1. Sound waves travel through an aluminum bar at a speed of 1400 m/s. If the frequency of the sound wave is 1000 Hz, what is the distance between crests on a sound wave traveling through aluminum?
The distance between crests is wave length. Hence, we have to determine wave length.
Wave length = velocity/frequency = 1400/1000 =1.4m Ans.
2. An airplane flies horizontally at a constant speed, piloted by rescuers who are searching for a disabled boat. When the plane is directly above the boat, the boat's crew blows a loud horn. By the time plane's sound detector receives the horn's sound, the plane has traveled a distance equal to one-half its altitude above the ocean. (The speed of sound is 343 m/s.) If it takes the sound
4.00s to reach the plane, determine a) the altitude of the plane, and b) its speed, in m/s. What is its speed in knots (nautical miles per hour)? a) We can draw the following figure.
Let the boat be at B and plane be at A when boat’ crew blows horn. Plane listens to horn’s sound at C after 4 secs of blowing the horn. It is given that AC = AB/2. Also BC = 343*4 = 1372m.
We have AB 2 +AC 2 = 1372 or AB 2 +AB 2 /4 = 1372 or AB = 1372*sqrt(4/5) = 1227.15m. Hence, plane is
1227.15m above the boat.
(b) Plane travels distance AC in 4 secs. We have AC = AB/2 = 1227.15/2 = 613.575m.
Hence, plane’s speed = 613.575/4 = 153.39 m/sec.
1 knot = 0.5144 m/sec. Hence, plane’s speed = 153.39/0.5144 = 298.19 knots.
3. Estimate the wavelength of a 20,000 Hz sound wave, which is near the upper limit of audibility for humans. Assume air at standard temperature and pressure.
At STP, sound velocity in air = 331m/sec
Wave length = velocity/frequency = 331*100/20000 = 1.655 cms Ans.

4) An organ pipe is a closed-ended air column. If its bottom note has a frequency of 20 Hz (close to the lowest audible frequency), then how long is the column? Assume standard temperature and pressure.
At STP, air velocity = 331m/sec.
Frequency = 20Hz. Hence, wave length = 331/20 = 16.55m
For close ended column, L = λ/4 = 16.55/4 = 4.14m.
5. The sound of a truck horn, observed when the truck is stationary, is 440 Hz. A drunk is sitting in the middle of the road at night. He sees the truck's headlights, and hears a 480 Hz horn. How rapidly is the truck approaching him, in m/s2? In mph? (As before, assume the speed of sound in air is 343 m/s.)
When observer is stationary and source is moving towards observer, the apparent frequency is given by f
2
 f v
1 v v s
, where f
2
= apparent frequency, v = sound velocity, v s
= source velocity and f
1
= source frequency.
Here, f
2
= 480Hz, f
1
= 440Hz, v = 343m/sec, v s
=?
Putting these values, we get
480

343
 v s

343 v s
480 v s
343

480
 
440
480



1 mph =

28.583 /

28.583
 
63.938
mph
The simulation at the source below allows you to experiment with many of the variables that produce sound when an object is hit or scratched. The website contains a good explanation of the variables, which include hardness, base frequency, duration, and pluck time.
University of British Columbia (n.d.). Sound simulation. Retrieved on March 1, 2008, from http://www.cs.ubc.ca/spider/kvdoel/sound_demo6.html
Because this simulation presents so many possibilities, this is open-ended. Choose one object
(for example, the circular plate) and experiment with different values of the parameters governing sound generation. Organize your work as follows.

Try striking or scraping the object at three different points. For the circular plate, this could be the edge, the center, and halfway in between.
Systematically vary the parameters (hardness, base frequency, etc.), one at a time. Summarize your conditions, and the results, in a table.
Be sure to click on the "Build" button every time you change the parameters, or the change won't go into effect. "Strike" or "scrape" the object by clicking on it. (Be sure your speakers are connected and the volume is turned up, but not too far up. You'll scare the family pets to death.) summarizing your results.
Hardness =200
Base frequency=500
Grid size =4
Duration = 5
Scrape time=0.5
Pluck time = 0.1
Hammer strength = 10000
Hit 1
In centre, we hear quite sharp sound; in middle, sound is sharper than in middle and at edge, sound is the sharpest.
Hit 2
Sound less sharp compared to hit 1 in three locations.
Pluck
Sound becomes less sharp compared to hit2.
Scrape (white noise)
As we go from centre to middle to edge, the sharpness of the sound increases.
Scrape (Gaussian noise)
Here also sound pattern is same as above but sound intensity increases as we go from centre to edge. Also sound is more sharp compared to the above.

Now we change the base frequency to 200.
Hardness =200
Base frequency=200
Grid size =4
Duration = 5
Scrape time=0.5
Pluck time = 0.1
Hammer strength = 10000
Sound pattern remains same as with base frequency 500 but intensity has decreased.
Now we change the base frequency to 100.
Hardness =200
Base frequency=100
Grid size =4
Duration = 5
Scrape time=0.5
Pluck time = 0.1
Hammer strength = 10000
Sound becomes more sharp compared to the above.
It was becoming difficult to accommodate in table. That’s why I have written observations as in text as given above.
If text is not acceptable, kindly make a table and forward so that I will fill it up.
Hit 1 Hit 2 Pluck Scrape (white Scrape

noise (Gaussian noise)
Cen Mid edge Cen Mid edge Cen Mid edge Cen Mid edge Cen Mid edge tre dle tre dle tre dle tre dle tre dle
Sha rp
>sh rper than mid dle
Mor e shar pe than centr e and midd le