Chapter 2
RQ 2 F = -k*x therefore units of k are Newtons/m
RQ 4 F = -k*x so factor of 2
RQ 5 E = 1/2 k*x*x => factor of 4
RQ 6 equation 2.6 is independent of mass
RQ 7 equation 2.8 so doubling volume decreases the frequency by sqrt(2)
doubling radius of neck (keeping volume constant) increases the frequency
by factor of 2
Q 1 Neglecting friction, the sum of kinetic and potential energy is
constant. When the kinetic energy is maximum, the potential energy is zero,
and vice versa, so the two maximums are equal (and equal to the total energy
...).
Q 4 Fig. 2.21 shows nodes of the "clang" mode. So that's where you
should strike the fork to maximize principal/clang ratio. Same picture
shows that mike position C is best.
E 1 a) k = F/x = (mg)/x = (1 kg * 9.8 m/s/s)/0.2 = 49 N/m
b) f = (1/2pi)sqrt(k/m) = 1.1 Hz
E 2 see Figure 1
E 4 equation 2.8 => f = 49 Hz
Chapter 3
RQ 1 v(light)/v(sound) = 300,000,000 (m/s)/(300 m/s) = 1 million times
faster
RQ 3 factor 1/2
RQ 4 fixed end: reflected pulse is inverted
free end: reflected pulse
is same phase as incident
RQ 5 size must be the same, shape must be mirror image
RQ 12 sound is bent up (see Fig. 3.18 c)
RQ 13 Dr. Huygens explains this very nicely
RQ 14 see RQ 1
RQ 15 Dr. Huygens explains thic very nicely, too
Q 4 yes, you could (and it would be a nice project, too !)
E2 eq. 3.7: f/fs = v/(v-vs) - v/(v+vs) = 343/342 - 343/344 =
0.6 % - they are out of tune by one tenth of a semitone.
E3 difference of temp. is say 38 C - 20 C = 18 C, then (eq. 3.5) dv
= 0.6 m/s * 18 = 10.8 m/s
then from eq. 3.1 df/f = dv/v = 10.8/343 =
3 % = half of a semitone!
E4 f = v(sound)/lambda => a) f = (343 m/s)/(15 in*.0254 m/in) = 900
Hz
b) f = (343)/(3*.0254) = 4.5 kHz
E5 eq. 3.3 => v = sqrt(56/0.00083) = 260 m/s
E6 lambda = v/f = (343/50 - 343/15,000) = 7m - 2 cm
E7 time = distance/speed = .63 m/260 m/s = 2.4 ms
Chapter 4
RQ 4 all are sharp peaks in the frequency spectrum. If frequencies
of partials are integer multiples of the lowest ("fundamental") frequency,
then they are called harmonics. First overtone is the second partial etc.
RQ 8 eq. 2.8 => sound hole (parameter a) must be smaller
Q 4 end correction of dL = 0.61r will affect all harmonics proportionately.
But if dL depends of lambda, then the overtones will get out of tune.
E 3 a) f1 = v/2L = 343/(2*16 ft*12 in/ft *.0254 m/in) = 35 Hz
f2 = 2*f1 = 70 Hz
b) f1 = v/4L = 17.5 Hz
f2 = 3*f1 = 53 Hz
E 5 f = v/2L = 343/2/2 = 85.8 Hz
two open ends => L' = L - 2*.61r = 2
- 2*.61*0.1 = 1.878 m f = v/2L = 343/2/1.878 = 91.3
E 6 eq. 4.2 => fundamental f = sqrt(t/mu)/(2L) = sqrt (56/.00083)/(2*0.65)
= 200 Hz
f2 = 2f1 = 400 Hz, f3 = 600 and f4 =
800 Hz
Problem A: Construct a log scale with ticks at 1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100 using a blank piece of paper, pencil, and log 2 = 0.3 see Figure 2Problem B: Draw on a) lin-lin scale b) log(vertical)-lin(horiz) scale
y1(x) = 10^-x
y2(x) = 1+10^-x
y3(x) = 2+10^-x
y4(x) = 1+sin(pi x) +10^-x
For the log scale, ticks at powers of ten will suffice, i.e. don't bother
subdividing the decades. See Figure 2.
Problem C: Use your calculator to determine log 850 = 2.93
and log 1150 = 3.06
Check your results against log 1000= 3
Problem D: Use your calculator to determine sin(31 degrees) = .515
and cos(46 degrees) = .695
Check your results against the "standard triangles" discussed in lecture
(recall: triangle with sides 1/1/sqrt(2) has angles 45, 45 and 90 degrees
triangle with sides 1/2/sqrt(3) has angles 30, 60 and 90 degrees
you figure out which side is against which angle in each of the two
cases ....)
sin(30 degrees) = 1/2 = .500
cos(45 degrees) = 1/sqrt(2) = .707
