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

1. Sketch the superposition of the following wave pulses.
   
   

   The superposition is the sum of the heights of each at each instant (A_{superposition} = A₁ + A₂)

   

   

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2. The waves on a certain string move at 50.0 m/s. The string is 7.5 m long. If both ends are fixed, what are the first five resonance or standing wave frequencies? If one end is free, what are the first five frequencies? Is there any way of telling that a string is fixed or free at one end by looking at a sequence of resonance frequencies?

   (i) The standing wave of resonance frequencies of a string fixed at both ends is given by the formula where n = 1,2,3,… Since we are given v and L, we have
   

   f_n = n(50 m/s)/(2 × 7.5 m)= n(3.33 Hz).
   

   The first five frequencies are thus: f₁ = 3.33 Hz, f₂ = 6.67 Hz, f₃ = 10.0 Hz, f₄ = 13.33 Hz, and f₅ = 16.67 Hz.
   

   (ii) The standing wave of resonance frequencies of a string fixed at only one end is given by the formula where n = 1,3,5,… Since we are given v and L, we have
   

   f_n = n(50 m/s)/(4 × 7.5 m) = n(1.67 Hz).

   The first five frequencies are thus: f₁ = 1.67 Hz, f₃ = 5.00 Hz, f₅ = 8.33 Hz, f₇ = 11.67 Hz, and f₉ = 15.00 Hz.
   

   (iii) If we have a sequence of resonance frequencies, we can tell if the string is fixed at both ends or open at one end by looking at the ratio of the frequencies. To find the ratio divide through by the greatest common divisor. For a string fixed at both ends, the ratio will have both even and odd numbers. A string fixed at only one end will have only odd numbers. For example, from part (i) above, the ratio is 1:2:3:4:5, where the common factor 3.33 Hz is factored out. In part (ii), the ratio is 1:3:5:7:9, where the common factor 1.67 Hz is factored out. Notice that the greatest common factor is the fundamental frequency.

   

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3. 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.

   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 f_n. 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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4. A string fixed at one end only is vibrating in its ninth harmonic mode. Sketch the wave. The speed of a wave on the string is v = 25.8 m/s and the string has a length of 8.25 m. What is the frequency of the wave? What is the wavelength of the wave? What is the fundamental frequency?

   The resonance frequencies of a string fixed at only one end is given by the formula where n = 1,3,5,… The first sentence tells us that n = 9 - the number of quarter wavelengths seen. We use this to sketch the standing wave.
   

   

   Using the formula, since we have n, v and L,
   

   f₉ = 9(25.8 m/s)/(4 × 8.25 m) = 7.04 Hz.

   The fundamental frequency occurs when n = 1, so

   f₁ = 1(25.8 m/s)/(4 × 8.25 m) = 0.78 Hz.

   The wavelength of a string fixed at both ends is given by the formula , where n = 1,3,5,… Thus we have
   

   λ₉ = 4(8.25 m)/9 = 3.67 m.

   Of course, we could just look at the sketch and see that one full wavelength is 4 quarters of the nine quarters displayed, so that the wavelength was just 4/9 times the length of the string.
   

   

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5. 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? Sketch the standing wave patterns. (e) If the speed of transverse waves on this string is 125 m/s, find the length of the string?
   

   (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, 9, … 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) The harmonics are the 5^th, 7^th, and 9^th.

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   .
   .

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

   .

   

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

   We see that this pattern is the same as the third harmonic of a string fixed at one end and free at the other. In the third harmonic, we see only 3/4^th of the length of the full wave, so λ = 4/3 × L = 4/3 × 1.05 m = 1.40 m.

   The frequency of vibration is given as f = 110 Hz.

   The speed v in the arrow (not the speed of the arrow!) is related to wavelegth and frequency by v = λf = (1.40 m)(110 Hz) = 154 m/s.

   

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7. 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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8. 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) = 0.204 m. The tube is 41 cm long.



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9. 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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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.

    

Questions? mike.coombes@kpu.ca