Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13

1. A bat produces a spherical sonar pulse with a power of 0.020 W. There is a bug 5.0 m away that has a cross sectional area of 100 mm². Assuming the bug also acts as a spherical source for the reflection, what will be the intensity of the reflection when it reaches the bat? Assume that the bug reflects all of the sound it intercepts and that the separation of bat and bug remains constant.

   Intensity drops as the square of the distance away, so the intensity of the sonar pulse reaching the bat is

   I_bug = P₀ / r² = (0.020 W) / (5.0 m)² = 0.00080 W/m².

   The total sound power that reflects off the bat (assuming none is absorbed) is 

   P_bug = I_bug × A_bug = (0.00080 W/m²) × (100 mm²) × (1 m / 1000 mm)² = 8.0 × 10^-8 W.

   Now this sound power radiates spherically back to the bat, so the intensity of the reflection is

   I_bat = P_bug / r² = (8.0 × 10^-8 W) / (5.0 m)² = 3.2 × 10^-9 W/m².



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2. Volcanic eruptions are loud. Big eruptions can be heard hundreds of kilometres away. Suppose you just hear an eruption that is 100 km away. Typically a person might notice a sound of about 1 × 10^-8 W/m² while outside. What would be the intensity at 100 m?

   Since intensity drops as the square of the distance (for spherical sound sources),  I_near = P₀ / r_near²  and I_far = P₀ / r_far² where P₀ is the sound power emitted by the volcano. Since  we don't know P₀ we elimate it from the two equations to get  I_near r_near² =  I_farr_far².  Solving for I_far yields

   I_near =  I_far r_far² / r_near² = (1 × 10^-8 W/m²) × (100 km)² / (100 m)² = 1 × 10^-2 W/m². 

   

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3. The human ear canal is approximately a cylindrical tube of length 2.5 cm. Find the fundamental resonant frequency? So we consider the ear as a thin tube open at the pinna and closed at the eardrum.

   The speed of sound is 340 m/s in air. The lowest frequency standing wave is open at both ends and is given by

   f = nv / 4L = (1)(340 m/s) / (4)(0.025 m) =  3400 Hz.

   

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4. Having no ruler at hand, you decide to determine the length of a hollow open-ended tube by holding one end next to a sound source and varying the frequency of the sound. The lowest frequency that causes resonance in the tube is f = 420 Hz. Assuming that the speed of sound is 343 m/s, what is the length of the tube? Sketch the standing wave.
   

   For a tube open at on end, the standing waves frequencies obey the formula where n = 1,3,5,… The lowest frequency corresponds to n = 1. Since we are given v and L, we have
   

   f₁ = 420 Hz = 1(343 m/s) /4L.

   Solving for L yields,
   

   L = (343 m/s) / 4(420 Hz) = 0.408 m. The tube is 41 cm long.



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5. A narrow hollow tube of length L = 3.50 m is open at both ends and sits on two sawhorses. A wind is blowing and the tube is "moaning". What is the lowest possible frequency for the sound produced? Sketch the standing wave. Take the speed of sound to be 340 m/s.
   
   

The open ends will be antinodes and the location of the sawhorses must be nodes, so the lowest frequency standing wave in the tube looks like



which is the n = 2 harmonic. The frequency of the standing wave is therefore

f = nv / 2L = (2)(340 m/s) / (2)(3.50 m) =  97.1 Hz.  



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6. As shown in the diagram below, a speaker plays a single frequency sound. On its way to you, some of the sound echoes (reflects) off a wall. As a result, the sound is quieter than it should be.
   (a) If the wall is flexible, what is the lowest frequency that creates the lowest sound intensity at your position?
   (b) If the wall is stiff, what is the lowest frequency that creates the lowest sound intensity at your position?
   
   

   The speed of sound is 340 m/s. The lowest sound intensity occurs for destructive interference.  If the wall is flexible, when the sound reflects the wave will not change phase. Thus the condition for destructive inference is
   

   D.I.        Δx / λ = ½n,        where n = 1, 3, 5, …

   We need to use the Pythagorean Theorem to get the difference in position, Δx = 2[(1.8 m) + (2.0 m)]^½ - 4.0 m = 1.381 m. We can use the relation v = λf, to eliminate λ in favour of f. Thus we get
   

   D.I.        f = nv / 2Δx = n(123 Hz),        where n = 1, 3, 5, …

   The lowest frequency is therefore 123 Hz.
   

   Now if the wall is stiff, the phase of one wave is flipped, that is it is moved out of phase by one half wavelength. We must account for this initial difference
   

   D.I.        Δx / λ + ½ = ½n,        where n = 1, 3, 5, …

   Or in terms of frequency

   D.I.        f = ½[n - 1]v / Δx = [n - 1](123 Hz),        where n = 1, 3, 5, …

   Here the n = 1 gives an unphysical solution, and the lowest frequency occurs for n = 3 and is 246 Hz.



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7. Two in-phase loudspeakers are located 3.00 m apart on the stage of an auditorium. A detector is placed 20.0 m from one speaker and 21.2 m from the other. A signal generator drives the two speakers in phase.   What is the lowest note for which destructive interference is a maximum?  What is the lowest note for which constructive interference is a maximum?  The speed of sound in air is 343 m/s. The signal generator now flips the phase of the nearer speaker by 180°, how do your answers above change?
   

   Since the sounds are in phase, destructive interference occurs when the difference in position is an odd number of half wavelengths,
   

   D.I.        Δx / λ = ½n,        where n = 1, 3, 5, …

   The difference in position is Δx = 21.2 m - 20.0 m = 1.2 m. We can use the relation v = λf, to eliminate λ in favour of f. Thus we get
   

   D.I.        f = nv / 2Δx = n(143 Hz),        where n = 1, 3, 5, …

   The lowest frequency is therefore 143 Hz.

   Similarly, constructive interference occurs when the  difference in position is an integral number of full wavelengths,
   

   C.I.        Δx / λ = n,        where n = 0, 1, 2, 3, …

   Using the difference in position is Δx = 1.2 m. and the relation v = λf, we get
   

   D.I.        f = nv / Δx = n(286 Hz),        where n = 0, 1, 2, 3, …

   The lowest frequency is therefore 286 Hz for n = 1. Note n = 0 giving f = 0 Hz is not a sensible physical result.
   

   Now the phase of one speaker is flipped, that is it is moved out of phase by one half wavelength. We must account for this initial difference
   

   D.I.        Δx / λ + ½ = ½n,        where n = 1, 3, 5, …

   Or in terms of frequency

   D.I.        f = ½[n - 1]v / Δx = [n - 1](143 Hz),        where n = 1, 3, 5, …
   Here the n = 1 gives an unphysical solution, and the lowest frequency occurs for n = 3 and is 286 Hz.

   We must account for this initial difference for constructive interference as well

   C.I.        Δx / λ + ½ = n,        where n = 0, 1, 2, 3, …

   Which becomes

   C.I.         f = [n - ½]v / Δx = [n - ½](286 Hz)       where n = 0, 1, 2, 3, …

   The lowest physical frequency occurs for n = 1 and is 143 Hz.
   

   Note that the effect of flipping the phase was to switch which frequencies exhibit constructive interference and which show destructive interference.

   

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8. When one student is doing an exam in an otherwise very quiet room, the sound level is 45 dB. What is the intensity of the noise produced by the student? If there are 30 equally noisy students in the room, and assuming that you are the same distance from all the students, what would the new sound level be?

   Sound intensity is related to sound level by the formula, I = I₀ × 10^-(β/10). So the intensity of one student is
   

   I_student = (1 × 10^-12 W/m²)10^(45/10) = 3.162 × 10^-8 W/m².

   For incoherent sound sources, intensity adds. So with the 30 students the sound intensity is
   

   I_group = 30 I_student = 9.487 × 10^-7 W/m².
   

   Using the formula for sound level,
   

   β = 10log(I/I₀) = 10log(9.487×10^-7 / 1×10^-12) = 59.8 dB.
   

   The sound level in the classroom reaches 60 dB.

   

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9. You've been out very late and when you come home your parents are very angry and start shouting at you. This upsets the family dog who starts howling. The diagram below shows their positions with you in the middle. The distances are r_DAD = 1.20 m, r_MOM = 1.35 m, and r_DOG = 2.00 m. The power in their voices are respectively, P_DAD = 1.25 mW, P_MOM = 0.85 mW, and P_DOG = 1.00 mW. Find the intensity of sound from each source at your position. Treat the sources as incoherent (in the physical sense) and find the sound level. Be sure to include the effects of the normal background sound level of 60 dB. I₀ = 10^-12 W/m².

   These are all incoherent sound sources, so the intensities add

   I_total = I_DAD + I_MOM + I_DOG + I_background .

   Assuming that we are dealing with spherical sound sources, I = P / 4πr², so
   

   I_DAD = (1.25 × 10^-3 W) / 4π(1.2 m)² = 6.908 × 10^-5 W/m²,
   I_MOM = (0.85 × 10^-3 W) / 4π(1.35 m)² = 3.711 × 10^-5 W/m², and
   I_DOG = (1.00 × 10^-3 W) / 4π(2 m)² = 1.989 × 10^-5 W/m² .

   The background intensity is found from the background sound level using
   

   I_background = I₀ × 10^β/10 = (1 × 10^-12 W/m²) × 10⁶ = 1 × 10^-6 W/m² .

   Thus the total intensity is I_total = 12.71 × 10^-5 W/m² and the sound level at your position is,
   

   β = 10log(I_total/I₀) = 10log[(12.71 × 10^-5)/(1 × 10^-12)] = 81.0 dB .

   

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10. The beat frequency between an unknown tuning fork and a 500 Hz tuning fork is 12 Hz. Compared with a 504 Hz tuning fork, the beat frequency is 16 Hz. What is the frequency of the unknown tuning fork?

    Let f be the unknown frequency, the beat frequency is defined as f_beat = |f - f_known|. Therefore, in the two stated cases, we have
    

    12 Hz = |f - 500 Hz|, and
    16 Hz = |f - 504 Hz|.

    Each of these equations has two possible solutions. However, since the beat frequency increased as the know frequency increases, the solution must be lower than 500 Hz. Hence
    

    f = 500 Hz - 12 Hz = 504 Hz - 16 Hz = 488 Hz . The unknown frequency is 488 Hz.

    

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11. A driver travels north on a highway at a speed of 25 m/s. A police car, driving south at a speed of 40 m/s, approaches with its siren sounding at a base frequency of 2500 Hz. (a) What frequency is heard by the driver as the police car approaches? (b) What frequency is heard by the driver after the police car passes him? If the driver had been travelling south, what would your results have been for (a) and (b)? The speed of sound in air is v = 340 m/s.

    The formula for the Doppler Shift is given by
    

    f_shift = f_siren [(1 ± u_driver/v)/(1 ± u_police /v].

    To use the above equation, we need to know how the driver and police constable are moving relative to one another, as is shown in the diagram below.
    

    

    (a) f_a = (2500 Hz)[1 + 25/340]/[1 - 40/340] = 3042 Hz .
    

    (b) f_b = (2500 Hz)[1 - 25/340]/[1 + 40/340] = 2072 Hz .
    

    (c) f_c = (2500 Hz)[1 - 25/340]/[1 - 40/340] = 2625Hz .
    

    (d) f_d = (2500 Hz)[1 + 25/340]/[1 + 40/340] = 2401 Hz .

    

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12. Owls, night hunters like bats, also use echolocation to find prey but it is passive not active as the owls are not emitting the sound. Discuss how this difference, and the requirement for some light to see, affect the hunting strategy of owls with respect to bats?

    Owls  have to quietly listen for squeaks from the small animals they hunt and then they swoop in for the kill. Owls have very keen night vision but it would be dangerous to fly in complete darkness in the woods or a barn. Bats use sonar to find flying prey, usually insects, that may be silent. With sonar, false reflections can be a problem but flying insects are usually some distance from other reflecting objects. On the other hand, a bat might have trouble locating an insect on the ground as there would be too many reflactions.  
    

    

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13. In sonar, an intermittent high frequency sound pulse is broadcast in all directions. The sound is reflected from solid objects and returns to broadcaster. The time it took for the echo to return and the direction from which the echo came are used to locate nearby objects. This is Echo Location. By measuring the Doppler Shift of the echo, the speed of the object can be found. There is an added complication in that the Doppler Shift occurs twice, once from the source to the receiver, and then from the receiver (now a source of the echo) back to the original source (which is now a receiver of the echo). The echo will have a frequency

    f' = f₀ [(1 ± u_r/v)/(1 ± u_s/v][(1 ± u_s/v)/(1 ± u_r/v)] .

    A submarine traveling at 17 km/h sends out pulses at 38.7 MHz. The delay in the echo off a second sub has been rapidly decreasing and is currently 75 ms. How far apart are the two subs? If the second sub is moving at 22 km/h, what is the frequency of the returned echo? The speed of sound in seawater is 1.54 km/s.
    

    The sound is emitted by the first sub, hits the second and returns to the first. Sound travels much faster than subs, so we may assume that the distance the sound travels is 2L.
    

    
    

    The distance the sound travels is related to t, by
    

    d = 2L = v_soundΔt .

    Thus the distance between the two sub is
    

    L = ½(1.54 × 10³ m/s) (75 × 10^-3 s) = 57.75 m .

    The altered frequency that the first sub hears is
    

    

    

Questions? mikec@kwantlen.bc.ca