1-A person has a hearing threshold 11 dB above normal at 100 Hz and 52 dB above

Question

1-A person has a hearing threshold 11 dB above normal at 100 Hz and 52 dB above normal at 4000 Hz. How much more intense must a 100 Hz tone be than a 4000 Hztone if they are both barely audible to this person?

2-An 8 hour exposure to a sound intensity level of 95.0 dB may cause hearing damage. What energy in joules falls on a 0.750 cm diameter eardrum so exposed?

3-Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00 105 kg on springs that have adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven—the driving force is transferred to the object, which oscillates instead of the entire building.

(a)

What effective force constant (in N/m) should the springs have to make the object oscillate with a period of 1.70 s? ____N/m

(b)

What energy (in J) is stored in the springs for a 2.30 m displacement from equilibrium?

4- What is the length (in m) of a tube that has a fundamental frequency of 168 Hz and a first overtone of 336 Hz if the speed of sound is 346 m/s?

Explanation / Answer

intensity level B1 = 10*log(I1/Io)

Io = threshold intensity = 10^-12 W/m^2

11 = 10*log(I1/10^-12)

I1 = 1.26*10^-11 W/m^2

B2 = 10*log(I2/Io)

52 = 10*log(I1/10^-12)

I2 = 1.58*10^-7 W/m^2

I2 / I1 = 12540 <<<<<-----------ANSWER

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2)

decibel level B = 10*log(I/I0)

95 = 10*log*(I/10^-12)

I = 0.0032 W/m^2

energy absorbed E = I*A*t

A =a rea of cross section = pi*r^2

r = radius = 0.75/2 = 0.375 cm = 0.00375 m

time t = 8 h = 8*3600 s

Energy = 0.0032*pi*0.00375^2*8*3600 = 0.0041 J      <<<<<-----------ANSWER

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3)

time period of spring T =2*pi*qrt(m/K)

T = 1.7 s

mass m = 4*10^5 kg

1.7 = 2*pi*sqrt(4*10^5/K)

K = 54.6*10^5 N/m     <<<<<-----------ANSWER

part(b)

energy stored E = (1/2)*K*y^2 = (1/2)*54.6*10^5*2.3^2 = 14441700 J <<<<<-----------ANSWER

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Q4)

fundamental frequency fo = 168 Hz

first overtone f1 = 336 Hz

f1/fo = 2

the pipe is open pipe

fo = v/(2L)

168 = 346/(2*L)

L = 1.03 m <<<<<-----------ANSWER

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