Pearson Physics Level 0 Unit IV Oscillatory Motion and Mechanical Waves: Chapter 7 Solutions Student Boo page 345 Exaple 7. Practice Probles. 60 s T 5.00 in in 300 s f T 300 s 3 3.33 0 Hz The frequency of the earthquae waves is 3.33 0 3 Hz.. T f 78Hz 0.03s The period of the huingbird s wing is 0.03 s. Student Boo page 347 7. Chec and Reflect Knowledge. For otion to be oscillatory, it ust repeat at regular intervals.. Equivalent units are revolutions/s and Hz. 3. Period: The tie required for one coplete oscillation (cycle) Frequency: The nuber of oscillations (cycles) per second 4. T. They are the inverse of one another. f 5. No, it is not possible. Frequency and period are inversely related to one another, so if the period increases, the frequency ust decrease. 6. Students answers will vary. Soe possible answers could include: cloc, wheel, bird/insect wings, planets, toys. Copyright 007 Pearson Education Canada
Applications 7. rev cycle s s 0.0rev 0.0cycles s s 0.0 Hz The frequency of the toy is 0.0 Hz. 8. The oars on a rowboat ove with oscillatory otion if the rowing otion is at a constant rate. 9. f T 0.004 00 s.50 0 Hz The guitar string has a frequency of.50 0 Hz. 0. T f 38Hz 0.06s The period of the dragonfly s wings is 0.06 s.. 4800 beats in f in 60 s 80.00 Hz T f 80.00 Hz 0.050 s The period of the bird s wings is 0.050 s.. (a) T f.50 Hz 0.400 s The period of the dog s tail is 0.400 s. Copyright 007 Pearson Education Canada
(b) 60 s # wags.50 Hz in.50 0 wags The dog will wag its tail.50 0 ties in in. Extensions 3. Students answers will vary. Possible insects researched could include: Insect Frequency (Hz) Mosquito 550 Housefly 330 Wasp 0 Ladybug (beetle) 46 90 Dragonfly 38 Butterfly 9 4. (a) This otion is periodic because the sater perfors the sae otion at the sae speed. (b) This otion can be considered periodic if the car s speed doesn t change. In practice, the otion is not periodic because the car s speed varies. (c) This otion can be considered periodic when a person is at rest, but not when he or she changes the level of physical exertion. 5. Frequency (Hz) (a) fluorescent light bulbs (b) overhead power lines (c) huan voice range (d) FM radio range (e) lowest note on a bass guitar 0 Hz 60 Hz Approx. 50 Hz 3000 Hz 88 MHz 08 MHz 4. Hz 3 Copyright 007 Pearson Education Canada
Student Boo page 35 Exaple 7. Practice Probles. Plotting the values gives the following graph: Force vs. Displaceent The slope is: slope (50 0.0) N (0.60 0.0).5 0 N/ The spring constant of this spring is.5 0 N/.. Deterine the slope (spring constant) by taing two values fro the line on the graph. Below is an exaple of a calculation. Actual values chosen ay vary. (30.0 0.0) N (.0 0.0) 5N/ The spring constant of this spring is 5 N/. 4 Copyright 007 Pearson Education Canada
Exaple 7.3 Practice Probles. Student Boo page 354 F x N 48.0 0.550 ( ) 6.4 N The restoring force of the spring is 6.4 N.. FA x 00.0 N 0.040 3.5 0 N 3 The spring constant is.5 0 N/. Exaple 7.4 Practice Probles. Given 75.0g x 0.0500 Student Boo page 357 5 Copyright 007 Pearson Education Canada
spring constant () All four springs are copressed a distance of 5.00 c by the force of gravity, so the weight on each spring is one-quarter of the cobined weight. Deterine the spring constant of each spring by dividing the total spring constant by 4. Fg 9.8 (75.0g) s 697.75N F x 697.75N 0.0500 53 955N/ for each spring 4 N 53 955 4 4.35 0 N/ 4 This spring constant of each spring in the car is.35 0 N/.. Given F g 40.0 N [down] A 5 N/ B 60.0 N/ displaceent (x) Both springs are affected by the weight of the ass attached to the end. Each spring will defor as if it has a 40.0 N ass attached to it. Assue that the springs are 6 Copyright 007 Pearson Education Canada
assless. Deterine the displaceent for each, and then add the displaceents together. F x Fg xa g A A A 40.0 N N 5.6 Fg BxB Fg xb B 40.0 N N 60.0 0.667 xa + xb (.6 ) + ( 0.667 ).3 The ass has a displaceent of.3 [down]. Concept Chec Student Boo page 359. The force-displaceent graph ust always intercept the origin because there will be no displaceent if no force is applied.. The line will becoe non-linear if the elastic liit of the spring is exceeded, and the spring peranently defors or breas. 3. The ass on the end of the spring will oentarily coe to a stop at its axiu displaceents the top or botto of its oveent. The student should tae a picture when the ass is in one of these positions. 7 Copyright 007 Pearson Education Canada
4. 5. The negative sign indicates that the restoring force is always opposite to the spring s displaceent. Exaple 7.5 Practice Probles. Student Boo page 36 F F sin θ g g (0.3000g) 9.8 (sin.0 ) s 0.6 N The restoring force on the pendulu is 0.6 N [right].. F F sinθ g θ g F g sin F g 4.00 N sin (0.500g) 9.8 s 54.6 An angle of 54.6 results in a restoring force of 4.00 N [left]. 8 Copyright 007 Pearson Education Canada
Student Boo page 365 7. Chec and Reflect Knowledge. Two factors affecting the restoring force of a pendulu are: (i) the angle through which it is displaced (ii) the ass of the bob. Massspring Displaceent Acceleration Velocity Restoring Force Syste ax x ax 0 ax ax a ax 0 ax ax v 0 0 0 in F 0 0 ax Pendulu Syste Displaceent Acceleration Velocity Restoring Force ax x ax 0 ax ax a ax 0 ax ax v 0 0 0 in F 0 0 ax 3. A pendulu is not a true siple haronic oscillator because the restoring force does not vary linearly with displaceent. Applications 4. Given.0 g 40.0 N/ x 0.5 restoring force ( F R ) Use Hooe s law to deterine the restoring force. Mass is not required in the solution. 9 Copyright 007 Pearson Education Canada
F R x N 40.0 0.5 ( ) 6.0 N The restoring force is 6.0 N in the opposite direction of the displaceent. 5. Given 4.0 g 5.0 N/ displaceent ( x ) Use the equation F x to deterine the displaceent. You do not need to use a negative sign ( F x ) because the force of gravity pulling the ass down is the applied force, not the restoring force. Fg ( 4.0g) 9.8 s 39.4 N Fg x Fg x 39.4 N N 5.0.6 The displaceent of the ass is.6 [down]. 6. Given F 5.0 N x 0.0 force required to pull the ass to a displaceent of +0.5 ( F ) 0 Copyright 007 Pearson Education Canada
Use the initial force and displaceent values to deterine the spring constant. Then use the spring constant to deterine the force necessary to displace the ass +0.5. Use Hooe s law without the negative sign because the force is applied. F x F x 5.0 N 0.0 5 N/ F x N 5 0.5 ( ).9 0 N The force required to ove the ass through a displaceent of 0.5 is 9 N. 7. Given Displaceent () Force (N) 0.00 0.00 0.0 0.5 0.0 0.33 0.30 0.4 0.40 0.60 spring constant () Using the data provided, plot a graph and find the slope. Force vs. Displaceent Copyright 007 Pearson Education Canada
You can choose any values fro the line of best fit. For exaple, the following calculation uses these values: (x, y ) (0.0, 0.0) and (x, y ) (0.34, 0.5) slope (0.5 0.0) N (0.34 0.0).5N/ The spring constant is.5 N/. 8. Fg Fgsin θ (0.400 g) 9.8 sin 5.00 s 0.34 N The restoring force on the pendulu is 0.34 N. 9. Given x 0.0 c [left] 0.000 g a 0.55 /s [right] spring constant () Use Newton s second law ( F a ) to deterine the restoring force. Deterine the spring constant fro the restoring force by using Hooe s law. F x a x a x ( 0.000g) 0.55 s 0.0 0.08 N/ The spring constant of the car s spring is 0.08 N/. Extension 0. Given Saple data: Ruler Deflection x (c) Mass (g) (N) Plastic 4.6 00.0.96 Wood. 500.0 4.9 Metal 3.0 40.0 0.39 F g Copyright 007 Pearson Education Canada
spring constant of each ruler () The force of gravity is different for each ruler, so you ust deterine it for each one. The values are recorded in the table. Use the equation F x to find for each ruler. Fg plasticx Fg plastic x.96 N 0.046 43N/ Fg woodx Fg wood x 4.9N 0.0. 0 N/ Fg etalx Fg etal x 0.39 N 0.030 3N/ The ran of the rulers fro highest to lowest spring constants is: wood. 0 N/ plastic 43N/ 3N/ etal Concept Chec Student Boo page 37. According to Hooe s Law, the restoring force always acts to return the object to its equilibriu position. To do this, it ust act opposite to the displaceent. Since the restoring force is siply the product of ass and acceleration, the acceleration ust also be opposite of the displaceent.. The object is accelerating, and the acceleration is not unifor. This results in a curved velocity-tie graph. Fro chapter, we now that the slope of the velocity-tie graph represents acceleration. The acceleration of the object is greatest at the object s axiu displaceent, which is why the velocity-tie graph is steep at the extrees of the object s otion. When the object is in the equilibriu position, it does not 3 Copyright 007 Pearson Education Canada
experience any acceleration, so the slope of the velocity-tie graph in this position is zero. 3. For the other half of the oscillator s journey, the acceleration-displaceent graph doesn t change. The object siply oves bacward fro (c) to (b) to (a). The velocity-displaceent graph is different. For the second half of the object s otion, the velocity is in the opposite direction, so it is negative and follows the path fro (c) to (b) to (a) as well. 4. The equation for axiu velocity is: vax A If the aplitude (A) is doubled, the axiu velocity will also be doubled as shown below: vax A or vax A The agnitude of the velocity would be twice as large. Student Boo page 37 Exaple 7.6 Practice Probles. Given 0.74 g 8. N/ a 4. /s [left] displaceent ( x ) Deterine the displaceent fro the acceleration equation. Tae right to be positive and left to be negative. 4 Copyright 007 Pearson Education Canada
F x a x a x (0.74g) 4. s N 8. + 0.36 The displaceent of the ass when it experiences an acceleration of 4. /s [left] is 0.36 [right].. vax A N 4.00. 0.0500 g 0.0 /s The axiu speed of the ass is 0.0 /s. 3. Given 6.05 N/ A 8.7 c v ax.05 /s ass of the oscillator () Find the ass of the oscillator by anipulating the axiu velocity equation. 5 Copyright 007 Pearson Education Canada
vax A vax A A v ax N (0.87) 6.05.05 s 0.96 g The ass of the oscillator is 0.96 g. Student Boo page 376 Exaple 7.7 Practice Probles. (a) Given.50 g 40.0 N/ A 0.800 (a) acceleration at a displaceent of 0.300 to the right ( a ) (b) axiu speed ( v ax ) (c) period (T) Tae left to be negative and right to be positive. 6 Copyright 007 Pearson Education Canada
a) b) F x a x x a N 40.0 0.300.50 g v ax ( ) 4.80/s A c) T T N 40.0 0.800.50 g 3.0 /s π.50g π N 40.0.57 s (a) The acceleration of the ass when it is 0.300 to the right is 4.80 /s. (b) The axiu speed of the ass is 3.0 /s. (c) The period of the ass is.57 s.. Given.60 g a 0.0 /s x 0.700 spring constant () 7 Copyright 007 Pearson Education Canada
Deterine the spring constant fro the acceleration equation. No directions are given, so tae the quantities of acceleration and displaceent as scalars. No negative sign is required in the equation for acceleration. F x a x x a a x ( 0.0060 g) 0.0 s 0.700 0.0743 N/ The spring constant of the spring is 0.0743 N/. 3. Given T.0 s 0.0 N/ ass () Manipulate the equation for the period of a siple haronic oscillator to solve for the ass. T π T 4π T 4π N (.0 s) 0.0 4π.0 g The ass of the oscillator is.0 g. 4. Given.07 N/ A 7.3 c v ax 7.0 /s period (T) 8 Copyright 007 Pearson Education Canada
You cannot deterine the period of the oscillator directly. Deterine the ass first; then use the ass to solve for the period. All values are scalars so you do not need to specify directions. Convert the aplitude to appropriate SI units. Solve for ass first: vax A vax A A v ax N (0.73 ).07 7.0 s 0.45 g Use the ass to solve for period: T π 0.45 g π N.07 0.638 s The period of the oscillator is 0.638 s. Student Boo page 377 Concept Chec. Doubling the displaceent has no effect on the period of oscillation of a siple haronic oscillator.. The two conditions that ust be satisfied are: (i) The period of the oscillator and the circular otion ust be the sae. (ii) The aplitude of the oscillator ust atch the radius of the circular otion. 3. The period of a ass-spring oscillator is given by the equation T π. If and are both doubled then we could write the equation as T π. Since both are increased by a factor of two, there is no change. This can be shown atheatically by 9 Copyright 007 Pearson Education Canada
cancelling the in the nuerator and the in the denoinator in the fraction: T π. This leaves the equation that we started with. 4. The equation for period is: T π Solving for gives: 4π T If T is doubled, then the equation can be written: 4π 4π or ( T) 4T The difference between the last equation and the initial equation for is a factor of one-quarter. The spring constant of the oscillator with twice the period is one-quarter of that of the other spring. Student Boo page 379 Exaple 7.8 Practice Probles. The gravitational field strength is always directed toward the centre of the planet. Solve for the agnitude of the field strength only. 4π l g T 4π ( 0.500 ) (.30 s) 3.73 /s The agnitude of the gravitational field strength on Mercury is 3.73 /s.. Only the scalar quantity of pendulu length is required. gt l 4π.63 5.00 s s ( ) 4π.03 The length of the pendulu is.03. 3. Convert the pendulu length to SI units. 0 Copyright 007 Pearson Education Canada
l T π g 0.300 π 3.7 s.79 s The period of the pendulu on Mars is.79 s. Concept Chec. The velocity of the swinging apple will be the slowest near the extrees of its otion. It will coe to an instantaneous stop at its axiu displaceents. It would be easiest to hit in these positions.. The factors that affect the accuracy of pendulu clocs are: teperature: A change in teperature will cause the etal ar to expand or contract, changing its length and changing its period. altitude and latitude: A pendulu cloc s period depends on the acceleration of gravity, which changes with these factors. Student Boo page 380 7.3 Chec and Reflect Knowledge. (a) Changing the aplitude has no effect. (b) Changing the spring constant has an inverse effect on the period as shown by the equation: T π If the spring constant gets larger, the period decreases. If the spring constant becoes saller, the period increases. (c) The period is directly related to the square root of the ass. If the ass increases, so does the period, as shown by this equation: T π. (a) The effect of changing the aplitude of a pendulu is the sae as the effect of changing it in a ass-spring syste; it has no effect. This is true for any siple haronic oscillator. (b) The gravitational field strength is inversely related to the period. Therefore, they will have the opposite effect on one another, just as the spring constant does for the ass-spring syste. (c) Mass has no effect on the period of a pendulu. 3. (a) At axiu acceleration, both systes are at their axiu displaceent. (b) The velocity of both systes is a axiu when the oscillator is in the equilibriu position. Copyright 007 Pearson Education Canada
(c) The restoring force varies as the acceleration does. Therefore, its axiu will be at the axiu displaceents. 4. The acceleration of a siple haronic oscillator is not unifor because the force varies with displaceent according to Hooe s law. The closer the ass is to the equilibriu position, the less force acts on it. Therefore, it experiences less acceleration than if it were farther away. The acceleration varies linearly with displaceent just as the force does. 5. Restoring force and acceleration act in the sae direction, so the acceleration is also positive. Applications 6. The bungee juper will experience the fastest speed when he is passing through the equilibriu position. Deterine his fastest speed using the following equation: vax A N 5.0 30.0 s 60.0 g 9.4 /s Copyright 007 Pearson Education Canada
The fastest the bungee juper will ove is 9.4 /s. 7. Given T 4.0 s g Mars 3.7/s length of the pendulu (l) The period of a pendulu varies with gravitational field strength. Modify the pendulu period equation to solve for length. T g l 4π ( 4.0 s ) 3.7 s 4π.5 The length of the pendulu is.5 8. Given 3.08 g T 0.33 s x.85 [right] acceleration ( a ) The oscillator s acceleration changes every half cycle, so we need to pay close attention to its vector nature. Acceleration to the right is positive, and to the left is negative. You cannot solve for the acceleration in one step. Deterine the spring constant first; then use the spring constant to find the acceleration. 3 Copyright 007 Pearson Education Canada
Solve for the spring constant: T π T 4π 4π T 3.08g 4π (0.33s) 65.48N/ Now solve for acceleration: F x a x x a N 65.48 ( +.85) 3.08g 3.08 0 /s 3 The acceleration of the ass is.08 0 /s [left]. 9. F x a x a x ( 50.0g).0 s 0.080 + 3.3 0 N/ The pogo stic has a spring constant of.3 0 3 N/. 4 Copyright 007 Pearson Education Canada
0. Fg Fgsinθ Fg sinθ Fg Fg θ sin g 0.468N sin (0.500g) 9.8 s.0 The pendulu bob is displaced through an angle of.0.. l T π g 0.500 π.64 s 3.47s The period of the pendulu is 3.47 s. Extensions. (a) Given Horizontal ass-spring syste: A.50 0.00 N/ Mass oving in circular path: r.50 v 5.00 /s 5 Copyright 007 Pearson Education Canada
period of the ass-spring syste (T) Both the ass oving in the circular path, and the ass-spring syste are synchronized, so they ust have the sae period. Deterine the period of the ass oving in the circular path, which gives the period of the ass-spring syste. π r v T π r T v π (.50 ) 5.00 s.88s The period of the ass-spring syste is the sae as the circular syste, so it is also.88 s. (b) Given Horizontal ass-spring syste: A.50 0.00 N/ T.88 s Mass oving in circular path: r.50 v 5.00 /s T.88 s ass of the ass-spring syste () The axiu speed of the ass-spring syste atches the speed of the ass oving in the circular path. Now that you now the period, deterine the ass using the equation for speed. vax A A v ax N (.50) 0.00 5.00 s 0.900g The ass of the ass-spring syste is 0.900 g. 6 Copyright 007 Pearson Education Canada
(c) Given Mass-spring syste: A.50 0.00 N/ T.88 s 0.900 g axiu acceleration of the ass-spring syste ( a ) The axiu acceleration for any siple haronic oscillator occurs at the axiu displaceent. You are solving only for the agnitude of the acceleration, as it could be in either direction. Therefore, oit the negative sign in the acceleration equation. x a N 0.00 (.50 ) 0.900g 6.7/s The agnitude of the acceleration is 6.7 /s. 3. Given f 0.0 Hz A 0.0500 axiu speed of quartz oscillator (v ax ) The quartz crystal is a siple haronic oscillator. Convert the frequency of its oscillations to period to solve for its ass. Once you now the ass, deterine the axiu speed. The speed is a scalar quantity, so the axiu displaceent (aplitude) does not require a direction. T f 4.00 0 Hz 4.00 0 s Solve for the spring constant: T π 7 Copyright 007 Pearson Education Canada
4π T 4 4 π (.00 0 g) 4 (.00 0 s) 5 7.896 0 N/ Find the axiu speed of the crystal: vax A 5 N 7.896 0 5 (5.00 0 ) 4.00 0 g 3.4 /s The quartz crystal oscillates with a axiu speed of 3.4 /s. 4. Given 0.00 g A 0.0 v ax 0.803/s [west] displaceent ( x ) You can deterine the displaceent for a specific acceleration only if you now the spring constant. Deterine the spring constant by using the equation for axiu speed, and solving for. Then use to deterine the displaceent. Tae east to be positive, and west to be negative. v ax A ax v A 0.803 (0.00g) s (0.0) 8.956 N/ 8 Copyright 007 Pearson Education Canada
Now deterine the displaceent using the spring constant: F x a x x a a x (0.00g) 3.58 s N 8.956 + 7.99c The displaceent of the ass when it experiences an acceleration of 3.58 /s [west] is 7.99 c [east]. 5. Given h.3 l 0.0 c period of the pendulu (T) The pendulu s period depends on the gravitational field strength at the height of.3. Deterine the gravitational field strength first, reebering to add Earth s radius to the altitude. Then deterine the period of the pendulu. GEarth g r N i 4 6.67 0 ( 5.98 0 g) g 6 4 6.37 0 +.3 0 ( ) 9.79 /s Now find the period of the pendulu: l T π g 0.00 π 9.79 s 0.898s The period of the pendulu aboard the plane is 0.898 s. 9 Copyright 007 Pearson Education Canada
Student Boo page 388 7.4 Chec and Reflect Knowledge. The wind causes a phenoenon called vortex shedding. It can create a forced vibration that atches the natural frequency of a building or bridge.. Forced frequency is the frequency at which an external force is applied to a resonating object. 3. Engineers use tuned ass dapers in buildings to iniize the aplitude of resonant vibration caused by the wind. The tuned ass dapers are coputer controlled so that a large ass (either a ass-spring syste or a pendulu) is set oscillating with the sae frequency as the building but in the opposite direction. 4. A force applied with the sae frequency as the resonant frequency of the object has the effect of increasing the aplitude of its oscillations. 5. The liitations were: (i) It only wored at specific elevations or altitudes due to the variations in gravity caused by Earth s shape. (ii) The etal ar of the pendulu expanded or contracted with changes in l teperature, changing its period. (Recall T π ). g (iii) It could not be used on a oving vessel (boat). 6. No, the radius of Earth is longer at the equator than at the poles. The teperature is very different as well. The pendulu cloc s period of oscillation would change significantly. A pendulu cloc built to operate at a specific geographic location ust reain there to be accurate. 7. Daping is the process where the aplitude of oscillations of an object is reduced. This can be accoplished by increasing or decreasing the tension on the object, or by providing a periodic force out of phase with the oscillations. An exaple could include the daping of a bridge by using stiffer aterials that require a large forced frequency to oscillate and reduce wind resistance. Applications 8. The waler would change the frequency of his or her waling step across the bridge. 9. The usical note is a specific frequency that atches the resonant frequency of the chapagne glass, which increases the aplitude of its vibration. If the energy (aplitude) of the forced vibration is large enough, the glass will brea. 0. The tuning fors vibration at a specific frequency creates a vibration in the piano strings. The piano tuner listens to see if the proper piano string begins to vibrate at the sae resonance as the for, and will adjust the tension on the string accordingly until it vibrates when the tuning for is held near.. a) A forced frequency that is different fro the resonant frequency will not change the resonant frequency but will liely dapen the vibrations. b) The friction created by the water changes the acceleration of the pendulu as it swings bac and forth and affects the period of the oscillations so its resonant frequency will change. The restoring force is the net force and will be the su of the perpendicular coponent of gravity and the force of friction. 30 Copyright 007 Pearson Education Canada
FR Fnet F F + F net g finetic Friction will also dapen the oscillations. c) Increasing the ass of the pendulu bob will not change the resonant frequency or dapen the oscillations. d) Moving the pendulu to a higher altitude will change the acceleration of gravity (it will be less). This will change the resonant frequency of the pendulu because the period will be greater. It will not dapen the oscillations. The size and shape of the crystal affect its resonant frequency. 3. a) Student answers will vary. Soe valid answers are: The quartz cloc is uch ore accurate than the pendulu cloc. The quartz cloc is not affected by teperature. The quartz cloc is not affected by latitude or elevation. The quartz cloc is not affected by otion so it can be used aboard oving vehicles. The quartz cloc is subject to uch less echanical friction. The quartz cloc does not require gravity to function. b) Student answers will vary. Soe valid answers are: The pendulu cloc does not require a battery. The pendulu cloc is ore pleasing to loo at. Extensions 4. Given g 9.8 /s f.00 Hz length of the pendulu (l) The period of a pendulu is deterined by the length of the ar and the acceleration of gravity. You can deterine the length of a pendulu with equation 9. To use this equation, you ust convert the frequency to period. l T π g l T 4π g T g l 4π (.00 s) 9.8 s 4π 0.48 A pendulu ar 0.48 long has a frequency of.00 Hz in Alberta. This pendulu would be ost accurate if the teperature reained constant, the pendulu was not oved, and friction was ept to a iniu. 3 Copyright 007 Pearson Education Canada
5. Autoobile anufacturers will do the following things: Place daping aterial in the doors. Use epoxies/glues to glue coponents together so they won t rattle (electronic circuitry in ignition systes is covered with an epoxy resin so the electrical coponents won t rattle and brea over tie). Use self-locing washers/nuts that aren t as susceptible to loosening fro vibration. 6. They use light, stiff aterials that will not vibrate easily. This is a for of daping. 7. They are placed in large ships to reduce the aount of rolling caused by the waves because the lateral bac-and-forth otion can cause seasicness. The tuned ass dapers perfor the sae role on a ship as they do in a building. 8. Orbital resonance in Saturn s rings occurs when the period of soe of the orbital debris (sall rocs) in a ring has the sae orbital period (or a ultiple of the period) as one of Saturn s oons. For exaple a oon ay go around once, and in that tie soe of the debris in the ring goes around exactly two ties. Over tie the gravitational pull of the oon wors to increase the aplitude of the debris orbital radius every tie the oon and the debris are aligned, creating a gap in the rings. Student Boo pages 390 39 Chapter 7 Review Knowledge. Oscillatory otion is unifor bac-and-forth otion.. A ball will bounce with oscillatory otion if it has a unifor period. This eans that it ust bounce to the sae height in each oscillation. 3. An elastic aterial will defor due to a stress. Once the stress is reoved, the aterial will return to its original shape. 4. The restoring force is the only force that acts on an isolated, frictionless, siple haronic oscillator. 5. The direction of the restoring force is always opposite to the displaceent. 6. Slope is the spring constant. 7. It is not in its equilibriu position. x 8. Acceleration depends on displaceent, as given by a. Therefore, it is not unifor. 9. The restoring force varies with the sine of the angle of displaceent. For sall angles, this relationship is alost linear. As the displaceent angle increases, the relationship is no longer linear. This difference begins to show at approxiately 5. 0. The forced frequency will increase the aplitude of the resonant frequency. 3 Copyright 007 Pearson Education Canada
Applications. (a) Note that the angles in the diagra have been exaggerated for illustration purposes. Fg Fgsinθ Fg 9.8 (.0g) sin5 s.5n The restoring force on the pendulu is.5 N [forward]. (b) Fg Fgsinθ Fg 9.8 (.0g) sin5 s 0.9 N The restoring force on the pendulu is 0.9 N.. f T 0.00400s 50Hz The frequency of the guitar string is 50 Hz. 3. T f 0.67 Hz.5 s The period is.5 s. 33 Copyright 007 Pearson Education Canada
4. f T 0.00 s 0.0 Hz The frequency is 0.0 Hz. 5. F F sinθ g g (.0g) 9.8 ( sin5 ) s 5.N The restoring force acting on the pendulu is 5. N [forward]. 6. Any two points fro the line of best fit can be used to deterine the slope. The following points were used for the solution below: (x, y ) (0.00, 0.00) and (x, y ) (0.75, 50) slope (50 0.00) N (0.75 0.0).0 0 N/ The spring constant of the spring is.0 0 N/. 7. Given height of spring fro floor.80 height of spring fro floor with 00.0-g ass attached.30 34 Copyright 007 Pearson Education Canada
(a) spring constant () (b) force needed to pull the 00.0-g ass through a displaceent of 0.0 c ( F ) (c) displaceent of the 300.0-g ass ( x ) Draw diagra of the situation described in each part the proble. You need to now the spring constant of the spring to deterine the force and displaceents in parts (b) and (c). To deterine the spring constant, use the force of gravity for the 00.0-g ass. The displaceent of the ass-spring is 50.0 c down. (a) x 0.500 [down] Fg x Fg x ( 0.00g) 9.8 s 0.500.00 N/ (b) Tae the displaceent of the 00.0-g ass (.30 ) fro the floor as the new equilibriu position, and deterine the force required to pull it down 0.0 c. x 0.00 [down] F x N (.00 )( 0.00 ) 0.400 N 35 Copyright 007 Pearson Education Canada
(c) Deterine the displaceent for the 300.0-g ass fro the spring s original equilibriu position (.80 ) fro the floor. Use the spring constant found in part (a) to deterine this displaceent. Fg x (0.300g) 9.8 s N.00.47 The height fro the floor is:.80.47 0.33 (a) The spring constant of the spring is.00 N/. (b) A person ust apply a force of 0.400 N to pull the hanging 00.0-g ass down through a displaceent of 0.0 c. (c) If the 00.0-g ass is reoved and a 300.0-g ass is attached, it will hang 0.33 fro the floor. 8. Given A 00.0 N/ B 50.0 N/ F 5.0 N total displaceent of both springs (x T ) Each spring will be displaced differently by the force. To deterine the total displaceent, calculate the displaceent of each spring; then add the two displaceents together. 36 Copyright 007 Pearson Education Canada
Spring A: A 00.0 N/ FA xa FA xa A 5.0 N N 00.0 0.50 Spring B: B 50.0 N/ FB xb FB xb B 5.0 N N 50.0 0.500 xt xa + xb 0.50 + 0.500 0.750 The cobined displaceent of both springs when a force of 5.0 N is applied is 0.750. 9. Force vs. Displaceent The graph shows that the elastic band does not obey Hooe s law because the graph is not linear. 37 Copyright 007 Pearson Education Canada
0. l T π g T g l 4π (.00 s ) 9.8 s 4π 0.48 or 4.8c The length of the pendulu ust be 4.8 c to swing with a period of s under the influence of a gravitational field of 9.8 /s.. T π 0.0 g π N 44.0 3.00s The period of the ass is 3.00 s.. vax A N 00.0 0.0 000 g 0.0379 /s The axiu speed of the crate is 0.0379 /s. 3. Given A 0.040 v ax 0.00 /s 0.480 g x 0.000 [upward] acceleration at displaceent of 0.000 [up] ( a ) To deterine the acceleration of the spider, find the spring constant of its thread. Do this first, and then use the spring constant in the acceleration equation. Mae sure to use the appropriate SI units for the spider s ass. Tae up as positive, and down as negative. 38 Copyright 007 Pearson Education Canada
vax A vax A vax A 4 0.00 (4.80 0 g) s (0.040 ) 3 3.0 0 N/ Use this spring constant to deterine the spider s acceleration at a displaceent of 0.000 [up]. F x a x x a 3 N 3.0 0 (0.000 ) 4 4.80 0 g 0.3 /s The spider experiences an acceleration of 0.3 /s [down], when it is at a displaceent of 0.000 [up]. 39 Copyright 007 Pearson Education Canada
4. l T π g 0.585 π 9.8 s.0 s The period of a pendulu with a length of 5.85 c is.0 s. 5. T f 0.8 Hz 5.49 s 4π l g T 4π 0.50 ( 5.49 s) 0.65 /s Pluto s gravitational field strength is 0.65 /s. 6. (a) X represents l g. (b) l X g l T g 4π (.79 s) 4π 0.086 s l 0.086 s 9.8 s 0.796 The length of the pendulu is 0.796. 40 Copyright 007 Pearson Education Canada
Extensions 7. (a) vax A N 0.0 0.00 0.50 g.6 /s The axiu speed of the ass is.6 /s. (b) T π 0.50 g π N 0.0 0.993 s The ass will have a period of vibration of 0.993 s. The period is independent of the aplitude. 8. (a) Given f 0.800 Hz frequency of the oscillator when the ass is doubled (f) To find the frequency when the ass is doubled, odify the equation for the period to solve for frequency. Substitute in place of in the equation to deterine the factor by which the frequency increases or decreases. Then deterine the actual change to the frequency by ultiplying this factor by the original frequency. 4 Copyright 007 Pearson Education Canada
T π f T π π Change to. f.44 π π (.44).44 π (0.707) π The frequency has changed by a factor of 0.707. Now deterine the new frequency. f 0.707 0.800 Hz 0.566 Hz The frequency of the oscillator will change fro 0.800 Hz to 0.566 Hz. (b) For any siple haronic oscillator, the aplitude of vibration does not affect the period, and therefore does not affect the frequency. The frequency will stay at 0.800 Hz. 9. (a) A bouncing ball is not a siple haronic oscillator because the force does not vary with displaceent. The force of gravity is constant regardless of the position of the ball. (b) The oveent of a puc on the ice cannot be considered SHM. The force applied by the stic is constant throughout its displaceent. (c) A pluced guitar string is an exaple of SHM. The guitar string s restoring force varies directly with its displaceent. 4 Copyright 007 Pearson Education Canada