Sound

Intensity = P/A
B = 10 log (I/I_o)

1.
Sound, form of energy vibration. The medium that the sound wave travels through, determines the speed. The energy is transmitted by waves. Sound waves need a medium to propagate. Sound waves are p-waves, longitudinal waves.

2.
To solve problems involving the intensity level of sound, you need to use the formula B = 10 log (I/I_o), where I_o is the reference intensity level, 10^-12. B is measured in decibels, and I is measured in W/m^2.

3.
Sound intensity is related to the ability to hear different frequencies since our ears are able to hear middle frequencies better, 2000-4000Hz. Sound intensity, which is measured in W/m^2 is different from sound intensity level, measured in dB. The sound intensity is derived into phons which are a unit for loudness, the intensity level in dB at 1000Hz.

4.
The fundamental harmonic, also known as the first harmonic of an open tube, can be found via f = v/2L, where v is the speed of the medium and f is the frequency. For a one end closed tube, it is found via f = v/4L.

5.
The Doppler effect describes the change in frequency due to the motion relative to the wave source. For example, a stationary siren is emitting a sound wave in a circular motion, and a duck flies towards the faint sound. As the duck nears the siren, the sound gets louder and softer, over and over.

6.
A shockwave occurs if a source of sound moves faster than the speed of sound. This is when the wave fronts it emits in the forward direction accumulate directly in front of it. If it moves faster than the speed of sound, then the wave fronts will occur on the sides of the object.

And a summarized version..
Intensity = P/A
B = 10 log (I/Io) where Io is the reference intensity 10^-12
B is measured in dB. I is measured in W/m^2.
Doppler Effect - Change in frequency due to motion of source.
Shockwave - caused when the speed of a wave source exceeds speed of wave in the medium. A sonic boom for example.

Problem 1
What is the intensity of sound at the pain level of 120dB? Compare it to that of a whisper at 20dB.

First, find the intensity of the 120dB
B = 10 log (I/Io)
120dB = 10 log (I/10^-12)
12dB = log (I/10^-12)
10^12 = I/10^-12
10^0 = I
1W/m^2 = I

Now, calculate the intensity of 20dB.
B = 10 log (I/Io)
20dB = 10 log (I/10^-12)
2 = log (I/10^-12)
10^2 = I/10^-12
10^-10W/m^2 = I

So, the intensity of sound at pain level (120dB) is 10^10 times that of a whisper at 20dB

Problem 2
A chicken standing a certain distance from an airplane with four equally noisy jet engines is feeling a sound level intensity bordering on pain, 120dB. What sound level intensity will this chicken experience if the duck pilot shuts down all but one engine(why would the duck do that??)

B = 10 log (I/Io)
120dB = 10 log (4I/10^-12)
10^12 = 4I/10^-12
1 = 4I.
0 .25w/m^2 = I
10 log .25/10^-12 = 113.9dB

Problem 3
A duck's boombox is said to have a signal-to-noise ratio of 58dB. What is the ratio of intensities of the signal and the background noise?

B = 10 log (I/Io)
58dB = 10 log I/10^-12
10^5.8 = I/10^-12
6.3 x 10^-7 W/m^2 = I
Background noise is 10^-12
Ratio is 6.3 x 10^5 : 1

Problem 4
A jet plane emits 5.0 x 10^5 j of sound energy per second. What is the intensity level 30 m away? Air absorbs sound at a rate of about 7.0dB/km. Calculate what the intensity level will be 1.0km and 5.0 km away from this jet plane, taking into account air absorption.

Intensity = P/A
A = 4pir^2, the surface area of a sphere.
B = 10 log (I/10^-12)

A) First solve for the intensity of the sound waves.
I = P/4pir^2
I = 5x10^2/(4pi(30)^2)
I = 44.218W/m^2

Now, solve for the intensity level, B.
B = 10 log (I/Io)
B = 10 log (44.218/10^-12)
b = 136.456 dB

B) I = P/4pir^2.
I = 5x10^5/(4pi(1000)^2)
I = 3.98x10^-2
B = 10 log(3.98x10^-2/10^-12)
B = 106dB.
Since air absorption is 7.0dB/km, and the soundwave has traveled 1km, then there will be 7.0dB off the top.
B = 99dB.

C) I = P/4pir^2
I = 5x10^5/(4pi(5000)^2)
I = 1.59x10^-3
B = 10 log (1.59x10^-3/10^-12)
B = 92dB.
Again, with the air absorption, but this time over 5km. So, 7.0dB/km*5km = 35dB off the top.
B = 57dB.

Problem 5
A tuning duck is set into vibration above a vertical open tube, filled with water. the water level is allowed to drop slowly. As it does so, the air in the tube above the water level is heard to resonate with the tuning duck when the distance from the tube opening to the water level is 0.125m and again at 0.395m. What is the frequency of the tuning duck?

0.395m - 0.125m = 0.27m (which is "L")
v = lambda*f
v/2L = f
343/(2(2.7)) = f
635Hz = f.s

Problem 6
A duck is flying above the Gulf of Mexico, when the duck suddenly spots a fast moving mallard below. the wake from the mallard forms an angle of 20 degrees on either side of the straight line motion. Knowing that the speed of the water waves is about 2.0m/s, what is the speed of the mallard?

sin theta = v sound/v mallard.
v sound/sin theta = v mallard.
2.0 / sin 20 = v mallard.
5.8476m/s = v mallard.

Problem 7
What will be the beat frequency if a duck quacks a middle C (262Hz) and another duck quacks a C# (277Hz) at the same time? Will this be audible to a stupid human? What if each is played two octaves lower?

277Hz - 262Hz = 15Hz
f beat = 15Hz no you stupid humans can't hear it.
277Hz/4 - 262Hz/4 = f beat = 3.75Hz, yes you stupid humans can hear it.

Problem 8
A duck travels at Mach 2.3 where the speed of sounds is 310m/s. A) What is the angle the shock wave makes with the direction of the duck's motion? B) If the duck is flying at a height of 7100m, how long after it is directly overhead will a chicken on the ground hear the shock wave?

A) sin theta = v sound/v duck
sin theta = 310/(2.3*310)
sin theta = 25.77 degrees.

B) v duck = 2.3*310m/s.
tan 25.77 = 7100/d
7100/ tan 25.77 = d.
14706.7m = d
d = vt
14706.7/(2.3*310) = t
20.6s = t

Problem 9
The frequency of a duck's quack as it flies at you is 522 Hz. After it passes you, its frequency is measured as 486 Hz. How fast was the duck flying? (Assume constant flying duck velocity.)

This problem is a good application of the doppler affect.
f` = f/(1-Vs/V) for wave source moving at you.
f` = f/(1+Vs/V) for a wave source moving away from you.
f` is the apparent frequency, f is the actual frequency, Vs is the velocity of the wave source, v is the velocity of the sound waves.
So, assuming 20 degrees celsius, the velocity of sound wave in air is 343m/s.

522 = f/(1-Vs/343) .. f = (522Vs - 179046)/343
486 = f/(1+Vs/343)
486 = (522Vs - 179046)/343/(1+Vs/343)
486 = -(522Vs - 179046)/(Vs+343)
49/4 = Vs
12.25m/s = Vs