Discussion Examples Chapter 13: Oscillations About Equilibrium

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1 Discussion Exaples Chapter 13: Oscillations About Equilibriu 17. he position of a ass on a spring is given by x 6.5 c cos t 0.88 s. (a) What is the period,, of this otion? (b) Where is the ass at t 0.5 s? (c) Show that the ass is at the sae location at 0.5 s seconds as it is at 0.5 s. Picture the Proble: A ass is attached to a spring. he ass is displaced fro equilibriu and released fro rest. he spring force causes the ass to oscillate about the equilibriu position in haronic otion. Strategy: he period can be obtained directly fro the arguent of the cosine function. Substituting the specific tie into the equation will yield the location at that tie. Substituting in the new tie will show that the ass is at the sae location one period later. Solution: 1. (a) Identify the period () fro the cosine equation: x Acos t, so here = 0.88 s.. (b) Substitute t = 0.5 s into the equation and evaluate x: x 6.5 ccos 0.5 s 1.4 c 0.88 s 3. (c) Substitute t = (0.5 s + ) into x Acos the equation and factor: 0.5 s Acos 0.5 s 4. Drop the phase shift, because cos x cos x : x Acos 0.5 s 5. Insert the nueric values: x 6.5 c cos 0.5 s 1.4 c sae location 0.88 s Insight: Increasing the tie by any ultiple of the period increases the arguent of the cosine function by the sae ultiple of, which has no effect upon the value of the cosine function. 44. IP he springs of a 511-g otorcycle have an effective force constant of 9130 N/. (a) If a person sits on the otorcycle, does its period of oscillation increase, decrease, or stay the sae? (b) By what percent and in what direction does the period of oscillation change when a 11-g person rides the otorcycle? Picture the Proble: If the otorcycle is pushed down slightly on its springs it will oscillate up and down in haronic otion. A rider sitting on the otorcycle effectively increases the ass of the otorcycle and oscillates also. Strategy: We can use the equation for the period of a ass on a spring. Writing this equation for the otorcycle without rider and again for the otorcycle with rider we can calculate the percent difference in the periods. Solution: 1. (a) he period increases, because the person s ass is added to the syste and.. (b) Write the equation for the period of the otorcycle without the rider: 3. Write the equation for the period M of the otorcycle with the rider: 4. Calculate the percent difference between the two periods: M Copyright 010 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist. No 14 1

2 5. Siplify by factoring out fro the nuerator and denoinator: M g % 511 g Insight: he percent change in the period does not depend on the spring force constant. It only depends on the fractional increase in ass. 50. IP A 0.40-g ass is attached to a spring with a force constant of 6 N/ and released fro rest a distance of 3. c fro the equilibriu position of the spring. (a) Give a strategy that allows you to find the speed of the ass when it is halfway to the equilibriu position. (b) Use your strategy to find this speed. Picture the Proble: A ass attached to a spring is stretched fro equilibriu position. Strategy: he wor done in stretching the spring is stored as potential energy in the spring until the ass is released. After the ass is released, the ass will accelerate, converting the potential energy into inetic energy. he energy will then transfer bac and forth between potential and inetic energies as the ass oscillates about the equilibriu position. Solve the conservation of echanical energy equation, E= K + U, for the inetic energy. hen use the equation for 1 inetic energy, K v, to solve for the velocity. Solution: 1. Set K E U and substitute expressions for each ter:. Solve for the speed and siplify: 3. Insert the nueric values: K E U v A A A A v g A N/ 0.03 v 0. /s Insight: When the displaceent is half the axiu displaceent, the speed is not half the axiu speed. 3 In fact, the speed is v ax, which is greater than half the speed. 61. United Nations Pendulu A large pendulu with a 00-lb gold-plated bob 1 inches in diaeter is on display in the lobby of the United Nations building. he pendulu has a length of 75 ft. It is used to show the rotation of the Earth for this reason it is referred to as a Foucault pendulu. What is the least aount of tie it taes for the bob to swing fro a position of axiu displaceent to the equilibriu position of the pendulu? (Assue that the acceleration due to gravity is g 9.81 /s at the UN building.) Picture the Proble: he pendulu ass is displaced slightly fro equilibriu and oscillates bac and forth through the vertical. Strategy: he tie the pendulu taes to ove fro axiu displaceent to equilibriu position is one-quarter of a period. Use equation 13-0 to deterine the tie. Solution: Insert the nueric values into L 75.0 ft s equation 13-0 and convert feet to eters: 4 g 9.81 /s ft Insight: he full period of this pendulu is 4(.4 s) = 9.6 seconds. A pendulu with only half this length would have a period of 6.8 s. Copyright 010 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist. No 14

3 Discussion Exaples Chapter 14: Waves and Sound 14. A brother and sister try to counicate with a string tied between two tin cans (Figure 14 33xx34). If the string is 9.5 long, has a ass of 3 g, and is pulled taut with a tension of 8.6 N, how uch tie does it tae for a wave to travel fro one end of the string to the other? Picture the Proble: he iage shows two people taling on a tin can telephone. he cans are connected by a 9.5-eter-long string weighing 3 gras. Strategy: Set the tie equal to the distance divided by the velocity, where the velocity is given by equation 14-. he linear ass density is the total ass divided by the length. Solution: 1. Set the tie equal to the distance divided by velocity:. Substitute d and insert nuerical values: d t d v F 0.03 g9.5 / d d t d F F 8.6 N Insight: he essage travels the sae distance in the air in 0.08 seconds, about 7 ties faster s 38. In a pig-calling contest, a caller produces a sound with an intensity level of 110 db. How any such callers would be required to reach the pain level of 10 db? Picture the Proble: We are given the sound intensity of one pig caller and are ased to calculate how any pig callers are needed to increase the intensity level by 10 db. Strategy: Multiply the intensity in equation 14-8 by N callers, setting the intensity level to 10 db and solve for N. NI I 10log 10log 10log I I Solution: 1. Write the intensity level for N callers: N. Insert the intensity levels and solve for N: N 10 db 10log N db 10log 110 db N 10/ callers Insight: Increasing the intensity level by 10 db increases the intensity by a factor of 10. herefore 10 callers, each with intensity level 110 db, would produce a net intensity level of 10 db. 100 callers (10 10 callers) would be needed to produce an intensity level of 130 db (10 db + 10 db). Copyright 010 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist. No 14 3

4 45. A train oving with a speed of 31.8 /s sounds a 136-Hz horn. What frequency is heard by an observer standing near the tracs as the train approaches? Picture the Proble: he train, a oving source, sounds its horn. We wish to calculate the frequency heard by a person standing near the tracs. Strategy: Solve equation for the observed frequency, using the negative sign because the train is oving toward the observer. Solution: Insert the given data into equation 14-10: 1 1 f f 1 uv /s 343 /s Hz 136 Hz Insight: If the train were oving away fro the observer, he would hear a frequency of 14 Hz. 6. IP wo violinists, one directly behind the other, play for a listener directly in front of the. Both violinists sound concert A (440 Hz). (a) What is the sallest separation between the violinists that will produce destructive interference for the listener? (b) Does this sallest separation increase or decrease if the violinists produce a note with a higher frequency? (c) Repeat part (a) for violinists who produce sounds of 540 Hz. Picture the Proble: wo violinists separated by a distance d, as shown in the figure, play a 440-Hz note. Strategy: We want to calculate the sallest distance d, for which the listener will hear destructive interference. Assue that the violins are in phase with each other. he sallest separation that will produce destructive interference occurs when the separation is equal to one-half of a wavelength. Set the distance to half a wavelength and use equation 14-1 to write the wavelength in ters of the frequency and speed of sound. Solution: 1. (a) Set the distance equal to half a wavelength: v 343 /s d f 440 Hz (b) he frequency is inversely proportional to the separation distance. herefore, higher frequency eans shorter iniu separation. 3. (c) Solve for the distance at 540 Hz: 343 /s d Hz Insight: In order for the destructive interference to occur, the violins notes ust be coherent and in phase with each other. his typically does not occur during a concert. Copyright 010 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist. No 14 4

5 71. IP BIO Standing Waves in the Huan Ear he huan ear canal is uch lie an organ pipe that is closed at one end (at the typanic ebrane or eardru) and open at the other (see the figure below). A typical ear canal has a length of about.4 c. (a) What is the fundaental frequency and wavelength of the ear canal? (b) Find the frequency and wavelength of the ear canal s third haronic. (Recall that the third haronic in this case is the standing wave with the second-lowest frequency.) (c) Suppose a person has an ear canal that is shorter than.4 c. Is the fundaental frequency of that person s ear canal greater than, less than, or the sae as the value found in part (a)? Explain. (Note that the frequencies found in parts (a) and (b) correspond closely to the frequencies of enhanced sensitivity in Figure 14 8.) Picture the Proble: he iage shows an ear canal of length.4 c. Strategy: reating the ear canal as a pipe closed at one end, we wish to calculate the fundaental and third haronic frequencies and wavelengths. Use equation to calculate the frequencies and wavelengths. For the fundaental use n = 1, and for the third haronic use n = 3. Solution: 1. (a) Calculate the fundaental frequency and wavelength using equation with n = 1:. (b) Calculate the third haronic frequency and wavelength using equation with n = 3: 1 nv /s f1 4L Hz 4Ln4.4 c c 3 nv /s f3 11 Hz 4L Ln4.4 c 3 3. c 3. (c) he fundaental frequency is inversely proportional to the length of the ear canal. herefore, if an ear canal is shorter than.4 c, the fundaental frequency of that person s ear canal is greater than the value found in part (a). Insight: For an ear canal of length. c the fundaental frequency will be 3.9 Hz. 73. IP A 1.5-g clothesline is stretched with a tension of.1 N between two poles 7.66 apart. What is the frequency of (a) the fundaental and (b) the second haronic? (c) If the tension in the clothesline is increased, do the frequencies in parts (a) and (b) increase, decrease, or stay the sae? Explain. Picture the Proble: he iage shows two clotheslines that are 7.66 long. One line is oscillating at the fundaental frequency and the other at the second haronic. Strategy: First use the tension and ass to calculate the speed of the waves, using equation 14-. hen use equation to calculate the frequencies. Solution: 1. (a) Solve equation 14- for the wave speed:. Set n = 1 in equation to calculate the fundaental frequency: 3. (b) Set n = in equation to calculate the second haronic frequency: F.1 N v /s g 7.66 nv /s f Hz L 7.66 f nv /s 15. Hz L (c) he wave speed is proportional to the square root of the tension, and the frequency is proportional to the wave speed. We conclude that if the tension in the clothesline is increased, the frequencies in parts (a) and (b) will increase. Insight: If the tension were doubled to 44. N, the frequencies would increase by a factor of to f Hz and f 1.5 Hz. We bent the rules for significant figures a little in step 1 in order to avoid rounding error. Copyright 010 Pearson Education, Inc. All rights reserved. his aterial is protected under all copyright laws as they currently exist. No 14 5