1.A person has severe hearing loss that reduces the intensity in his inner ear b

Question

1.A person has severe hearing loss that reduces the intensity in his inner ear by 25.0 dB. To compensate, a hearing aid can amplify the sound pressure amplitude by a certain factor to bring the intensity level back to the value for a healthy ear. What amplification factor is needed for this person?

2.The distance between a node and the nearest antinode of a standing sound wave in air is 7.30 m. What is the frequency of this wave? (Use 343 m/s as the speed of sound in air.)

3.A pipe is open at both ends. The pipe has resonant frequencies of 520 Hz and 624 Hz (among others). Find the two lowest possible values for the length of the pipe. (Enter your answers from smallest to largest.)

4.A violin string tuned to 485.0 Hz is found to exhibit 3.3 beats per second with a guitar string. The guitar string is then tightened, and the beat frequency is found to decrease. What was the original frequency of the guitar string? (Enter your answer to one decimal place.)

5.A bat is pursuing an insect and using echolocation of 60-kHz sound waves to track the insect. The insect is initially 16 cm from the bat, but then moves to a distance of 48 cm as it tries to escape. What is the difference in the two echo times the bat measures?

Explanation / Answer

1) 25 = 10log(I2/i1)

so amplification is 10^2.5= 316

2) so wavelength/4 = 7.3

wavelength = 4*7.3

f = v/wavelength = 343/(4*7.3)=11.75 Hz

3. so here f = odd v/4L

so lowest if seperation is one step

so v/2L = 104

L = 343/(2*104)=1.65 m

next is when sepeation is two steps

v/2L = 52

L = 343/(2*52)=3.30 m

4) so tightening will raise its frequency so if that decreased beat that means it was below it

so 485-3.3 = 481.7 Hz

5) t1 = 2*.16/343=9.33E-4

t2 = 2*.48/343=2.80E-3

so dt = 2.80E-3-9.33E-4=1.87E-3 s

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