Pearson Physics Level 0 Unit IV Oscillatory Motion and Mechanical Waves: Unit IV Review Solutions Student Book pages 440 443 Vocabulary. aplitude: axiu displaceent of an oscillation antinodes: points of interaction between waves on a spring or other ediu at which only constructive interference occurs; in a standing wave, antinodes occur at intervals of ½; in an interference pattern, antinodes occur at path difference intervals of whole wavelengths closed-pipe air colun: an air colun found in a pipe closed at one end; the longest wavelength that can resonate in a closed pipe is four ties the length of the pipe constructive interference: overlap of pulses to create a pulse of greater aplitude crest: region where the ediu rises above the equilibriu position destructive interference: overlap of pulses to create a pulse of lesser aplitude diverging: ter applied to a ray that spreads out as it oves away fro the origin Doppler effect: apparent change in frequency and wavelength of a wave that is perceived by an observer oving relative to the source of the wave equilibriu: ter applied to the rest position or the position of a ediu fro which the aplitude of a wave can be easured forced frequency: the frequency at which an external force is applied to an oscillating object frequency: the nuber of oscillations (cycles) per second fundaental frequency: lowest frequency produced by a particular instruent; corresponds to the standing wave having a single antinode, with a node at each end of the string Hooke s law: relationship where the stretch produced by a force applied to a spring is proportional to the agnitude of the force incident wave: wave front oving out fro the point of origin toward a barrier in phase: condition of crests or troughs fro two waves occupying the sae point in a ediu; produces constructive interference interference: effect of two pulses or waves crossing within a ediu; the ediu takes on a shape that is different fro the shape of either pulse alone longitudinal wave: wave with the otion of the ediu being parallel to the otion of the wave axiu: (line of antinodes) line of points linking antinodes that occur as the result of constructive interference between waves echanical resonance: the increase in aplitude of oscillation of a syste as a result of a periodic force whose frequency is equal or very close to the resonant frequency of the syste ediu: aterial, for exaple air or water through which waves travel; the ediu does not travel with the wave iniu: (nodal line) line of points lining nodes that occur as the result of destructive interference between waves Copyright 009 Pearson Education Canada
nodes or nodal points: point on a spring or other ediu at which only destructive interference occurs; in a standing wave, nodes occur at intervals of ½; in an interference pattern, nodes occur at path difference intervals of ½ open-pipe air colun: air colun present in a pipe opened at both ends; the longest wavelength that can resonate in an open pipe is twice the length of the pipe oscillation: a repetitive back-and-forth otion oscillatory otion: otion in which the period of each cycle is constant overtone: any frequency of vibration of a string that ay exist siultaneously with the fundaental frequency period: the tie required for an object to ake one coplete oscillation (cycle) phase shift: result of waves fro one source having to travel farther to reach a particular point in the interference pattern than waves fro another source principle of superposition: displaceent of the cobined pulse at each point of interference is the su of the displaceents of the individual pulses pulse: disturbance of short duration in a ediu; usually seen as the crest or trough of a wave ray: line that indicates only the direction of otion of the wave front at any point where the ray and the wave front intersect reflected wave: wave front oving away fro a barrier resonance: increase in the aplitude of a wave due to a transfer of energy in phase with the natural frequency of the wave resonant frequency: the natural frequency of vibration of an object that will produce a standing wave restoring force: a force acting opposite to the displaceent to ove an object back to its equilibriu position shock wave: strong copression wave produced as an aircraft exceeds the speed of sound siple haronic otion: (SHM) oscillatory otion where the restoring force is proportional to the displaceent of the ass siple haronic oscillator: an object that oves with siple haronic otion sonic boo: the sound like an explosion, experienced by an observer, of the shock wave fored at the front of a plane travelling at or above the speed of sound sound barrier: ter applied to the increase in aerodynaic resistance as an aircraft approaches the speed of sound spring constant: the aount of stiffness of a spring (k) standing wave: condition in a spring or other ediu in which a wave sees to oscillate around stationary points called nodes; wavelength of a standing wave is the distance between alternate nodes or alternate antinodes transverse wave: wave with the otion of the ediu being perpendicular to the otion of the wave trough: region where the ediu is lower than the equilibriu position two-point-source interference pattern: a pattern of axia and inia that results fro the interaction of waves eanating fro two point sources that are in phase; the pattern results fro the overlapping of crests and troughs while the waves ove through each other Copyright 009 Pearson Education Canada
wave: disturbance that oves outward fro its point of origin, transferring energy through a ediu by eans of vibrations wave front: iaginary line that joins all points reached by the wave at the sae instant wave train: series of waves foring a continuous series of crests and troughs wave velocity: the rate of change in position of a wave oving outward fro its point of origin wavelength: distance between two points on a wave that have identical status; usually easured fro crest to crest or fro trough to trough Knowledge Chapter 7. Frequency and period are the inverse of one another. Frequency is the nuber of cycles per second, and period is the nuber of seconds per cycle. 3. Two other accepted units are: rp: revolutions per inute cycles/s: cycles (or oscillations) per second 4. (a) The velocity is zero when the object is at its farthest displaceent fro the equilibriu position. (b) The restoring force is greatest at its farthest displaceent fro the equilibriu position. 5. For aplitudes greater than 5, a pendulu cannot be considered a siple haronic oscillator because the restoring force no longer varies directly with its displaceent. 6. The sign is necessary because the restoring force always acts in the opposite direction to the displaceent. 7. To easure the spring constant of the wood, a ethod siilar to extension question 0 in 7. Check and Reflect on page 365 of the Student Book could be used. A length of this wood could be fastened to a horizontal surface so it extends beyond the edge of the surface. A weight could be attached to this free end, and its displaceent easured. F The equation F kx would be anipulated to solve for the spring constant k x. The spring constant would depend on the type of wood, the grain of the particular piece, and its shape. l 8. The equation for the period of oscillation of a pendulu ist ". According to g this equation, the period varies inversely with the square root of the gravitational field strength. This eans that as gravity decreases, the period gets larger. At a higher altitude where gravity is less, the period will be longer. Therefore, the pendulu will oscillate ore slowly in Jasper (higher altitude) than it would in Calgary. 9. The sound fro a tuning fork is a result of it oscillating at its resonant frequency. If a siilar tuning fork with the sae resonant frequency were held nearby, the sound waves fro the first tuning fork becoe the forced frequency that induces the second tuning fork to resonate. 0. (a) The length of the pendulu ar affects the resonant frequency of the pendulu. 3 Copyright 009 Pearson Education Canada
(b) The acceleration of gravity affects the resonant frequency of a pendulu. Since Earth is not perfectly round, its gravitational field changes with latitude and so does the resonant frequency of a pendulu. (c) The longitude of the pendulu s position will not affect its resonant frequency because the acceleration of gravity is not affected by longitudinal position. (d) The elevation of the pendulu does affect its resonant frequency because the acceleration of gravity changes with elevation. (e) The restoring force is a function of the pendulu bob s position fro its equilibriu position. For sall angles, the aplitude of a pendulu s otion does not affect its resonant frequency, so neither does the restoring force. Chapter 8. In the upper spring, there are ½ wavelengths shown. In the lower spring, there are 3½ wavelengths shown.. Reflection of a sound wave fro a wall: (a) does not affect the speed; (b) does not affect the wavelength; (c) generally produces a reduced aplitude at the reflecting surface, unless the surface is perfectly sooth; (d) directs the reflected wave away fro the wall, such that the angle between the incident wave and the wall is equal to the angle between the reflected wave and the wall. 3. In a longitudinal wave, the copressions and rarefactions will have the greatest potential energy and the points at their equilibriu position will have the greatest kinetic energy. 4. When a circular wave is reflected fro a straight barrier, the shape of the wave is not affected; only its direction of otion changes. 5. The energy stored in a pulse depends on the aplitude of the pulse and on its shape. 6. If waves ove fro a region of a given velocity to one where the velocity is reduced, then (a) the frequency is unchanged, (b) the wavelength is reduced (by the sae factor as the speed) and (c) the direction of the wave changes. 7. In the interference pattern created by two in-phase point sources, the line along which destructive interference occurs is called a iniu or a nodal line. 8. The tension and the ass density of the spring deterine the speed of a wave in a spring. 9. When two wave trains of equal wavelength and aplitude ove through each other in opposite directions, they create a standing wave. Thus, if a wave is generated at one end of a spring and reflected fro the other end, when the reflected and incident waves pass through each other, they will generate a standing wave. The antinodes and nodes of the standing wave are regions where, respectively, constructive and destructive interference occur. These regions occur at fixed points along the spring aking it appear that the spring is oscillating back and forth in a wave pattern but no waves actually ove along the spring. 4 Copyright 009 Pearson Education Canada
0.. When you place your finger at various positions along a violin string, you effectively change the length of the string without changing the tension in the string. Thus the length of the longest standing wave that can be created in the string is also changed. Since the speed of the standing wave is the sae (the tension of the string is not affected) the universal wave equation states that the fundaental frequency for the string is increased as the string is shortened.. If you strike a tuning fork with different forces you will alter the aplitude of the vibrations of its ars. This results in an increase in the volue of the sound produced by the tuning fork. 3. (a) The central axiu of an interference pattern created by two in-phase point sources is located at points where the waves have travelled equal distances fro their sources. In geoetry, the set of points equidistant fro two fixed points is the perpendicular bisector of the line segent joining the fixed points. Since the waves started out in-phase and have travelled the sae distance fro their respective sources, they arrive in-phase, producing a series of points at which only constructive interference occurs. (b) The second order axiu is produced along the set of points (a line) that are wavelengths farther fro one source than the other. Since the waves began their respective journeys in-phase, the wave fro the source that is farther away will arrive in-phase at these points wavelengths after the wave fro the closer source. 4. In an open-pipe, the longest wavelength that can result in resonance is equal to twice the length of the pipe. 5. If you were near the intersection, as the car (oving eastward) approached, you would hear the siren at a pitch that is higher than its true pitch, as well as hear the siren get louder. When the car turned north on the street in front of you, the pitch would suddenly get lower and the sound would start to becoe less loud as the car oved northward away fro you. On the other hand, if you were soe distance south of the intersection that the car was approaching, even though the siren would get louder as it approached the intersection, the pitch of the siren would appear to be the true pitch of the siren. This is because as the car oves eastward, it is oving across your path rather than toward you. When the car turned north, you would hear the sae decrease in pitch and loudness of the sound as described above. 6. According to the universal wave equation, the velocity of a wave is the product of its frequency and its wavelength. 5 Copyright 009 Pearson Education Canada
7. As a sound oves toward you, the wavelengths are copressed by the otion of the source. That is, the distance between crests is reduced by the distance the source oves during one period of the wave's frequency. The universal wave equation states that if the speed of a wave is constant (as is true for this sound) then the frequency varies inversely as the wavelength. Since the wavelength is decreased by the otion of the source, the frequency is increased by the sae factor that the wavelength is decreased. Applications 8. F kx # N $ %.55 &.0 ' ( ) * 3.06 N The force necessary to stretch the spring.0 is 3.06 N. 9. Given f 400.0 Hz 4 k 5.0 + 0 N/ ass of the string () The string behaves as a siple haronic oscillator, so you can deterine its ass by changing its frequency to period, and applying the equation T ". k T f 400.0 Hz 0.00500 s T " k T k 4", 3 # 4 N $ ).500+ 0 s* % 5.0+ 0 & ' ( 4" +, 3 7.9 0 kg 7.9 g The ass of the guitar string is 7.9 g. 30. Given - 90.0. proportion of the force of gravity that is the restoring force 6 Copyright 009 Pearson Education Canada
Deterine the restoring force for the angle given. Then create a siple ratio with the force of gravity. F g/ Fg sin - F sin 90.0. g F g F g )* This equation shows that the restoring force is equal to the force of gravity. The entire force of gravity is the restoring force. 3. Given Displaceent () Force (N) 0. 0.38 0..5 0.3 3.4 0.4 6.08 0.5 9.5 0.6 3.68 spring constant (k) If the elastic obeys Hooke s law, then a force-displaceent graph should be linear, and you can deterine the spring constant. 7 Copyright 009 Pearson Education Canada
Force vs. Displaceent The graph is not linear for the displaceent of the elastic band. The elastic band does not obey Hooke s law as the force-displaceent graph is not linear. You cannot deterine a spring constant. 3. Given F = 40.0 N x = 80.0 c = 0.800 acceleration of the ass when the displaceent is 5.0 c (a) Deterine the spring constant first. Then use Hooke s law to deterine the acceleration at the displaceent of 5.0 c. Reeber to convert all easureents to appropriate SI units. 8 Copyright 009 Pearson Education Canada
33. F kx F k x 40.0 N 0.800 50.0 N/ F, kx a, kx, kx a # N $,% 50.0 &, 0.5 ' ( 0.0 kg 0.5 N/kg ) * 0.5 /s The acceleration of the ass is 0.5 /s. Displaceent (c) Force (N).5 0.0 5.0.0 7.5 3.0 0.0 39.0.5 49.0 spring constant (k) If the spring obeys Hooke s law, then a force-displaceent graph should be linear, and you can deterine the spring constant. Force vs. Displaceent 9 Copyright 009 Pearson Education Canada
The graph is linear, so you can deterine the slope. k slope y x (47.5, 0.0) N (.0, 0.0) c 4.0 N/c The spring constant of the spring is 4.0 N/c. 34. (a) Given = 50.0 g = 0.0500 kg k = 5.0 N/ a = 50.0 /s aplitude of vibration (A) The aplitude is the axiu displaceent. At this position, the ass experiences axiu acceleration. Since you know this value, you can deterine the aplitude fro Hooke s law. F, kx F, kx When the acceleration has a positive value, the displaceent has a negative value. It follows that x, A So, F, k(, A) a ka a A k # $ % 50.0 & ' s ( ) 0.0500kg* N 5.0 0.00 The aplitude refers to the axiu displaceent and is a scalar quantity, so it is 0.00. (b) Given = 50.0 g = 0.0500 kg k = 5.0 N/ a = 50.0 /s axiu velocity of the ass ( v ax ) 0 Copyright 009 Pearson Education Canada
Deterine the axiu velocity using the aplitude you calculated in part (a). k vax A N 5.0 0.00 0.0500kg.4 /s The axiu velocity of the ass is.4 /s. 35. Given, 5 =.0+ 0 kg, 3 T 4.5+ 0 s A.0 c = 0.00 axiu wing speed ( v ax ) Deterine the axiu wing speed by first deterining its spring constant k. Then apply the equation for axiu velocity. Reeber to convert all easureents to appropriate SI units. T " k 4" k T vax A, 5 ) + *, 3 ) 4.5+ 0 s* 4".0 0 kg 9.496 N/ k N 9.496 0.00.0 0, 5 kg + 5 /s The axiu wing speed of the bee is 5 /s. 36. Given 4 = 0.0 t =.00+ 0 kg A =.50 v ax.40 /s period of the daper s oscillations (T) Copyright 009 Pearson Education Canada
To find the period, deterine the spring constant first by using the equation for axiu speed. Then use the spring constant in the period equation. vax A 4.40 (.00 0 kg) # $ % & + s ' ( (.50 ) 87. N/ " k vax k A T " k 4.00 0 kg + N 87. 6.73 s The period of the daper is 6.73 s. 37. Given A 0.80 x 0.60 v ax.5 /s speed of the branch when its displaceent is 0.60 (v) You can deterine the speed of any siple haronic oscillator using the law of conservation of energy. The total energy of a siple haronic oscillator is the su of the kinetic and potential energy of the oscillator in any position. Therefore, the kinetic energy in any position is the total energy inus the potential energy in that position. Furtherore, the total energy is the sae as the axiu potential energy (when the object is at its axiu displaceent). Therefore: E E, E k pax p You can use this equality to write an expression that does not require the ass or spring constant to find the speed of the oscillator. Here is the derivation: Copyright 009 Pearson Education Canada
E E, E k pax p v ka kx, v k( A, x ) k v ( A, x ) k A x ( A, ) A A k x A k (, ) since v ax A x vax (, ) A x v vax, A Now substitute the values into the equation. (0.60 ) v.5, s (0.80 ) 0.99 /s The speed of the branch is 0.99 /s. 38. Given f = 0.5 Hz length of cable in the ass daper to atch the resonance of the building (l) The daper s frequency ust atch the resonant frequency of the building to dapen it. Convert the frequency to period and use that to solve for the length of the cable. 3 Copyright 009 Pearson Education Canada
T f 0.5 Hz 8.00 s T " T g l g l 4" ) 8.00 s * # $ % 9.8 s & ' ( 4" 5.9 The length of the pendulu daper cable ust be 5.9. 39. When a wave slows down, the property of the wave that is not changed is its frequency. The universal wave equation states that when the frequency is constant the wavelength varies directly as the velocity. Thus the decrease in speed results in a proportionate decrease in wavelength. [NOTE: Even though it is not a specified outcoe in the curriculu, you ight ention to the students that the change in speed also results in a partial reflection of the wave. This eans that the aplitude of the wave that continues on at a reduced speed ust be less than that of the incident wave, since soe of its energy has been reflected away. This concept will be dealt with in ore detail in the section on light in Physics 30.] 40. Because of the elastic nature of the ediu, the wave sets up a series of otions that ove, in sequence, through the ediu, with each subsequent segent of the ediu repeating the otion of the previous segent. The energy is carried through the ediu by this sequence of otions. After the energy passes, the segents will be approxiately in their original positions. The energy has oved through the ediu leaving the ediu, for the ost part, unchanged. 4. Given v 3.00+ 0 8 s, 7 4.30+ 0, 7 7.50 + 0 frequency range fro the given range of wavelengths. For each wavelength, the frequency can be calculated using the universal wave equation. 4 Copyright 009 Pearson Education Canada
v f v f 8 3.00+ 0 s, 7 4.30 + 0 4 6.98+ 0 s 4 6.98+ 0 Hz v f 8 3.00+ 0 s, 7 7.50 + 0 4 4.00+ 0 s 4 4.00 + 0 Hz For visible light, the range of frequencies is fro 4.00 0 4 Hz to 6.98 0 4 Hz. 4. Given 8 v 3.00 + 0 s f 6 6 50 khz.50+ 0 Hz.50+ 0 s wavelength for the radio waves Use the universal wave equation. v f v f 8 3.00 + 0 s 6.50+ 0 s 40 Radio waves with a frequency of 50 khz have a wavelength of 40. 43. Given T 0.350 s v 0.840 s wavelength of the sine curve that the pendulu traces on the paper 5 Copyright 009 Pearson Education Canada
Calculate the frequency of the pendulu then use the universal wave equation to calculate the wavelength of the sine curve. f T 0.350 s.857 s.857 Hz v f v f 0.840 s.857 s 0.94 A ore succinct solution would be: v f v f v T vt # $ % 0.840 & ) 0.350 s* ' s ( 0.94 The wavelength of the sine curve traced by the pendulu is 0.94. 44. Given f 545 Hz 545 s.60 ) * 4 d 5.50 k.0 k.0+ 0 6 Copyright 009 Pearson Education Canada
tie required for the sound to travel a distance of 5.50 k and back. First use the universal wave equation to calculate the speed of the sound under water. Then use the relationship for distance, speed, and tie to calculate the tie. v f # $ % 545 &.60 ' s ( 3.47 + 0 s d vt d t v ) * 4.0 + 0 3.47 + 0 s 7.76 s The sound travels at.4 0 3 /s so that it takes 7.76 s for it to travel.0 k. 45. Given l 3.00 f 480 Hz 480 s n 4 speed of the wave in the wire Since each wavelength contains two antinodes, the 4 antinodes ean that there are wavelengths along the wire. Divide the length of the wire by to find the wavelength of the standing wave. Use the wavelength and the frequency to calculate the speed of the wave in the wire. l n 3.00 (4) 0.50 v f # $ % 480 & ) 0.50 * ' s ( 0 s The speed at which a wave oves along the wire is 0 /s. 7 Copyright 009 Pearson Education Canada
46. Given l 5.40 v 3.00 s (a) f.50 Hz.50 s (a) nuber of nodes and antinodes in the standing wave (b) next lower frequency for which a standing wave could exist (a) Calculate the wavelength of the standing wave. Divide the length of the spring by the wavelength to calculate the nuber of wavelengths along the spring. Multiply the nuber of wavelengths by two to give the nuber of antinodes. The nuber of nodes is one greater than the nuber of antinodes. v f v f 3.00 s.50 s.0 Nuber of antinodes: l n # 5.40 $ % & '.0 ( 9 Nuber of nodes: n + = 0 (b) As the frequency increases, the nuber of antinodes in a standing wave also increases. Thus, the standing wave for the next lower frequency has one less node. Reduce the nuber of nodes by one; calculate the new wavelength and the new frequency. The speed of the wave in the spring is the sae. Nuber of nodes at the next lower frequency = 8 n 8 # l $ n % & ' ( l n (5.40 ) 8.35 8 Copyright 009 Pearson Education Canada
v f v f 3.00 s.35. s. Hz Alternative arguent: Since the speed of the wave in the spring is constant, the frequency varies inversely as the wavelength. The wavelength varies inversely as the nuber of nodes. Thus the frequency varies directly as the nuber of nodes. Hence, f n f n f n f n # 8 $.50 Hz% & ' 9 (. Hz (a) The standing wave contains 0 nodes and 9 antinodes. (b) To decrease the nuber of antinodes by one, the frequency ust decrease to. Hz. 47. Given f 93 Hz 93 s l 33.0 c 0.330 l = l 3 (a) speed of the waves in the string (b) frequency when the string is shortened to /3 its original length (a) For the fundaental frequency, the string is ½ wavelength long. Multiply the length of the string by two to give the wavelength. Use the wavelength and the frequency to find the speed using the universal wave equation. l ) * 0.330 0.660 9 Copyright 009 Pearson Education Canada
v f # $ % 93 & ) 0.660 * ' s ( 93.4 s 93 s (b) Calculate the new length of the string, then double that length to give the wavelength. Since the speed of the wave in the spring is unchanged when you press the string at a new position, use the wave speed calculated in (a) to find the new frequency. l l 3 (0.330 ) 3 0.0 l (0.0 ) 0.440 v f f v 93.4 s 0.440 440 s 440 Hz (a) The speed of the waves in the string is 93 /s. (b) When the string is shortened to /3 its original length, the frequency increases fro 93 Hz to 440 Hz. 48. Given f 54 Hz 54 s v 340 s (a) length of the shortest closed-pipe for which resonance would occur (b) length of the next longer closed-pipe for which resonance would occur (a) The shortest closed-pipe that produces resonance for a given frequency is ¼ the wavelength. Calculate the wavelength using the universal wave equation then divide the answer by four. 0 Copyright 009 Pearson Education Canada
v f v f 340 s 54 s.08. l 4 ).08 * 4 0.55 (b) The next longer closed-pipe for which resonance occurs is ¾-wavelength long. Multiply the wavelength fro part (a) by ¾ to find the pipe length. 3 l 4 3 ).08 * 4.66 (a) The shortest closed-pipe for which resonance would occur is 0.55 long. (b) The next longer closed-pipe resonator for this frequency would be.66 long. 49. Given fs 875 Hz fd 870 Hz vw 500 s velocity of the source (subarine) Since your easureent of the frequency for the subarine's sonar is lower than its true frequency, the subarine ust be oving away fro you. Using the Doppler effect equation for an object oving away fro you, isolate the variable for the speed of the source and solve. # vw $ fd % & fs ' vw 0 vs ( # fs, fd $ vs % & v f ' d ( w # 875 Hz, 870 Hz $ % & 500 ' 870 Hz ( s 8.6 s The subarine is oving away fro you at 8.6 /s. Copyright 009 Pearson Education Canada
50. Given k # 5 $# h $ vs 44 %.44+ 0 &% 40.0 3 & h ' h (' 3.6 + 0 s ( s fs 0 Hz 0 s vw 30 s (a) apparent frequency of the siren when the car is oving toward you (b) apparent frequency of the siren when the car is oving away fro you (a) Use the equation for the Doppler effect for an object oving toward you. # vw $ fd % & fs ' vw, vs ( # $ % 30 s & % & 0 30 40.0 s %, & ' s s ( 3.8+ 0 s 3.8+ 0 Hz (b) Use the equation for the Doppler effect for an object oving away fro you. # vw $ fd % & fs ' vw 0 vs ( # $ % 30 s & % & 0 30 40.0 s % 0 & ' s s ( 9.96 + 0 s 9.96 + 0 Hz 3 (a) When the car is approaching you, you hear its siren at a frequency of.8+ 0 Hz rather than 0 Hz. (b) When the car is receding fro you, you hear its siren at a frequency of 9.96 + 0 Hz rather than 0 Hz. Extensions 5. The frequency of vibration of a ass-spring syste is inversely related to the square root of the ass. Copyright 009 Pearson Education Canada
T " T k f " or k f k f or f 3 " If the ass gets larger, the frequency gets saller. Conversely, if the ass gets saller, the frequency gets larger. 5. Given T = 3.7 s l =.0 the planet on which the alien landed find the gravitational field strength of the planet (g) A pendulu s period depends on the gravitational field strength. Deterine the gravitational field strength of the planet the alien landed on fro the period given so you can copare it with the gravitational field strengths of Mercury, Venus, and Earth. You know that Earth s gravitational field strength is 9.8 N/kg, so you need to deterine the gravitational field strength for Mercury and Venus. Mercury s gravitational field strength can be deterined using Newton s equation fro chapter 4: G g r 6 Mercury s radius, r M.44 + 0 3 Mercury s ass, M 3.30 + 0 kg GM gm r #, N " $ 3 % 6.67 + 0 (3.30 0 kg) & + kg ' ( 6 (.44+ 0 ) 3.70 N/kg Venus s radius, Venus s ass, g V V r 6.05+ 0 V 4 V 4.87 + 0 kg G r #, N " $ 4 % 6.67 + 0 (4.87 0 kg) & + kg ' ( 6 (6.05+ 0 ) 8.87 N/kg 6 3 Copyright 009 Pearson Education Canada
A table of the gravitational field strengths of the planets appears below: Planet g (N/kg) Mercury 3.70 Venus 8.87 Earth 9.8 The gravitational field strength deterined by the pendulu is: l T " g 4" l g T " 4 (.0 ) ) 3.7s* 3.7 /s 3.7 N/kg This gravitational field strength atches Mercury s. The gravitational field strength deterined by the pendulu is 3.7 N/kg, which atches Mercury. The alien landed on Mercury. 53. Given Two in-phase point sources separated by 3.5 diagra of the interference pattern 4 Copyright 009 Pearson Education Canada
In the diagra in the text, the sources are 3 apart so that a 3 rd order axiu lies on the line through the sources (S S ) extended. In the diagra above, that location is occupied by a 4 th order iniu. Because of the wider separation of the sources (3 ½ ) the third order axia have oved so that they lie roughly at an angle of 59º to either side of the central axiu. This diagra and the text diagra both have a central axiu. The text diagra has three axia and three inia to the right and to the left of the central axiu. This diagra, because the sources are farther apart, has three axia and four inia to each side of the central axiu. That eans the pattern has behaved uch like a hand fan that has been closed slightly, in that all the axia and inia are a bit closer together. 54. The interference patterns, such as those on pages 45 or 46, are the result of inphase sources. Since the sound fro speakers is not of a single frequency and not in phase, it should not be expected that a stationary interference pattern would be generated. In fact, as the frequencies vary, the interference pattern shifts. Hence, for a set of speakers, the interference pattern is not detectable because it is rapidly shifting ore or less randoly. 55. Given approaching sound source with speed increasing fro zero to Mach graph of ratio of frequencies vs. speed of sound source If the speed of the source is Mach (the speed of sound) the ratio of the frequencies is indeterinate as it requires division by zero. 5 Copyright 009 Pearson Education Canada
Skills Practice 56. Period vs. Frequency The graph of period versus frequency is an inverse relation. 57. Students answers will vary. A suitable procedure ight be:. Hang a spring fro a ceiling.. Measure the height of the free end fro the ceiling. 3. Attach a ass of a predeterined aount to the free end of the spring, and allow it to coe to rest in a stretched position. 4. Measure the height of the end of the spring to the ceiling. 5. Deterine the displaceent of the spring by subtracting the unstretched distance fro the stretched distance. 6. Deterine the applied force on the ass by using F g. F 7. Use Hooke s law ( F kx ) to solve for k : k x. 8. Repeat steps 3 to 7 for different asses and average the values of k found. Students ight include a diagra as follows: 6 Copyright 009 Pearson Education Canada
58. 59. Students answers will vary. A possible procedure is:. Deterine the spring constant of the spring. (Use a force eter to pull the ass F through a easured displaceent; then apply k x ).. Pull the ass attached to the spring through soe displaceent and hold it. 3. Using a stopwatch, release the ass and let it vibrate on the spring. Tie 0 coplete oscillations. 4. Deterine the period of an oscillation (T) by dividing the tie for 0 oscillations by 0. 5. Use the equation T " to solve for. k 60. No, it will not because ass has no relevance to the period of oscillation of a pendulu. The period of a pendulu is only deterined by its length and g. 7 Copyright 009 Pearson Education Canada
6. 6. Huygens pendulu clock was the first clock of this type. It relied on the echanical resonance of a swinging pendulu. If cut to the right length, the pendulu ar would swing at a specific frequency that could be used to keep tie reliably. This is its resonant frequency. To prevent the ar fro succubing to friction, a sall force was iparted on the ar on each swing. This sall push given to the ar was the forced frequency. The clock had liitations, however. It was only accurate for specific geographical regions as the period of the pendulu swing depended on g, which varies across Earth due to variations in the gravitational field strength at different latitudes and elevations. The etal ar would expand or contract in hot or cold weather, affecting its period. The pendulu could not be used on ships because of the rolling otion of the vessel. 63. Students should find that the ter red shift in astronoy refers to the Doppler effect applied to light. If the light source is travelling away fro the observer, then the apparent frequency of the light will be lower than the actual frequency of the light eitted by the source. The effect of decreasing frequency and increasing wavelength is to ake the colour of the visible light received fro the source appear to be shifted toward the red end of the spectru. This is given as evidence that the universe is expanding as all other galaxies see to be oving away fro ours. 64. If two springs with different elastic constants are connected end-to-end, a wave will change speed when it passes fro one spring to the other. In this anner, one can observe the changes in a pulse as it oves fro a ediu of high (low) speed into a ediu of low (high) speed. 8 Copyright 009 Pearson Education Canada
65. The following are several points that the students ay include in their arguents. Objects The object carries the energy fro place to place. Energy is transferred fro one object to another when the objects interact. Masses ay carry energy through space, including a vacuu. Energy oves fro place to place at the speed of the ass, which also is a factor in the quantity of the energy the ass carries. The energy stays concentrated in the object until it interacts with other objects or its surroundings. The interaction of the object with its surroundings ay cause it to lose energy to the. Mechanical Waves The ediu transits the wave which carries the energy. The wave interacts with objects or another ediu to transfer energy. Mechanical waves can only ove through a space in which a ediu exists to enable their otion. The nature of the ediu deterines the properties of the waves including the speed at which they propagate. This deterines the speed at which energy can ove through fro place to place. The wave ay spread out (diverge) in order for its energy to becoe ore and ore diluted along its wave front, so that at any point along the wave front, the energy available for transfer ay be diinished. Self-assessent Students' answers in this section will vary greatly depending on their experiences, depth of understanding, and the depth of the research that they do into each of the topics. 9 Copyright 009 Pearson Education Canada