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Physics 1120: Standing Waves and Sound Level Solutions

    Sound Level

  1. A wire of length 4.35 m and mass 137 g is under a tension of 125 N. A standing wave has formed which has seven nodes including the endpoints. What is the frequency of this wave? Which harmonic is it? What is the fundamental frequency? The maximum amplitude at the antinodes is 0.0075 m, write an equation for this standing wave.

    2. First we sketch the standing wave.

    The equations for a string fixed at both ends are and. Examining the sketch , we see that n = #node - 1 = 6, so that this is the sixth harmonic. We are given L, so we need the speed of the wave v to determine fn. The speed of the wave can be found from the formula, where μ is the linear density given by . Using the given data, the speed may be computed

    .

    Hence, 

    The fundamental, or n = 1, frequency is f1 = 7.24 Hz.

    The equation of a standing wave is given by  In this case, we have

    ,

    where  has been used.

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  2. A string fixed at one end only is vibrating in its third harmonic. The wave function is y(x,t) = 0.02sin(3.13x)cos(512t), where y and x are in metres and t is in seconds. (a) What is the wavelength of the wave? (b) What is the length of the string. (c) What is the speed of the transverse wave in the string?


    3. The equation of a standing wave is given by We are told that this is the third harmonic, so n = 3.

    (a) Since  for strings with one free end, we have . From this we determine that the wavelength,λ3 = 2.01 m.

    (b) We also find the length of the string to be L = 1.51 m.

    (c) The speed of the wave is given by. Comparing the equation of the standing wave with the particular given wave, we see that. Thus

    .

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  3. Three successive resonance frequencies for a certain string are 175, 245, and 315 Hz. (a) Find the ratio of these three modes. (b) How can you tell that this sting has an antinode at one end? (c) What is the fundamental frequency? (d) Which harmonics are these resonance frequencies? (e) If the speed of transverse waves on this string is 125 m/s, find the length of the string?


    4. (a) The ratio of these three frequencies is 175:245:315, or 535:735:935, or 5:7:9.

    (b) For a string fixed at both ends, the resonant frequencies are given by , where n = 1, 2, 3, 4, ? For a string fixed at only one end, the resonant frequencies are given by, where n = 1, 3, 5, 7, … Since the given sequence has only odd numbers, we may conclude that the string is fixed at only one end.

    (c) The fundamental frequency is the greatest common factor of the sequence, so f1 = 35.0 Hz.

    (d) Since , we can rearrange this equation to find the length,

    .

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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.

    5. 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. In an archery contest, a 1.15-m arrow sticks in the target and vibrates. If the speed of the waves in that arrow is 150 m/s, what are the three lowest resonance frequencies for the that vibration? Sketch the standing waves.

    The patterns will be of a standing wave with one fixed and one free end.

    Such patterns obey the formula , where n = 1, 3, 5, 7, …. Solving fn = n(31.61 Hz). so the lowest three frequencies are f1 = 31.61 Hz, f3 = 97.83 Hz, and f5 = 163.04 Hz.

    The first thee patterns are shown below.

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  6. 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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  7. While watching an archery show you taped earlier, you notice that when the 1.05 m arrows hit a target they vibrate with the following pattern. Using frame advance on your VCR you deduce that the arrow vibrates 110 times per second. What is the speed of the waves travelling through the arrow?

    The patterns will be of a standing wave with one fixed and one free end. This is the third harmonic.

    Such patterns obey the formula , where n = 1, 3, 5, 7, …. Here n = 3. Solving for the speed of the wave v = 4Lfn/n = (4)(1.05 m)(110 Hz)/3 = 154 m/s.

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    Sound Level

  8. A typical speaker diaphragm vibrates with a maximum displacement of 2.00 mm. Assuming that this is also the maximum displacement of the nearby air molecules, find the maximum pressure amplitude. Take the frequency of the speaker to be f = 3000 Hz, the density of the air to be ρ = 1.29 kg/m3, and the speed of sound to be 340 m/s.

    maximum pressure amplitude is given by the formula

    p0 = ρωvδ0 = (1.29 kg/m3)(2π × 3000 Hz)(340 m/s)(0.002 m) = 16.5 × 103 Pascal.

    Since ordinary atmosphere pressure is approximately 1 × 105, this is about a 16% change in pressure.

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  9. 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?

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

    Istudent = (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

    Igroup = 30 Istudent = 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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  10. 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 rDAD = 1.20 m, rMOM = 1.35 m, and rDOG = 2.00 m. The power in their voices are respectively, PDAD = 1.25 mW, PMOM = 0.85 mW, and PDOG = 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 40 dB. I0 = 10-12 W/m2.

    11. These are all incoherent sound sources, so the intensities add

    Itotal = IDAD + IMOM + IDOG + Ibackground .

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

    IDAD = (1.25 × 10-3 W) / 4π(1.2 m)2 = 6.908 × 10-5 W/m2,

    IMOM = (0.85 × 10-3 W) / 4π(1.35 m)2 = 3.711 × 10-5 W/m2, and

    IDOG = (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

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

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

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

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    Doppler Shift

  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.


    12. The formula for the Doppler Shift is given by

    fshift = fsiren [(1 ± udriver/v)/(1 ± upolice /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) fa = (2500 Hz)[1 + 25/340]/[1 − 40/340] = 3042 Hz .

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

    (c) fc = (2500 Hz)[1 − 25/340]/[1 − 40/340] = 2625Hz .

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

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  12. 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


    13. f´ = f0 [(1 ± ur/v)/(1 ± us/v][(1 ± us/v)/(1 ± ur/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 = vsoundΔt .

    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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  13. A siren with a frequency f0 = 2000 Hz is attached to a block. The siren and block together have a mass of 2.00 kg. The block is attached to a spring of unknown spring constant K. The spring is compressed an unknown distance and then released. The siren and block oscillate back and forth. A listener hears the source as emitting a varying frequency since the siren is moving. The highest frequency that the listener hears is 2060 Hz. The listener also determines that he hears this highest frequency repeat every 1.50 seconds. The speed of sound in air is 343 m/s.
    (a) How fast is the block moving when the listener hears the highest frequency?
    (b) Where in its motion is the block when the listener hears the highest frequency?
    (c) What is the lowest frequency that the listener would hear?
    (d) Where in its motion is the block when the listener hears the lowest frequency?
    (e) What is the period and angular frequency of the block?
    (f) What is the amplitude of the displacement of the block?
    (g) What is the spring constant of the spring?

    14. This problem combines SHM with the Doppler Effect.

    (a) The moving source is approaching the stationary observor when the observed frequency is highest. The Doppler Effect equation for this case

    fobs = fsource ( 1 + vsource/vsound),

    We can rewrite the above equation to solve for the velocity of the source and use the given numerical values.

    vsource = vsound ( fobs/fsource − 1)

    With the numbers we have, we get

    vsource = (343 m/s)(2060/2000 − 1) = 10.29 m/s
    .

    (b) Accoding to SHM, the block is moving fastest towards the observor when it passes to the equilibrium moving to the left.

    (c) & (d) The Doppler Effect produces the lowest observed frequency when the block is moving away from the observor and passing through equilibrium. The block will have the same speed as in part (a). The equation for this is

    fobs = fsource ( 1 − vsource/vsound),

    and given a result which is 60 Hz below the source frequency,fobs = 1940 Hz.

    (e) The peak observed frequency occurs periodically and must match the period of the motion, so T = 1.50 s. The angular frequency is ω = 2π/T = 4.189 Hz.

    (f) Maximum velocity is given by the formula vmax = ωA. So A = vmax / ω = (10.29 m/s) / (4.189 Hz) = 2.457 m.

    (g) Conservation of Energy says ½mvmax2 = ½KA2 where K is the spring constant or stiffness. Using vmax = ωA, this equation reduces to mω2 = K. With our values, K = (2.00 kg)(4.189)2 = 35.10 N/m.

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    Beats

  14. 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 fbeat = |f - fknown|. 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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  15. You have two 400-Hz tuning forks which you ring together. You drop one fork down a well. Before the fork hits bottom and stops ringing you hear a beat frequency of 15 Hz. How deep is the well?

    You hear a beat frequency because the falling tuning fork is Doppler shifted to a lower frequency as it is moving away. In fact it must be at 400 − 15 = 385 Hz.

    Using the Doppler Shift equation for the case when the source is moving away,

    fobs = fsource ( 1 − vsource/vsound),

    we can find the speed of the moving tuning fork by rearranging the above equation.

    vsource = vsound ( 1 − fobs/fsource)

    With the numbers we have and taking vsound = 340 m/s, we get

    vsource = (343 m/s)( 1 − 385/400) = 12.75 m/s
    .

    Using simple kinematics for an object dropped from rest under the influence of gravity, the distance the tuning fork fell was

    d = vsource2 / 2g = (12.75 m/s)2/2/(9.81 m/s2) = 8.29 m.

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