Chapter 4.1 Q1 n oscillation is any otion in which the displaceent of a particle fro a fixed point keeps changing direction and there is a periodicity in the otion i.e. the otion repeats in soe way. In siple haronic otion, the displaceent fro an equilibriu position and the acceleration are proportional and opposite each other. Q In a SHM oscillation the period is independent of the aplitude and the acceleration is opposite and proportional to the displaceent fro equilibriu. Q It is an oscillation since we ay define the displaceent of the particle fro the iddle point and in that case the displaceent changes direction and the otion repeats. The otion is not siple haronic however since there is no acceleration that is proportional (and opposite) to the displaceent. Q4 It is an oscillation since the otion repeats. The otion is not siple haronic however since the acceleration is constant and is not proportional (and opposite) to the displaceent. Q5 We know that a ass at the end of a spring when displaced will perfor siple haronic oscillations with a period given by T =!. Hence easuring the spring k constant and the period of oscillations allows deterination of the ass. Q6 They are not siple haronic because as shown in the textbook the restoring force whereas opposite to, is not proportional to the displaceent away fro the equilibriu position. If however the aplitude of oscillations is sall the force does becoe approxiately proportional to the displaceent and the oscillations are then approxiately siple haronic. Q7 t t = the equation says that x = cos!. The next tie x assues this value is at a tie given by cos(! T + ") = cos". Thus we ust solve the equation cos(! T + ") = cos". This eans that the angles! T + " and! differ by! and so! solutions are " T + # = # +! $ T =. " Q8 (a) t t = we have y = 5. cos() = 5.. (b) t t = 1. s we use the calculator (in radian ode) to find y = 5.cos(! 1.) = ".7. (c)!. = 5. cos( t) " t = cos (! ) = 1.98 " t =.99 s. (d) Use v = ±! " x. We 5! 1 know that! =. s. Therefore, 6. = ±. 5! x " 5! x = 9. " x = ± 4.. Q9 (a) The equation is siply y = 8.cos(! 4 t) = 8.cos(8! t). (b) The velocity is therefore v = " 8.# 8! sin(8! t) and the acceleration is a = " 8. #(8! ) cos(8! t). t
t =.5 s we evaluate (in radian ode) y = 8.cos(8! ".5) = # 4.7 c, v!! = " 8.# 8 sin(8 #.5) = " 5.7 s and a!! " = " 8. #(8 ) cos(8 #.5) =.6# 1 s. Q1 The angular frequency is! = " f = " # 46 = 9". The axiu velocity is! " # # 1 = 9 $ 5.$ 1 = 14 s and the axiu acceleration is # 4 #! = (9" ) $ 5.$ 1 = 4.$ 1 s. Q11 (a) The equation of the string ay be rewritten as ( ) y = 6.sin(! x) cos(! " 5 t) fro which we deduce that the frequency of all points is 5 Hz and that the phase of all points is zero. (b) Fro (a) the aplitude is = 6.sin(! x) and so is different for different points on the string. (c) The axiu aplitude is obtained when sin(! x) = 1, i.e. the axiu aplitude is 6.. (d) The displaceent is always zero at the ends of the string, in particular at the right end where x = L, the length of the string. The displaceent is zero all the tie when 6.sin(! x) = i.e. when! x =! " x = 1.. (e) L When x = =.75 the aplitude is 6.sin(! x) = 6.sin(.75! ) = 4.. 4 Q1 cceleration: t the acceleration is axiu in agnitude and is directed to the right. t B it is directed to the right and has saller agnitude than at. t C we have the equilibriu position and so the acceleration is zero. t D the acceleration is to the left and has the least agnitude copared to and B. Net force: since F = a whatever si true for acceleration is also true for net force. Q1 (a) t axiu displaceent we will have the axiu acceleration i.e. 45 # 1 aax =!. The aplitude is = 4.5 c and! = " f = " s. Therefore, 6 # 45 $ " 4 " a ax = '! ( % 4.5% 1 = 999 & 1.% 1 s. (b) s the piston oves past its ) 6 * equilibriu position it has the axiu speed and so 45 " v ax =! # 4.5# 1 = 1 s. (c) Fax = aax =.5! 999 ".5! 1 N. 6 Q14 (a) The area is approxiately.5 c (the exact value is.51 c) (ideally the graph should have been given on graph paper and we would then count squares). (b) This is the displaceent fro when the velocity is zero to when it is zero again i.e. fro one extree position until the other i.e. twice the aplitude. (c) The period is.4 s and so the equation for displaceent is! t x = ".5sin( ) = ".5sin( 5! t).. 4
!! # 1 Q15 We need to graph the equation a = "! x where " = = = 1.57 s. The slope T.5 " would be! = 158 s or just 1.58 since we are plotting c on the horizontal axis. This gives the graph in the answers in the textbook. Q16 (a) The defining relation for SHM is that a = "! x which iplies that a graph of acceleration versus displaceent is a straight line through the origin with negative slope just as the given graph. (b) The slope of the graph gives "!. The easured slope is 1.5!! =! 15 s and so! = 15 =.87 s. Thus the period is T = = 1.6 s. (c) The.1.87 axiu velocity is! =.87#.1 =.9 s. (d) The axiu net force is a 1 1.15.87.1 ET =! = " " =.1 J. =! =.15".87 ". 1 =. N. (e) The total energy is Q17 (a) The forces on the ass when the plate is at the top are shown below: N g The net force is g! N = a. Since we have siple haronic otion a =! x = 4" f x in agnitude, and the largest acceleration is obtained when x =, the aplitude of the oscillation. The frequency is 5. Hz. The critical point is when N =. g 9.8 I.e. g = 4! f and so =.99 4! f = 4! " 5 =. The aplitude ust not exceed this value. (b) t the lowest point: N " g = a = 4! f # = + N g 4! f N = $ + $ $ $ $.1 9.8.1 4! 5.99 N =.5 N Q18 (a) We have siple haronic oscillations about a ean value of scale reading of 7 kg which is the ass of the passenger. The period is 1 s. (b) t the top point of the oscillations the reading is 5 kg and at the botto it is 9 kg. t the lowest point (see Q17) we have that
N = g + 4! f 4! = g + T 7 4 9" 9.8 = 7" 9.8 + 1 = 7.1 " "! " (This passenger has probably regretted buying a ticket on this particular cruise.) Q19 (a) The volue within the sphere of radius x is 4! x and that of the entire sphere Mx 4! R x GMx is therefore the ass enclosed is the fraction M. (b) F = G R =. R x R GMx GM (c) The acceleration of the ass is given by a =! a x R " =! R which is the GM! R condition for SHM with! =. (d) T = =!. (e) R " GM 6 (6.4# 1 ) T =! = 585 s = 85 in. (f) Fro gravitation we know that 1 4 6.67# 1 # 6.# 1 "! # = $ v = = $ T =! v GM GM % R & R R R R ' T ( GM as in (d). Q (a) The tension in each string is T. Taking coponents of the two tension forces we see that the horizontal coponents T cos! cancel out leaving a downward net fro the vertical coponents of T sin!. (b) Fro Fnet = a we get a = " T sin! " g and " T sin! hence a = " g. Since the tension is very large and the ass is sall the first ter is bound to doinate and so we ay neglect the acceleration of gravity in the T sin! x x expression above. Hence, approxiately, a = ". Now sin! " tan! = = L L 4T and finally a =! x. This eans we will have siple haronic oscillations with L 4T! =. (c) The period is given by! L! L " = 4T. Q1 The period of oscillation does not depend on the acceleration of gravity. ccelerating ay be thought of as changing the acceleration of gravity (increasing it for upwards accelerations and decreasing for downward accelerations). Hence the period will be unaffected.
Q The displaceent is in general y = cos(! t + ") and so the velocity is v = #! sin(! t + "). So the initial displaceent is x = cos! and the initial velocity is v = #! sin". In other words, x cos! = v sin! = # " But cos! sin! 1 + =, i.e. + = 1$ = x + " x # " v # % & % & v ' ( '! (!. Q (a) In this case the putty will stick to the ass because of frictional forces between the and so soe energy will be lost reducing the aplitude of oscillation. The ass will increase, increasing the period of oscillations. (b) When the block is oentarily at rest the putty will rest on it without any frictional forces acting and so the energy of the spring ass syste will be unaffected. Hence the aplitude of oscillation will also be unaffected. Since the ass has increased however the period will increase as well. Q4 (a) The aplitude is.6. The frequency is! f = 6.8 " f = 1.8 # 1.8 Hz 1 1 and the period is T =.94 s f = 1.8 =. (b) The total energy is 1 ET =!. Now # 1 1! = " f = 6.8 s and so E T =! 1.8! (6.8)! (.6) = 5.4 J. (c) The elastic 1 1 potential energy is EP =! x =.8 " (6.8) " (.15) =.65 J. The kinetic energy is therefore E = 5.4!.65 = 4. 75 J. K Q5 When extended by an aount x the force pulling back on the body is kx and so k k # 1 a =! kx " a =! x and so! = = = 1.95 s giving a period of.!! T = = =.57 s. (b) With the springs connected this way, and the ass pulled to.95 the side by sall aount one spring will be copressed and the other extended. Hence the net force on the ass will still be kx so the period will not change. Q6 (a) t the top the woan s total energy is gravitational potential energy equal to gh where h is the height easured fro the lowest position that we seek. t the lowest position all the gravitational potential energy has been converted into elastic
1 energy kx and so 1 gh = kx. Since h = 15 + x we have that ust now solve for the height h : 1 6! 1! h =!!( h 5) 6h = 11( h " h + 5) 11h " 9h + 475 = 11h " 9h + 475 = 1 ( 15) gh = k h!. We The physically eaningful solution is h = 7.. (b) The forces on the woan at the position in (a) are her weight vertically downwards and the tension in the spring upwards. Hence the net force is F = T! g = kx! g = "(7.! 15)! 6 = 84 N hence net F net 84! = = = 4.7 s (c) Let x be the extension of the spring at soe arbitrary a 6 position of the woan. Then the net force on her is Fnet = T! g = kx! g directed upwards i.e. opposite to the direction of x. So a =!( kx! g). The acceleration is not proportional to the displaceent so it looks we do not have SHM. But we ust easure displaceent fro an equilibriu position. This is when the extension of the spring is x and kx = g. In other words call the displaceent to be y = x! x. Then a k y x g ky kx g ky =!( ( + )! ) =!! + =! since kx = g. Hence we do have the k k condition for SHM. nd so a =! y so that! = #! = 1.91s and finally! T = =.8 s. (d) She will coe to rest when the tension in the spring equals her " g 6! 1 weight i.e. when kx = g " x = = =.7. Hence the distance fro the top k is 15 +.7 = 17.7. (e) Since she will perfor any oscillations before coing to rest the oscillations are underdaped. (f) It has been converted to other fors of energy ainly theral energy in the air and at the point of support of the spring. Q7 SHM oscillation is characterized by the equation a = "! x. With daping there will be, in addition, a force opposing the otion, which is usually proportional to the speed of the particle. Depending on the agnitude of this force we can have three distinct cases: (i) very sall opposing force in which case the oscillations die out slowly. (ii) a specific larger force which akes the particle return to its equilibriu position as fast as possible but without oscillations. (iii) Finally, any force larger than the force in (ii) in which case the particle again returns to its equilibriu position without oscillations but in a tie longer than that in (ii). Q8 Desirable: (i) shock absorbers in cars. It is desirable because without daping very tie the car hits a bup the car (and its passengers) would suffer uncofortable oscillations up and down. (ii) Scales that are used in weighing things. If the oscillations
of the pointer were not daped, the pointer would oscillate for a long tie before settling down to the true reading of the ass of the object being weighed. Undesirable: (i) the oscillations in your grandfather s clock hanging on the wall. (ii) the oscillations of a child in a swing when the father is too lazy to push! n 1 Q9 (a) the ratio of successive aplitudes does appear to be constant at about +!.6. n! n+ 1 n + 1 (b) The fraction of energy lost is!.6 =. 64 ". 6. (c) The ratio of n n successive aplitudes would now be saller than.6 and the period would get very slightly longer. Q See Figures 1.17, 1.18.and 1.19 in the textbook on pages 7 and 8. Q1 They should ideally be critically daped so that the passengers do not go on oscillating for a long tie after the car hits a bup in the road as they would in the case of underdaped oscillations. They should not be overdaped since in that case the car would return to its noral (and safe) condition for driving in a longer tie. Q s hinted at in Q1, overdaped shock absorbers ight ake the passenger feel the bup less but it would take a longer tie to bring the car into its noral driving condition. This is not safe as the brakes will take a longer distance to bring the car to a stop in this case. Q See Figures 1.17, 1.18.and 1.19 in the textbook on pages 7 and 8. k Q4 (a) The angular frequency is! = and the frequency is found fro! = " f. (b) The displaceent is in general y = cos(! t + ") and so the velocity is v = #! sin(! t + "). Hence x cos(! t + ") = v sin(! t + ") = #! " x # " v # But cos (! t + ") + sin ((! t + ") = 1, i.e. $ % + $ % = 1. (c) The graph is an ellipse & ' &! ' centered at the origin. (d) The area of the ellipse in (c) is! ". (e) With daping the energy of the syste will be reduced and the aplitude will get saller. Hence the area of the ellipse will get saller. This will happen as the graph in (c) becoes an elliptical spiral that closes on to the origin. n
Q5 Free oscillations are oscillations in which no other force, other than the restoring force that is opposite and proportional to displaceent, act on the syste. In the case of forced oscillations there are other additional forces acting on the syste. Q6 ny structure has its own natural frequency of oscillation. When the structure is subjected to an external periodic force the aplitude of the resulting oscillations will be the largest when the natural frequency of the syste and that frequency of the external force are the sae. This is called a state of resonance. n exaple of resonance is the icrowave oven in which icrowaves of frequency equal to that of the vibrating water olecules are directed at food. The water olecules oscillate with the greatest aplitude in this case, which eans that the teperature of the water olecules in the food goes up and the food is wared. Q7 In case the periodic arching gets the bridge in resonance with dangerous large aplitude oscillations that ight break the bridge. 1 k Q8 (a) f =!. (b) The net force on the ass is the tension in the spring, kx and so v kx L + x =! R where L is the natural length of the spring. Now, v = =! Rf and so T (! Rf ) L + x = kx. We ust solve this equation to find the extension x : 4! ( L + x) f = kx 4! L f = x( k " 4! f ) 4! L f x = k " 4! f 1 k (c) The denoinator in (b) is k " 4! f = k " 4! = which is ipossible. (d) 4! We have a case of resonance where a displaceent becoes uncontrollably large. This is an idealized situation though where daping has been copletely neglected. In a real situation with friction and daping this would not happen.