Physics of Musical Sound , 33-114 Homework #6, Spring, 2014
SOLUTIONS
Problem 1) The trick here is often expressed as In music, 6+5=10. The point is that terms like
fth count both the starting and ending notes. Lets start with C. Then the major sixth take
Physics of Musical Sound, 33-114 Homework #1, Spring, 2014
SOLUTIONS
Problem 1) Hall 1-6
Reasonable chest area estimates might range from (0.1 - 0.3) m2 . Lets use S = 0.2 m2 .
(Careful: 0.1 m2 = 1000 cm2 , and NOT 10 cm2 . Make sure you can see why.)
The
Physics of Musical Sounds, 33-114 Homework #7, Spring, 2014
SOLUTIONS
Problem 1) If we pick any number of harmonics from a series nf1 , then the period of the sum is
1/f1 . The period of the nth harmonic alone would be n times smaller, but an integral num
Physics of Musical Sound, 33-114 Homework #5, Spring, 2014
SOLUTIONS
In the case of calculations, round-o errors of 1 cent are acceptable. But you should have exact
values in 12-tet (simply multiples of 100) and not have round-o errors for the most basic
Physics of Musical Sound, 33-114 Homework #4, Spring, 2014
SOLUTIONS
Problem 1) Modied Hall problem
a) The amplitude will vary as the square-root of the intensity.
There are really two steps to this; I dont expect you to give this level of detail! But it
Physics of Musical Sounds, 33-114 Homework #11, Spring, 2014
SOLUTIONS
Problem 1) Hall 13-8
A clarinet has notes (with all holes closed) of f, 3f, 5f, . The same pattern applies to any other
ngering as well; the fundamental f simply changes. So we get 3 t
Physics of Musical Sounds, 33-114 Homework #9, Spring, 2014
SOLUTIONS
Problems 1) and 2) are good examples of being careful to distinguish variables with subscripts, so
we avoid confusion. You are of course free to be a bit less explicit, but if you are w
Physics of Musical Sounds, 33-114 Homework #10, Spring, 2014
SOLUTIONS
Problem 1) Chapter 11, Problem 15.
The key point is that air is very light and easily displaced as the heavier glass vibrates. Liquid is
1000 times denser, and is harder to move. The n
Physics of Musical Sounds, 33-114 Homework #8, Spring, 2014
SOLUTIONS
Problem 1) Hall 9-18
The top pair has a node only at the outer edge, and an anti-node in the center; it is (0 1). The next
pair has an additional circular node between the edge and the
Physics of Musical Sound, 33-114 Homework #3, Spring, 2014
SOLUTIONS
Problem 1) Hall, Ch. 5, Problem 3
The total power P = IS = (106 )(7 105 ) = 7 1011 W.
E = P t, so in 10 seconds, we would have 7 1010 J.
Problem 2) Hall, Ch. 5, Problem 5
SIL = 10 log(10
Physics of Musical Sound, 33-114 Homework #2, Spring, 2014
SOLUTIONS
Problem 1) Hall 4-7
= v/f = (344 m/s)/(688 Hz) = 0.5 m.
Constructive: |L1 L2 | = n for integral n,
so |L1 L2 | = 0.0, 0.5, 1.0, 1.5, . m.
Destructive: |L1 L2 | = (n + 1/2) for integral