solve for the ratio I 2 / I 1 : () 2 1 30 3 2 10 10 1 log 10 10 10 10 I I I I = == = The sound intensity increases by a factor of one thousand. The intensity is proportional to the square of the amplitude. So, the amplitude increases only by a factor of 1000 31.6 = . This means the amplitude of the sound-wave is increased 31.6 times. I is proportional to A 2 , In order to make I increase to 1000 times, A needs to increase to 1000 31.6 = times. A is proportional to I Attention: Please carefully review log functions. If log(x) = 2, x = 10 2 ;If log(y)=b , y= 10 b Also, log(A/B) = log A- Log B Please check your old math book or found those online…. Problem 2. Two sound waves in air (call them X and Y) have wavelengths of 1.5 m and 3 m, and amplitudes of 0.01 Pa and 1 Pa, respectively. a.) Find the frequency of each wave. f X = Hz, f Y = Hz Which sound wave has a higher pitch? b.) Which sound wave is louder? Fill in the following: The intensity of Y is that of X by times. Soundwave Y is dB soundwave X. (dB=decibels) Solution. (a) We will use 343 m/s for the speed of sound. Then ( ) 343 m/s 229 Hz 1.5 m 343 m/s 114 Hz 3 m X X Y Y v f v f λ = = Thus X has the higher pitch. Whichever has shorter wavelength has higher frequency. (b) As for loudness, intensity and so forth, the wave with the higher pressure amplitude ( Y ) is louder.
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hw11b - HW#11b Note: numbers used in solution steps are...

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