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

1. A wire of length 4.35 m and mass 137 g is under a tension of 125 N. What is the speed of a wave in this wire? If the tension is doubled, what is the speed? If the mass is doubled?

   The speed of a wave on a string is given by the formula , where is the linear density given by . Thus the speed is

   .

   If we double the tension, v = 89.1 m/s .

   If we double the mass, v = 44.5 m/s .
   

   

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2. A wave on a string has the formula y = 0.030sin(0.55x - 62.8t). What is the wavelength, frequency, period, and speed of the wave. The string has a linear density μ = 0.020 kg/m. What is the tension in the string? What is the rate of energy flow of the wave?

   First recall that the formula for a wave on a string is given by Thus, by inspection, we have A = 0.030 m, , and . Solving for λ and T yields λ = 11.4 m and T = 0.100 s.

   We know that frequency is given by f = 1/T = 10.0 Hz.

   As well, the speed of the wave is given by v = λ/T = 114 m/s.

   To find the tension in the string, we take and rewrite it as . For the given values, F_tension = 260 N.

   The rate of energy flow, or energy per unit time, or power, is given by the formula , where ω = 2πf. For the given values, P = 4.05 Watts.
   

   

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3. On a point on a string, a peak of an harmonic wave is observed to pass every 0.050 s. The distance between peaks is 0.75 m. The height of the peak is 0.025 m. What is the equation of this wave? Assume that the wave is moving to the left. What is the speed of the wave? The string has a linear density μ = 0.020 kg/m. What is the tension in the string? What is the rate of energy flow of the wave?

   Since the peak of the harmonic wave is observed to pass every 0.050 s, T = 0.050 s. Since the distance between successive peaks is one wavelength, λ = 0.75 m. The amplitude of a wave is given by the height of the peak, so A = 0.025 m.

   Next recall that the formula for a wave on a string is given by For waves moving to the left, we need the + sign. So the required equation must be

   

   The speed of the wave is given by v = λ/T = 1.50 m/s.

   From the previous question, we saw that could be rearranged to give . For the given values, F_tension = 0.045 N.

   The rate of energy flow, or energy per unit time, or power, is given by the formula , where ω = 2πf. For the given values, P = 0.296 Watts.
   

   

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4. 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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5. 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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6. 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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7. 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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8. 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
   

   

   

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9. 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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10. 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.  
    

    

Questions? mike.coombes@kpu.ca