Worksheet #12 1. The second harmonic sinusoidal standing wave in a pipe with both ends open has a wavelength of 5.00 m. The sound speed is 343 m/s. (a) How many nodes and anti-nodes are there in the displacement wave? How many nodes are there in the pressure wave? Draw the positions of these nodes and anti-nodes. (b) What is the length of the pipe? (c) What is the frequency of the third harmonic? (d) What is the frequency of the fundamental mode in this pipe? (e) If the same pipe was closed at one end and open at the other (stopped pipe), what would be the frequency of the fundamental (first) and third harmonics? 1
Worksheet #12 2. As shown in the figure, a weight is attached to the end of a string draped over a pulley; the other end of the string is attached to a mechanical oscillator with a set frequency f = 80.0 Hz. The length of the string is L = 1.00 m, and the linear mass density of the string is µ = 75.0 g/m. For certain values of the weight, the string will resonate. What weights are needed for each of the first three normal modes of the string? 2
Worksheet #12 3. A whistle with a fundamental frequency of 300 Hz is aboard a train traveling at 35.0 m/s as it approaches a tunnel in the smooth, vertical face of a rock cliff. The engineer sounds the whistle. A rider in the last car at the rear of the train simultaneously hears the train whistle and also the echo from the wall. (a) How many beats per second does the rider hear? (b) What would be the beat frequency heard by an observer on the ground near the end of the train? 3
Worksheet #12 4. The Wind Sculpture. Your friend, an artist, would like to use wires to hang a sculpture as sort of a string instrument. She decides to set it up as shown in the diagram below. Her basic design involves attaching two 75 cm pieces of wire from two eye-hooks on the ceiling that are approximately 50 cm apart and then hanging the 25.0 kg sculpture from a 50 cm horizontal bar with insignificant mass (which also does not vibrate much in the wind) from some point along the bar. The aural effect that she would like to achieve is that when the wind blows across the two strings, they play a perfect fifth, i.e. the ratio of the fundamental frequencies of the two sounds is 3 : 2. Your friend tells you that she has been successful in hanging the sculpture but not in choosing the point along the bar to hang the sculpture giving the desired sound. Desperate for success, she knows that you are taking physics 111 and asks you for help. Before you tackle the analysis, you use your knowledge of waves to gather some more information. You take a sample of the wire back to your lab and measure its linear mass density to be µ = 1.00 g/m. What is your advice concerning the design of the sculpture? What notes will the two strings play in your design? 4
1. The second harmonic sinusoidal standing wave in a pipe with both ends open has a wavelength of 5.00 m. The sound speed is 343 m/s. (a) How many nodes and anti-nodes are there in the displacement wave? How many nodes are there in the pressure wave? Draw the positions of these nodes and anti-nodes. Answer: From the figures to the right, there are 3 nodes and 4 antinodes for the sound wave, while there are 4 nodes and 3 antinodes for the pressure wave. (b) What is the length of the pipe? Answer: λ n = 2L n L = n 2 λ n. Given the harmonic mode 3, and its wavelength as λ 3 = 5.00 m, we get: L = 3 (5.00 m) = 7.50 m. 2 (c) What is the frequency of the third harmonic? Answer: f 3 = v v = λ 3 (2/3) L = 3 ( ) 343 m/s = 68.6 Hz 2 7.50 m (d) What is the frequency of the fundamental mode in this pipe? Answer: f 1 = v = v λ 1 (2) L = 1 ( ) 343 m/s = 22.9 Hz 2 7.50 m sound, mode 3 pressure mode 3 (e) If the same pipe was closed at one end and open at the other (stopped pipe), what would be the frequency of the fundamental (first) and third harmonics? Answer: If the pipe is closed at one end, sound wave will have sound, stopped pipe, 1st mode a node at the closed end. From the figure to the right: f 1 = v λ = v 1 4L = 1 ( ) 343 m/s = 11.4 Hz 4 7.50 m f 3 = v v λ = 3 (4/3) L = 3 ( ) 343 m/s = 34.3 Hz 4 7.50 m sound, stopped pipe, 3rd mode 1
2. As shown in the figure, a weight is attached to the end of a string draped over a pulley; the other end of the string is attached to a mechanical oscillator with a set frequency f = 80.0 Hz. The length of the string is L = 1.00 m, and the linear mass density of the string is µ = 75.0 g/m. For certain values of the weight, the string will resonate. What weights are needed for each of the first three normal modes of the string? Answer: Normal modes of a string with two fixed ends have the following wavelengths λ n = 2L n where n = 1, 2,... also: λ n = v f = 1 f FT µ = 1 f Mg To generate the required frequency, f, the mass M needs to be: µ n = 1 ( ) M n = (λ n f) 2 µ 2Lf 2 g = µ 196 kg = n g n 2 Plugging n = 1, 2 and 3, we get: and M 1 = 196 kg, M 2 = 49.0 kg, M 3 = 21.8 kg. n = 2 L = 1 λ 2 L = 2 λ 2 n = 3 L = 3 λ 2 2
3. A whistle with a fundamental frequency of 300 Hz is aboard a train traveling at 35.0 m/s as it approaches a tunnel in the smooth, vertical face of a rock cliff. The engineer sounds the whistle. A rider in the last car at the rear of the train simultaneously hears the train whistle and also the echo from the wall. (a) How many beats per second does the rider hear? Answer: The passenger hears sound directly from the whistle and its echo reflected by wall. The frequency he hears from the whistle is same as whistle frequency f 0 because he does not move relative to the whistle. The wall is stationary while the train is approaching the wall, so the velocity of the train is negative and, hence, he hears the wave reflected off the wall with the following frequency: f wall = v sound v sound v train f 0 = 343 m/s (300 Hz) = 334 Hz. 343 m/s 35.0 m/s Now the wall is the source to the passenger who is approaching this source with a speed 35.0 m/s f pass = v sound + v train f wall = v sound + v train 343 m/s + 35.0 m/s f 0 = (300 Hz) = 368 Hz v sound v sound v train 343 m/s 35.0 m/s Therefore, the beet frequency is the difference of the frequencies the passenger hears: f pass,beat = f pass f 0 = 368 Hz 300 Hz = 68.2 Hz (b) What would be the beat frequency heard by an observer on the ground near the end of the train? Answer: The train is moving away from the observer, therefore the train s velocity is now positive. The frequency the observer hear directly from the whistle is: f ground = v sound v sound + v train f 0 = 343 m/s = 272 Hz. 343 m/s + 35.0 m/s Now the observer does not move relative to the wall therefore he hears the echo from the wall at the same frequency the wall hears! The beat frequency is then: f ground,beat = f wall f ground = 334 Hz 272 Hz = 61.9 Hz 3
4. The Wind Sculpture. Your friend, an artist, would like to use wires to hang a sculpture as sort of a string instrument. She decides to set it up as shown in the diagram below. Her basic design involves attaching two 75 cm pieces of wire from two eye-hooks on the ceiling that are approximately 50 cm apart and then hanging the 25.0 kg sculpture from a 50 cm horizontal bar with insignificant mass (which also does not vibrate much in the wind) from some point along the bar. The aural effect that she would like to achieve is that when the wind blows across the two strings, they play a perfect fifth, i.e. the ratio of the fundamental frequencies of the two sounds is 3 : 2. Your friend tells you that she has been successful in hanging the sculpture but not in choosing the point along the bar to hang the sculpture giving the desired sound. Desperate for success, she knows that you are taking physics 111 and asks you for help. Before you tackle the analysis, you use your knowledge of waves to gather some more information. You take a sample of the wire back to your lab and measure its linear mass density to be µ = 1.00 g/m. What is your advice concerning the design of the sculpture? What notes will the two strings play in your design? Answer: The two strings have same length, therefore the standing waves of modes of same order will have same wavelength. The system of sculpture-rod is at equilibrium. Since we like the ratio between the strings standing wave frequencies of modes of same order to be 3:2 and wavelength are the same, then then the ratio between the speeds of involved waves are also of ration 3:2. f R f L = 3 2 v R f L = 3 2 But v 2 of waves in a string is proportional to T, the tension in the string. Therefore: T R = 9 4 T L (1) 4
Using Newtons second law for the forces: T L + T R = mg (2) T L + 9 4 T L = mg T L = 4 mg = 75.4 N (3) 13 At this point it is useful to draw a free-body diagram for the rod. See figure to the right. Using Newton s second law for torques about the right end of the rod: T L mgx = T L L x = 4 L = 15.4 cm (4) 13 where m is the mass of the sculpture and x is the distance from the sculpture to the right end. Back substituting for T R : T R x The frequencies are then: T R = 9 mg = 169.6 N (5) 13 f R = 1 2L f R = 1 2L TR µ TL µ = 183 Hz = 275 Hz mg My advice is that she needs to hang the sculpture 15.4 cm from the right end of the bar to achieve the correct tone ratio. In particular she will get the left tone at 183 Hz and the right tone at 275 Hz 5