PHYS207    Spring2013  Set 1 Solutions:

HW from Handout: see Figure

Chapter 1
RQ 3:  Longitudinal wave: displacement is parallel to wave propagation. Example: sound.
           Transverse wave: displacement is perpendicular to wave propagation. Example: light.
RQ 6: velocity (i.e. rate of change of position with time)
RQ 7: force F, mass m and the resulting acceleration a
RQ 10: pressure is the "normal" (perpendicular) force per unit area
RQ 12: 1 atm = 100,000 Pascals = 100,000 N/m/m = 15 lbs/in/in
RQ 16: equation 1.12
          y = transverse displacement (in meters) of the string from flat (equilibrium)
          PE = potential energy, i.e. work needed to displace the string from flat (from y=0)
          (and also work the string can do when released ...)
          L = length of the string, and T = tension
E 4 55 mi/h*1.610 km/mi = 89 km/h
       89 km/h*(1000 m/km)/3600( s/h) = 25 m/s
E 5 a = v/t = 50 mi/h*1610 m/mi/3600(s/h)/12 (s) = 1.9 m/s/s
      compare this with acceleration of free fall g = 9.8 m/s/s
E 9 picture a (typical?) body as a cube of side 2 ft.
    Then Force = pressure*surface = 15 (lbs/in/in)*6*(24 in)(24 in) =52,000 lbs!
   (the answer in Text assumes a slightly better model of human body ...)
E 11 1 hp = 746 W => efficiency = 746/2/450 = 83 %. The remaining 17 % gets converted into heat.

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 An end correction of dL = 0.61r would affect all harmonics proportionately. But if dL depends on lambda (and it does, for lambda smaller than the pipe diameter) then the frequency spectrum of the resonance peaks of the pipe will get inharmonic. However, as we shall see later, the sound produced by a blown pipe is still periodic, and so its spectrum is harmonic, and the higher harmonics are no longer reinforced by the pipe resonances. As a net result, a narrow pipe sounds bright, large-diameter pipe sounds dull. (Note: We will talk some more about this fairly complex issue, and it will NOT come up at the Exam).
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 = 2.122 m  f = v/2L = 343/(2*2.122) = 80.8 Hz
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


Chapter 5
RQ1: 10^12 = 1,000,000,000,000
RQ2: 20,000/20 = 1000
RQ3: about 740nm/400 nm = slightly less than 2
RQ9: electrical signals are generated upon hair cell bending
RQ10: far from window (i.e. the end)
Ex.1:  f = v/4L = 340/(4*.03) = 2.8 kHz
Ex2:  say lambda = distance between ears = 20 cm; then f=v/lambda=340/.2 = 1700 Hz
At this or higher frequencies, the phase information is therefore ambiguous. See discussion of localization on p. 90.
Ex3:  F = 10^-2 N/m^2 *(0.55 10-4 m^2) = 5.5 10^-7 N ( recall that 1 N = 0.2 lb)
Ex6:  dt = dl/v  dl = extra path to one ear = distance between ears*cos(45 degrees)
 => dt = .2/(sqrt(2)*340) = 0.4 ms

Chapter 6
RQ5: it goes like 1/r^2 => by factor of 4
RQ10:  add intensities. Since log 2 = 0.3, the result will be 55+3 = 58 dB
Q1: Fig 6.4 => Lp = 40 dB at 2000 Hz has 41 phons
                    Lp = 65 dB at 50 Hz has about 40 phons
Ex1:  read off Fig. 6.9: 42, 25, 8, 4, 0, 18 dB
Ex2:  Lp = 30 dB at 3000 Hz has 35 phons. To match this at 100 Hz we need about 48 dB.
Ex3:  violinists are incoherent (sic) => 50 + 10 log4 = 56 dB
Ex5:  eq. 6.2: sound power level = 10 log (5W*0.1/10^-12) = 117 dB
   intensity I = P/(4pi r^2) = 0.5 W/4pi = 0.04 W/m^2 at 1m
               => SPL = 10 log( .04/10^-12) = 106 dB
  at 4 m: I  = 0.5/(4pi 4^2) = .0025 W/m^2 => SPL = 10 log (.0025/10^-12) = 94 dB

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 2

Problem B: Draw on  a) lin-lin scale  b) log(vertical)-lin(horiz)  scale
for x in the range from -6 to +6, the graphs of the functions:

     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/sqrt(3)/2 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  

Problem E: You are listening to your favored music at the threshold of pain (110 phons) at 50 Hz as well as at 1 kHz and at 6 kHz. When neighbours complain, you reduce the sound intensity level by a factor of million (independent of frequency). Determine the resulting loudness (sones) at the three frequencies, and compare with the original.

reading off the graphs, you get original loudness of about 120 sones at all three frequencies. The reduced levels will be 35 phons(0.5 sones) at 50 Hz, 50 phons (2.3 sones) at 1 kHz, and 45 phons (1.5 sones) at 6 kHz. So the sound will lack the bass (as well as some treble) and will need equalizing.

Problem F: Two incoherent sources (e.g. violinists) emit sound of same power, and frequencies f1 and f2, respectively. What can you say about the resulting sound as compared to the two separate sounds, if
a) f1 = 1000 Hz f2 = 1000 Hz:
within critical band (and incoherent) => add intensities => add 3 dB to SIL and/or LL
b) f1 = 1000 Hz f2 = 1500 Hz :
Fig. 5.10 => this is outside the critical band =>
    add the loudnesses (more or less - see item 2 in box on p. 111)
c) f1 = 150 Hz f2 = 3500 Hz: the brain will perceive two independent sounds

Problem G:  If one violin produces SIL of 75 dB
a) what will you get from two violins (playing unison)? 75+10log2 = 78 dB
b) how much from 10 violins? 75+10log10 = 85 dB
c) how many violins do you need to get to 95 dB? 100 violins (since 75+10log100 = 95)

Concepts you may be asked to explain:

Longitudinal vs. transverse waves
reflection / refraction / absorption
interference / superposition
diffraction / Huygens' Principle
oscillator: simple / damped / driven
resonance
standing waves; resonance; modes and nodes (including derivation of fn = n f1 for a string)
waveform / wavefront / wavelength / period / frequency / speed of propagation
amplitude / phase / spectrum
mechanism of hearing: (eardrum, Eustachian tube, cochlea, basilar membrane, hair cells): what is the basic design principle?
loudness vs. intensity:  Sound Intensity, Sound Intensity Level (and Sound Pressure Level), Loudness level, Loudness
critical band, addition of sounds