Chapter 5. (a) During siple haronic otion, the speed is (oentarily) zero when the object is at a turning point (that is, when x = +x or x = x ). Consider that it starts at x = +x and we are told that t = 0.5 second elapses until the object reaches x = x. To execute a full cycle of the otion (which taes a period T to coplete), the object which started at x = +x, ust return to x = +x (which, by syetry, will occur 0.5 second after it was at x = x ). Thus, T = t = 0.50 s. (b) Frequency is siply the reciprocal of the period: f = /T =.0 Hz. (c) The 36 c distance between x = +x and x = x is x. Thus, x = 36/ = 8 c.. (a) The acceleration aplitude is related to the axiu force by Newton s second law: F ax = a. The textboo notes (in the discussion iediately after Eq. 5-7) that the acceleration aplitude is a = x, where is the angular frequency ( = f since there are radians in one cycle). The frequency is the reciprocal of the period: f = /T = /0.0 = 5.0 Hz, so the angular frequency is = 0 (understood to be valid to two significant figures). Therefore, (b) Using Eq. 5-, we obtain F ax = x = 0. g 0 rad / s 0.085 =0 N. b gb g b g 0.g 0 rad/s. 0 N/. 3. The textboo notes (in the discussion iediately after Eq. 5-7) that the acceleration aplitude is a = x, where is the angular frequency ( = f since there are radians in one cycle). Therefore, in this circustance, we obtain a x ( f ) x 6.60 Hz 0.00 37.8 /s. 4. (a) Since the proble gives the frequency f = 3.00 Hz, we have = f = 6 rad/s (understood to be valid to three significant figures). Each spring is considered to support one fourth of the ass car so that Eq. 5- leads to 450g 6 rad/s.9 0 5 N/. / 4 4 car 700
70 (b) If the new ass being supported by the four springs is total = [450 + 5(73)] g = 85 g, then Eq. 5- leads to new fnew / 4 (85/ 4) g total 5.90 N/.68Hz. 5. THINK The blade of the shaver undergoes siple haronic otion. We want to find its aplitude, axiu speed and axiu acceleration. EXPRESS The aplitude x is half the range of the displaceent D. Once the aplitude is nown, the axiu speed v is related to the aplitude by v = x, where is the angular frequency. Siilarly, the axiu acceleration is a x. ANALYZE (a) The aplitude is x = D/ = (.0 )/ =.0. (b) The axiu speed v is related to the aplitude x by v = x, where is the angular frequency. Since = f, where f is the frequency, (c) The axiu acceleration is v 3 = fx = 0 Hz.0 0 = 0.75 /s. a x f x 3 = = = 0 Hz.00 = 5.7 0 /s. LEARN In SHM, acceleration is proportional to the displaceent x. 6. (a) The angular frequency is given by = f = /T, where f is the frequency and T is the period. The relationship f = /T was used to obtain the last for. Thus = /(.00 0 5 s) = 6.8 0 5 rad/s. (b) The axiu speed v and axiu displaceent x are related by v = x, so x v.00 0 = = 5 6.8 0 3 / s rad / s 3 =.59 0. 7. THINK This proble copares the agnitude of the acceleration of an oscillating diaphrag in a loudspeaer to gravitational acceleration g. EXPRESS The agnitude of the axiu acceleration is given by a = x, where is the angular frequency and x is the aplitude.
70 CHAPTER 5 ANALYZE (a) The angular frequency for which the axiu acceleration has a agnitude g is given by g / x, so the corresponding frequency is f g 9.8 /s 498 Hz. 6 x.00 (b) For frequencies greater than 498 Hz, the acceleration exceeds g for soe part of the otion. LEARN The acceleration a of the diaphrag in a loudspeaer increases with, or equivalently, with f. 8. We note (fro the graph in the text) that x = 6.00 c. Also the value at t = 0 is x o =.00 c. Then Eq. 5-3 leads to = cos (.00/6.00) = +.9 rad or 4.37 rad. The other root (+4.37 rad) can be rejected on the grounds that it would lead to a positive slope at t = 0. 9. (a) Maing sure our calculator is in radians ode, we find F HG b g I K J x = 6.0cos 3.0 + = 3.0. 3 (b) Differentiating with respect to tie and evaluating at t =.0 s, we find (c) Differentiating again, we obtain b g b g dx F v = = 3 6.0 3.0 + = 49. dt H G I sin 3 K J / s b g b g b g dv F a = = 3 6.0 3.0 + =.7 0. dt H G I cos 3 K J / s (d) In the second paragraph after Eq. 5-3, the textboo defines the phase of the otion. In this case (with t =.0 s) the phase is 3(.0) + /3 0 rad. (e) Coparing with Eq. 5-3, we see that = 3 rad/s. Therefore, f = / =.5 Hz. (f) The period is the reciprocal of the frequency: T = /f 0.67 s.
703 0. (a) The proble describes the tie taen to execute one cycle of the otion. The period is T = 0.75 s. (b) Frequency is siply the reciprocal of the period: f = /T.3 Hz, where the SI unit abbreviation Hz stands for Hertz, which eans a cycle-per-second. (c) Since radians are equivalent to a cycle, the angular frequency (in radians-persecond) is related to frequency f by = f so that 8.4 rad/s.. When displaced fro equilibriu, the net force exerted by the springs is x acting in a direction so as to return the bloc to its equilibriu position (x = 0). Since the acceleration a d x / dt, Newton s second law yields d x = x. dt Substituting x = x cos(t + ) and siplifying, we find /, where is in radians per unit tie. Since there are radians in a cycle, and frequency f easures cycles per second, we obtain (7580 N/) f = = 39.6 Hz. 0.45 g. We note (fro the graph) that v = x = 5.00 c/s. Also the value at t = 0 is v o = 4.00 c/s. Then Eq. 5-6 leads to = sin (4.00/5.00) = 0.97 rad or +5.36 rad. The other root (+4.07 rad) can be rejected on the grounds that it would lead to a positive slope at t = 0. 3. THINK The ass-spring syste undergoes siple haronic otion. Given the aplitude and the period, we can deterine the corresponding frequency, angular frequency, spring constant, axiu speed and axiu force. EXPRESS The angular frequency is given by = f = /T, where f is the frequency and T is the period, with f = /T. The angular frequency is related to the spring constant and the ass by. The axiu speed v is related to the aplitude x by v = x. ANALYZE (a) The otion repeats every 0.500 s so the period ust be T = 0.500 s. (b) The frequency is the reciprocal of the period: f = /T = /(0.500 s) =.00 Hz. (c) The angular frequency is = f = (.00 Hz) =.6 rad/s.
704 CHAPTER 5 (d) We solve for the spring constant and obtain = = (0.500 g)(.6 rad/s) = 79.0 N/. (e) The aplitude is x =35.0 c = 0.350, so the axiu speed is v = x = (.6 rad/s)(0.350 ) = 4.40 /s. (f) The axiu force is exerted when the displaceent is a axiu. Thus, we have F = x = (79.0 N/)(0.350 ) = 7.6 N. LEARN With the axiu acceleration given by a x, we see that the agnitude of the axiu force can also be written as F x x a. Maxiu acceleration occurs at the endpoints of the path of the bloc. 4. Equation 5- gives the angular velocity: 00 N/ 7.07rad/s..00 g Energy ethods (discussed in Section 5-4) provide one ethod of solution. Here, we use trigonoetric techniques based on Eq. 5-3 and Eq. 5-6. (a) Dividing Eq. 5-6 by Eq. 5-3, we obtain v = tanbt + x so that the phase (t + ) is found fro v 3.45 /s t tan tan. x 7.07 rad/s0.9 With the calculator in radians ode, this gives the phase equal to.3 rad. Plugging this bac into Eq. 5-3 leads to 0.9 x cos(.3) x 0.500. (b) Since t + =.3 rad at t =.00 s, we can use the above value of to solve for the phase constant. We obtain = 8.38 rad (though this, as well as the previous result, can have or 4 (and so on) added to it without changing the physics of the situation). With this value of, we find x o = x cos = 0.5. (c) And we obtain v o = x sin = 3.06 /s. g
705 5. THINK Our syste consists of two particles undergoing SHM along a coon straight-line segent. Their oscillations are out of phase. EXPRESS Let A t x cos T be the coordinate as a function of tie for particle and x A t cos T 6 be the coordinate as a function of tie for particle. Here T is the period. Note that since the range of the otion is A, the aplitudes are both A/. The arguents of the cosine functions are in radians. Particle is at one end of its path (x = A/) when t = 0. Particle is at A/ when t/t + /6 = 0 or t = T/. That is, particle lags particle by onetwelfth a period. ANALYZE (a) The coordinates of the particles 0.50 s later (that is, at t = 0.50 s) are and A 0.50 s x cos 0.5A.5 s A 0.50 s x cos 0.43 A..5 s 6 Their separation at that tie is x = x x = 0.5A + 0.43A = 0.8A. (b) The velocities of the particles are given by and v v dx A t dt T T sin dx A t sin. dt T T 6 We evaluate these expressions for t = 0.50 s and find they are both negative-valued, indicating that the particles are oving in the sae direction. LEARN The plots of x and v as a function of tie for particle (soid) and particle (dashed line) are given below.
706 CHAPTER 5 6. They pass each other at tie t, at x x x where x x cos( t ) and x x cos( t ). Fro this, we conclude that cos( t ) cos( t ), and therefore that the phases (the arguents of the cosines) are either both equal to /3 or one is /3 while the other is /3. Also at this instant, we have v = v 0 where v x sin( t ) and v x sin( t ). This leads to sin(t + ) = sin(t + ). This leads us to conclude that the phases have opposite sign. Thus, one phase is /3 and the other phase is /3; the wt ter cancels if we tae the phase difference, which is seen to be /3 ( /3) = /3. 7. (a) Equation 5-8 leads to Therefore, f = / = 5.58 Hz. a 3 /s a x 35.07 rad/s. x 0.00 (b) Equation 5- provides a relation between (found in the previous part) and the ass: 400 N/ = 0.35g. (35.07 rad/s) (c) By energy conservation, x (the energy of the syste at a turning point) is equal to the su of inetic and potential energies at the tie t described in the proble. x = + v x x = +. v x Consequently, x (0.35 g / 400 N/)(3.6 /s) (0.00 ) 0.400.
707 8. Fro highest level to lowest level is twice the aplitude x of the otion. The period is related to the angular frequency by Eq. 5-5. Thus, x d and = 0.503 rad/h. The phase constant in Eq. 5-3 is zero since we start our cloc when x o = x (at the highest point). We solve for t when x is one-fourth of the total distance fro highest to lowest level, or (which is the sae) half the distance fro highest level to iddle level (where we locate the origin of coordinates). Thus, we see t when the ocean surface is at x x 4 d. With x x cos( t ), we obtain d d cos 0.503t 0 cos(0.503 t) 4 which has t =.08 h as the sallest positive root. The calculator is in radians ode during this calculation. 9. Both parts of this proble deal with the critical case when the axiu acceleration becoes equal to that of free fall. The textboo notes (in the discussion iediately after Eq. 5-7) that the acceleration aplitude is a = x, where is the angular frequency; this is the expression we set equal to g = 9.8 /s. (a) Using Eq. 5-5 and T =.0 s, we have F I HG T K J gt x = g x = = 0.5. 4 (b) Since = f, and x = 0.050 is given, we find = g =. Hz. x f x g f 0. We note that the ratio of Eq. 5-6 and Eq. 5-3 is v/x = tan(t + ) where =.0 rad/s in this proble. Evaluating this at t = 0 and using the values fro the graphs shown in the proble, we find v 4.00 c/s 0 tan tan.03 rad (or 5.5 rad). x0 (.0 c)(.0 rad/s) One can chec that the other root (4.7 rad) is unacceptable since it would give the wrong signs for the individual values of v 0 and x 0.. Let the spring constants be and. When displaced fro equilibriu, the agnitude of the net force exerted by the springs is x + x acting in a direction so as to return the bloc to its equilibriu position (x = 0). Since the acceleration a = d x/d, Newton s second law yields
708 CHAPTER 5 d x =. x x dt Substituting x = x cos(t + ) and siplifying, we find + = where is in radians per unit tie. Since there are radians in a cycle, and frequency f easures cycles per second, we obtain = = f. The single springs each acting alone would produce siple haronic otions of frequency f= 30 Hz, f = 45 Hz, respectively. Coparing these expressions, it is clear that f f f (30 Hz) +(45 Hz) 54 Hz.. The stateent that the spring does not affect the collision justifies the use of elastic collision forulas in section 0-5. We are told the period of SHM so that we can find the ass of bloc : T T 4 0.600 g. At this point, the rebound speed of bloc can be found fro Eq. 0-30: 0.00 g 0.600 g v f 8.00 /s 4.00 /s. 0.00 g 0.600 g This becoes the initial speed v 0 of the projectile otion of bloc. A variety of choices for the positive axis directions are possible, and we choose left as the +x direction and down as the +y direction, in this instance. With the launch angle being zero, Eq. 4- and Eq. 4- (with g replaced with +g) lead to h (4.90 ) x x0 v0t v0 (4.00 /s). g 9.8 /s Since x x 0 = d, we arrive at d = 4.00.
709 3. THINK The axiu force that can be exerted by the surface ust be less than the static frictional force or else the bloc will not follow the surface in its otion. EXPRESS The static frictional force is given by fs sfn, where µ s is the coefficient of static friction and F N is the noral force exerted by the surface on the bloc. Since the bloc does not accelerate vertically, we now that F N = g, where is the ass of the bloc. If the bloc follows the table and oves in siple haronic otion, the agnitude of the axiu force exerted on it is given by F = a = x = (f) x, where a is the agnitude of the axiu acceleration, is the angular frequency, and f is the frequency. The relationship = f was used to obtain the last for. ANALYZE We substitute F = (f) x and F N = g into F µ s F N to obtain (f) x µ s g. The largest aplitude for which the bloc does not slip is x = s g = f b g b0.50gc9.8 / s h 0. 03..0 Hz b LEARN A larger aplitude would require a larger force at the end points of the otion. The bloc slips if the surface cannot supply a larger force. 4. We wish to find the effective spring constant for the cobination of springs shown in the figure. We do this by finding the agnitude F of the force exerted on the ass when the total elongation of the springs is x. Then eff = F/x. Suppose the left-hand spring is elongated by x and the right-hand spring is elongated by x r. The left-hand spring exerts a force of agnitude x on the right-hand spring and the right-hand spring exerts a force of agnitude x r on the left-hand spring. By Newton s third law these ust be equal, so x x r. The two elongations ust be the sae, and the total elongation is twice the elongation of either spring: x x. The left-hand spring exerts a force on the bloc and its agnitude is F x. Thus, x / eff x r /. The bloc behaves as if it were subject to the force of a single spring, with spring constant /. To find the frequency of its otion, replace eff in f a / f eff / with / to obtain = f. With = 0.45 g and = 6430 N/, the frequency is f = 8. Hz. g
70 CHAPTER 5 5. (a) We interpret the proble as asing for the equilibriu position; that is, the bloc is gently lowered until forces balance (as opposed to being suddenly released and allowed to oscillate). If the aount the spring is stretched is x, then we exaine force-coponents along the incline surface and find g sin (4.0 N)sin 40.0 x g sin x 0.0750 0 N/ at equilibriu. The calculator is in degrees ode in the above calculation. The distance fro the top of the incline is therefore (0.450 + 0.75) = 0.55. (b) Just as with a vertical spring, the effect of gravity (or one of its coponents) is siply to shift the equilibriu position; it does not change the characteristics (such as the period) of siple haronic otion. Thus, Eq. 5-3 applies, and we obtain 4.0 N 9.80 /s T 0.686 s. 0 N/ 6. To be on the verge of slipping eans that the force exerted on the saller bloc (at the point of axiu acceleration) is f ax = µ s g. The textboo notes (in the discussion iediately after Eq. 5-7) that the acceleration aplitude is a = x, where / ( M) is the angular frequency (fro Eq. 5-). Therefore, using Newton s second law, we have a = g = + M x g s s which leads to x sg( M ) (0.40)(9.8 /s )(.8 g 0 g) 0.3 3 c. 00 N/ 7. THINK This proble explores the relationship between energies, both inetic and potential, with aplitude in SHM. EXPRESS In siple haronic otion, let the displaceent be x(t) = x cos(t + ). The corresponding velocity is v( t) dx / dt x sin( t ). Using the expressions for x(t) and v(t), we find the potential and inetic energies to be
7 U ( t) x ( t) x cos ( t ) K t v t x t x t ( ) ( ) sin ( ) sin ( ) where = is the spring constant and x is the aplitude. The total energy is E U( t) K( t) x cos ( t ) sin ( t ) x. ANALYZE (a) The condition x( t) x / iplies that cos( t ) /, or sin( t ) 3 /. Thus, the fraction of energy that is inetic is (b) Siilarly, we have (c) Since E x x /. K 3 3 sin ( t ) E. 4 U cos ( t ). E 4 x and U x() t, U/E = x x. Solving x x = / for x, we get LEARN The figure to the right depicts the potential energy (solid line) and inetic energy (dashed line) as a function of tie, assuing x(0) x. The curves intersect when K U E /, or equivalently, cos t sin t /. 8. The total echanical energy is equal to the (axiu) inetic energy as it passes through the equilibriu position (x = 0): v = (.0 g)(0.85 /s) = 0.7 J. Looing at the graph in the proble, we see that U(x = 0) = 0.5 J. Since the potential 3 function has the for U( x) bx, the constant is b 5.0 0 J/c. Thus, U(x) = 0.7 J when x = c. (a) Thus, the ass does turn bac before reaching x = 5 c. (b) It turns bac at x = c.
7 CHAPTER 5 9. THINK Knowing the aplitude and the spring constant, we can calculate the echanical energy of the ass-spring syste in siple haronic otion. EXPRESS In siple haronic otion, let the displaceent be x(t) = x cos(t + ). The corresponding velocity is v( t) dx / dt x sin( t ). Using the expressions for x(t) and v(t), we find the potential and inetic energies to be where = U ( t) x ( t) x cos ( t ) K( t) v ( t) xsin ( t ) xsin ( t ) is the spring constant and x is the aplitude. The total energy is E U( t) K( t) x cos ( t ) sin ( t ) x. ANALYZE With.3 N/c 30 N/ and x.4 c 0.04, the echanical energy is c ha f E = x =.3 0 0.04 = 3.7 0 N / J. LEARN An alternative to calculate E is to note that when the bloc is at the end of its path and is oentarily stopped ( v 0 K 0), its displaceent is equal to the aplitude and all the energy is potential in nature ( E U K U ). With the spring potential energy taen to be zero when the bloc is at its equilibriu position, we recover the expression E x /. 30. (a) The energy at the turning point is all potential energy: E x where E =.00 J and x = 0.00. Thus, = E = 00 N /. x (b) The energy as the bloc passes through the equilibriu position (with speed v =.0 /s) is purely inetic: E E v.39 g. v (c) Equation 5- (divided by ) yields
73 f = 9. Hz. 3. (a) Equation 5- (divided by ) yields f = 000 N / 5. 00 g. 5Hz. (b) With x 0 = 0.500, we have U 0 x 5 J. 0 (c) With v 0 = 0.0 /s, the initial inetic energy is K 0 v 50 J. 0 (d) Since the total energy E = K 0 + U 0 = 375 J is conserved, then consideration of the energy at the turning point leads to = E E x x = 0.866. 3. We infer fro the graph (since echanical energy is conserved) that the total energy in the syste is 6.0 J; we also note that the aplitude is apparently x = c = 0.. Therefore we can set the axiu potential energy equal to 6.0 J and solve for the spring constant : x = 6.0 J = 8.3 0 N/. 33. The proble consists of two distinct parts: the copletely inelastic collision (which is assued to occur instantaneously, the bullet ebedding itself in the bloc before the bloc oves through significant distance) followed by siple haronic otion (of ass + M attached to a spring of spring constant ). (a) Moentu conservation readily yields v = v/( + M). With = 9.5 g, M = 5.4 g, and v = 630 /s, we obtain v. /s. (b) Since v occurs at the equilibriu position, then v = v for the siple haronic otion. The relation v = x can be used to solve for x, or we can pursue the alternate (though related) approach of energy conservation. Here we choose the latter: v M v x M x M which siplifies to 3 v (9.50 g)(630 /s) x 3.30. 3 M (6000 N/)(9.50 g 5.4g)
74 CHAPTER 5 34. We note that the spring constant is = 4 /T =.97 0 5 N/. It is iportant to deterine where in its siple haronic otion (which phase of its otion) bloc is when the ipact occurs. Since = /T and the given value of t (when the collision taes place) is one-fourth of T, then t = / and the location then of bloc is x = x cos(t + ) where = / which gives x = x cos(/ + /) = x. This eans bloc is at a turning point in its otion (and thus has zero speed right before the ipact occurs); this eans, too, that the spring is stretched an aount of c = 0.0 at this oent. To calculate its after-collision speed (which will be the sae as that of bloc right after the ipact, since they stic together in the process) we use oentu conservation and obtain v = (4.0 g)(6.0 /s)/(6.0 g) = 4.0 /s. Thus, at the end of the ipact itself (while bloc is still at the sae position as before the ipact) the syste (consisting now of a total ass M = 6.0 g) has inetic energy and potential energy U = x = K = (6.0 g)(4.0 /s) = 48 J (.97 05 N/)(0.00 ) 0 J, eaning the total echanical energy in the syste at this stage is approxiately E = K + U = 58 J. When the syste reaches its new turning point (at the new aplitude X ) then this aount ust equal its (axiu) potential energy there: E = (.97 05 N/) X. Therefore, we find E (58 J) X 0.04. 5.970 N/ 35. The textboo notes (in the discussion iediately after Eq. 5-7) that the acceleration aplitude is a = x, where is the angular frequency and x = 0.000 is the aplitude. Thus, a = 8000 /s leads to = 000 rad/s. Using Newton s second law with = 0.00 g, we have F c a fh a f I H K F = a = acos t + = 80 N cos 000t 3 where t is understood to be in seconds.
75 (a) Equation 5-5 gives T = / = 3. 0 3 s. (b) The relation v = x can be used to solve for v, or we can pursue the alternate (though related) approach of energy conservation. Here we choose the latter. By Eq. 5-, the spring constant is = = 40000 N/. Then, energy conservation leads to (c) The total energy is x x = v v = x 4.0 /s. v 0. 080 J. (d) At the axiu displaceent, the force acting on the particle is F x 4 3 (4.0 0 N/)(.0 0 ) 80 N. (e) At half of the axiu displaceent, x.0, and the force is F x 4 3 (4.0 0 N/)(.0 0 ) 40 N. 36. We note that the ratio of Eq. 5-6 and Eq. 5-3 is v/x = tan(t + ) where is given by Eq. 5-. Since the inetic energy is v and the potential energy is x (which ay be conveniently written as x ) then the ratio of inetic to potential energy is siply (v/x) / = tan (t + ), which at t = 0 is tan. Since = /6 in this proble, then the ratio of inetic to potential energy at t = 0 is tan () = /3. 37. (a) The object oscillates about its equilibriu point, where the downward force of gravity is balanced by the upward force of the spring. If is the elongation of the spring at equilibriu, then g, where is the spring constant and is the ass of the object. Thus g and a f a f. f g Now the equilibriu point is halfway between the points where the object is oentarily at rest. One of these points is where the spring is unstretched and the other is the lowest point, 0 c below. Thus 5. 0c 0. 050 and 9.8 /s f. Hz. 0.050
76 CHAPTER 5 (b) Use conservation of energy. We tae the zero of gravitational potential energy to be at the initial position of the object, where the spring is unstretched. Then both the initial potential and inetic energies are zero. We tae the y-axis to be positive in the downward direction and let y = 0.080. The potential energy when the object is at this point is U y gy. The energy equation becoes We solve for the speed: 0 y gy v. g 9.8 /s v gy y gy y 9.8 /s 0.080 0.080 0.050 0.56 /s (c) Let be the original ass and be the additional ass. The new angular frequency is / ( ). This should be half the original angular frequency, or. We solve / ( ) / for. Square both sides of the equation, then tae the reciprocal to obtain + = 4. This gives = /3 = (300 g)/3 = 00 g = 0.00 g. (d) The equilibriu position is deterined by the balancing of the gravitational and spring forces: y = ( + )g. Thus y = ( + )g/. We will need to find the value of the spring constant using = = ( f ). Then 0.00 g 0.300 g9.80 /s + g y = 0.00. f 0.00 g.4 Hz This is easured fro the initial position. 38. Fro Eq. 5-3 (in absolute value) we find the torsion constant: 0.0 N 0.35 N /rad. 0.85 rad With I = R /5 (the rotational inertia for a solid sphere fro Chapter ), Eq. 5 3 leads to 5 R 5 95 g 0.5 T s. 0.35 N/rad
77 39. THINK The balance wheel in the watch undergoes angular siple haronic oscillation. Fro the aplitude and period, we can calculate the corresponding angular velocity and angular acceleration. EXPRESS We tae the angular displaceent of the wheel to be t = cos(t/t), where is the aplitude and T is the period. We differentiate with respect to tie to find the angular velocity: = d/dt = (/T) sin(t/t). The sybol is used for the angular velocity of the wheel so it is not confused with the angular frequency. ANALYZE (a) The axiu angular velocity is rad 39.5 rad/s. T 0.500 s (b) When = /, then / = /, cos(t/t) = /, and t T t T sin cos 3 where the trigonoetric identity cos +sin = is used. Thus, a f I KJ = 3 F ti H K = F I F sin H = 34.. 0.500 K rad H G rad / s T T s During another portion of the cycle its angular speed is +34. rad/s when its angular displaceent is / rad. (c) The angular acceleration is When = /4, or 4 rad/s. d cos t/ T. dt T T 0.500 s 4 = 4 rad/s, LEARN The angular displaceent, angular velocity and angular acceleration as a function of tie are plotted next.
78 CHAPTER 5 40. We use Eq. 5-9 and the parallel-axis theore I = I c + h where h = d, the unnown. For a eter stic of ass, the rotational inertia about its center of ass is I c = L / where L =.0. Thus, for T =.5 s, we obtain T L / d gd L d. gd g Squaring both sides and solving for d leads to the quadratic forula: a f a f 4 g T / d T / L / 3 d =. Choosing the plus sign leads to an ipossible value for d (d =.5 L). If we choose the inus sign, we obtain a physically eaningful result: d = 0.056. 4. THINK Our physical pendulu consists of a dis and a rod. To find the period of oscillation, we first calculate the oent of inertia and the distance between the centerof-ass of the dis-rod syste to the pivot. EXPRESS A unifor dis pivoted at its center has a rotational inertia of Mr, where M is its ass and r is its radius. The dis of this proble rotates about a point that is displaced fro its center by r+ L, where L is the length of the rod, so, according to the parallel-axis theore, its rotational inertia is Mr M( L r). The rod is pivoted at one end and has a rotational inertia of L /3, where is its ass. ANALYZE (a) The total rotational inertia of the dis and rod is I Mr M ( L r) L 3 (0.500g)(0.00) (0.500g)(0.500 0.00) (0.70g)(0.500) 3 0.05g.
79 (b) We put the origin at the pivot. The center of ass of the dis is d = L+ r = 0.500 +0.00 = 0.600 away and the center of ass of the rod is r L / ( 0. 500 ) / 0. 50 away, on the sae line. The distance fro the pivot point to the center of ass of the dis-rod syste is a fa f a fa f M + 0.500 0.600 + 0.70 0.50 d g g r d = = M + 0.500 g + 0.70 g (c) The period of oscillation is = 0.477. T I 0.05 g.50 s. M gd (0.500 g 0.70 g)(9.80 /s )(0.447 ) LEARN Consider the liit where M 0 (i.e., unifor dis reoved). In this case, I L /3, d L/and the period of oscillation becoes r T I L / 3 L gd g( L / ) 3g which is the result given in Eq. 5-3. 4. (a) Coparing the given expression to Eq. 5-3 (after changing notation x ), we see that = 4.43 rad/s. Since = g/l then we can solve for the length: L = 0.499. (b) Since v = x = L = (4.43 rad/s)(0.499 )(0.0800 rad) and = 0.0600 g, then we can find the axiu inetic energy: v = 9.40 0 4 J. 43. (a) Referring to Saple Proble 5.5 Physical pendulu, period and length, we see that the distance between P and C is h 3 L L 6 L. The parallel axis theore (see Eq. 5 30) leads to F H = + = + = I L h L L. 36 9 I K Equation 5-9 then gives I L / 9 T gh gl / 6 L 3g which yields T =.64 s for L =.00.
70 CHAPTER 5 (b) We note that this T is identical to that coputed in Saple Proble 5.5 Physical pendulu, period and length. As far as the characteristics of the periodic otion are concerned, the center of oscillation provides a pivot that is equivalent to that chosen in the Saple Proble (pivot at the edge of the stic). 44. To use Eq. 5-9 we need to locate the center of ass and we need to copute the rotational inertia about A. The center of ass of the stic shown horizontal in the figure is at A, and the center of ass of the other stic is 0.50 below A. The two stics are of equal ass, so the center of ass of the syste is h (0.50 ) 0.5 below A, as shown in the figure. Now, the rotational inertia of the syste is the su of the rotational inertia I of the stic shown horizontal in the figure and the rotational inertia I of the stic shown vertical. Thus, we have I = I + = + = 5 I ML ML ML 3 where L =.00 and M is the ass of a eter stic (which cancels in the next step). Now, with = M (the total ass), Eq. 5-9 yields 5 ML T Mgh 5L 6g where h = L/4 was used. Thus, T =.83 s. 45. Fro Eq. 5-8, we find the length of the pendulu when the period is T = 8.85 s: L = gt. 4 The new length is L = L d where d = 0.350. The new period is which yields T = 8.77 s. T L L d T d g g g 4 g 46. We require T L g o I gh siilar to the approach taen in part (b) of Saple Proble 5.5 Physical pendulu, period and length, but treating in our case a ore general possibility for I. Canceling, squaring both sides, and canceling g leads directly to the result; L o = I/h.
7 47. We use Eq. 5-9 and the parallel-axis theore I = I c + h where h = d. For a solid dis of ass, the rotational inertia about its center of ass is I c = R /. Therefore, R / d R d (.35 c) +(.75 c) T 0.366 s. gd gd (980 c/s )(.75 c) 48. (a) For the physical pendulu we have T = I gh = Ico h gh. If we substitute r for h and use ite (i) in Table 0-, we have a b T r. g r In the figure below, we plot T as a function of r, for a = 0.35 and b = 0.45. (b) The iniu of T can be located by setting its derivative to zero, dt / dr 0. This yields a b (0.35 ) (0.45 ) r 0.6. (c) The direction fro the center does not atter, so the locus of points is a circle around the center, of radius [(a + b )/] /. 49. Replacing x and v in Eq. 5-3 and Eq. 5-6 with and d/dt, respectively, we identify 4.44 rad/s as the angular frequency Then we evaluate the expressions at t = 0 and divide the second by the first: d / dt = tan. at t0
7 CHAPTER 5 (a) The value of at t = 0 is 0.0400 rad, and the value of d/dt then is 0.00 rad/s, so we are able to solve for the phase constant: = tan [0.00/(0.0400 x 4.44)] = 0.845 rad. (b) Once is deterined we can plug bac in to o = cos to solve for the angular aplitude. We find = 0.060 rad. 50. (a) The rotational inertia of a unifor rod with pivot point at its end is I = L / + L = /3ML. Therefore, Eq. 5-9 leads to 3 ML 3gT 3(9.8 /s )(.5 s) T L 0.84. Mg L 8 8 (b) By energy conservation Ebotto of swing Eend of swing K U where U Mg( cos ) with being the distance fro the axis of rotation to the center of ass. If we use the sall-angle approxiation ( cos with in radians (Appendix E)), we obtain L U 0.5 g9.8 /s where = 0.7 rad. Thus, K = U = 0.03 J. If we calculate ( cos) directly (without using the sall angle approxiation) then we obtain within 0.3% of the sae answer. 5. This is siilar to the situation treated in Saple Proble 5.5 Physical pendulu, period and length, except that O is no longer at the end of the stic. Referring to the center of ass as C (assued to be the geoetric center of the stic), we see that the distance between O and C is h = x. The parallel axis theore (see Eq. 5-30) leads to L I L h x. Equation 5-9 gives L I x L x T gh gx gx c h c h. (a) Miniizing T by graphing (or special calculator functions) is straightforward, but the standard calculus ethod (setting the derivative equal to zero and solving) is soewhat
73 awward. We pursue the calculus ethod but choose to wor with gt / instead of T (it should be clear that gt / is a iniu whenever T is a iniu). The result is gt d L e d x x j d i L 0 dx dx x which yields x L/ (.85 )/ 0.53 as the value of x that should produce the sallest possible value of T. (b) With L =.85 and x = 0.53, we obtain T =. s fro the expression derived in part (a). 5. Consider that the length of the spring as shown in the figure (with one of the bloc s corners lying directly above the bloc s center) is soe value L (its rest length). If the (constant) distance between the bloc s center and the point on the wall where the spring attaches is a distance r, then rcos = d/, and rcos = L defines the angle easured fro a line on the bloc drawn fro the center to the top corner to the line of r (a straight line fro the center of the bloc to the point of attachent of the spring on the wall). In ters of this angle, then, the proble ass us to consider the dynaics that results fro increasing fro its original value o to o + 3º and then releasing the syste and letting it oscillate. If the new (stretched) length of spring is L (when = o + 3º), then it is a straightforward trigonoetric exercise to show that (L) = r + (d/ ) r(d/ )cos( o + 3º) = L + d d cos(3º)+ Ldsin(3º) since o = 45º. The difference between L (as deterined by this expression) and the original spring length L is the aount the spring has been stretched (denoted here as x ). If one plots x versus L over a range that sees reasonable considering the figure shown in the proble (say, fro L = 0.03 to L = 0.0 ) one quicly sees that x 0.00 is an excellent approxiation (and is very close to what one would get by approxiating x as the arc length of the path ade by that upper bloc corner as the bloc is turned through 3º, even though this latter procedure should in principle overestiate x ). Using this value of x with the given spring constant leads to a potential energy of U = x = 0.0096 J. Setting this equal to the inetic energy the bloc has as it passes bac through the initial position, we have K = 0.0096 J = I where is the axiu angular speed of the bloc (and is not to be confused with the angular frequency of the oscillation, though they are related by = o if o is expressed in radians). The rotational inertia of the bloc is I = 6 Md = 0.008 g. Thus, we can solve the above relation for the axiu angular speed of the bloc:
74 CHAPTER 5 K (0.0096 J).8 rad/s. I 0.008 g Therefore the angular frequency of the oscillation is = / o = 34.6 rad/s. Using Eq. 5-5, then, the period is T = 0.8 s. 53. THINK By assuing that the torque exerted by the spring on the rod is proportional to the angle of rotation of the rod and that the torque tends to pull the rod toward its equilibriu orientation, we see that the rod will oscillate in siple haronic otion. EXPRESS Let = C, where is the torque, is the angle of rotation, and C is a constant of proportionality, then the angular frequency of oscillation is C / I and the period is I T, C where I is the rotational inertia of the rod. The plan is to find the torque as a function of and identify the constant C in ters of given quantities. This iediately gives the period in ters of given quantities. Let 0 be the distance fro the pivot point to the wall. This is also the equilibriu length of the spring. Suppose the rod turns through the angle, with the left end oving away fro the wall. This end is now (L/) sin further fro the wall and has oved a distance (L/)( cos ) to the right. The length of the spring is now ( L/ ) ( cos ) [ ( L/ )sin ]. 0 If the angle is sall we ay approxiate cos with and sin with in radians. Then the length of the spring is given by 0 L / and its elongation is x = L/. The force it exerts on the rod has agnitude F = x = L/. Since is sall we ay approxiate the torque exerted by the spring on the rod by = FL/, where the pivot point was taen as the origin. Thus, = (L /4). The constant of proportionality C that relates the torque and angle of rotation is C = L /4. The rotational inertia for a rod pivoted at its center is I = L / (see Table 0-), where is its ass. ANALYZE Substituting the expressions for C and I, we find the period of oscillation to be I L / T. C L / 4 3 With = 0.600 g and = 850 N/, we obtain T = 0.0653 s. LEARN As in the case of a siple linear haronic oscillator fored by a ass and a spring, the period of the rotating rod is inversely proportional to. Our result indicates
75 that the rod oscillates very rapidly, with a frequency f / T 5.3 Hz, i.e., about 5 ties in one second. 54. We note that the initial angle is o = 7º = 0. rad (though it turns out this value will cancel in later calculations). If we approxiate the initial stretch of the spring as the arclength that the corresponding point on the plate has oved through (x = r o where r = 0.05 ) then the initial potential energy is approxiately x = 0.0093 J. This should equal to the inetic energy of the plate ( I where this is the axiu angular speed of the plate, not the angular frequency ). Noting that the axiu angular speed of the plate is = o where = /T with T = 0 s = 0.0 s as deterined fro the graph, then we can find the rotational inertial fro I = 0.0093 J. Thus, I 5.30 g. 55. (a) The period of the pendulu is given by T I / gd, where I is its rotational inertia, =. g is its ass, and d is the distance fro the center of ass to the pivot point. The rotational inertia of a rod pivoted at its center is L / with L =.0. According to the parallel-axis theore, its rotational inertia when it is pivoted a distance d fro the center is I = L / + d. Thus, T ( L / d ) gd L d gd. Miniizing T with respect to d, dt/d(d) = 0, we obtain d L /. Therefore, the iniu period T is T L ( L / ) L (.0 ) in.6 s. g( L / ) g (9.80 /s ) (b) If d is chosen to iniize the period, then as L is increased the period will increase as well. (c) The period does not depend on the ass of the pendulu, so T does not change when increases. 56. The table of oents of inertia in Chapter, plus the parallel axis theore found in that chapter, leads to I P = MR + Mh = (.5 g)(0. ) + (.5 g)(0.97 ) =.4 g ² where P is the hinge pin shown in the figure (the point of support for the physical pendulu), which is a distance h = 0. + 0.76 away fro the center of the dis.
76 CHAPTER 5 (a) Without the torsion spring connected, the period is T = I P Mgh =.00 s. (b) Now we have two restoring torques acting in tande to pull the pendulu bac to the vertical position when it is displaced. The agnitude of the torque-su is (Mgh + ) = I P where the sall-angle approxiation (sin in radians) and Newton s second law (for rotational dynaics) have been used. Maing the appropriate adjustent to the period forula, we have T = I P Mgh +. The proble stateent requires T = T + 0.50 s. Thus, T = (.00 0.50)s =.50 s. Consequently, 4 = T I P Mgh = 8.5 N /rad. 57. Since the energy is proportional to the aplitude squared (see Eq. 5-), we find the fractional change (assued sall) is E E E de E dx = = xdx dx =. x x x Thus, if we approxiate the fractional change in x as dx /x, then the above calculation shows that ultiplying this by should give the fractional energy change. Therefore, if x decreases by 3%, then E ust decrease by 6.0%. 58. Referring to the nubers in Saple Proble 5.6 Daped haronic oscillator, tie to decay, energy, we have = 0.5 g, b = 0.070 g/s, and T = 0.34 s. Thus, when t = 0T, the daping factor becoes e e b gb gb g b g 039.. bt 0. 070 0 0. 34 / 0. 5 59. THINK In the presence of a daping force, the aplitude of oscillation of the assspring syste decreases with tie. EXPRESS As discussed in 5-8, when a daping force is present, we have bt / x( t) xe cos( t ) where b is the daping constant and the angular frequency is given by b 4.
77 ANALYZE (a) We want to solve e bt/ = /3 for t. We tae the natural logarith of both sides to obtain bt/ = ln(/3). Therefore, Thus, (b) The angular frequency is t = (/b) ln(/3) = (/b) ln 3. a f =.50 g t 3 = 4.3. 0.30 g / s ln s b 8. 00 N / 0. 30 g / s. 3 rad / s. 4 50. g 4 50. g The period is T = / = ()/(.3 rad/s) =.7 s and the nuber of oscillations is a a f f t/t = (4.3 s)/(.7 s) = 5.7. LEARN The displaceent x(t) as a function of tie is shown below. The aplitude, bt / xe, decreases exponentially with tie. 60. (a) Fro Hooe s law, we have 500 g9.8 /s 0c = 4.9 0 N/c. (b) The aplitude decreasing by 50% during one period of the otion iplies e bt where T. Since the proble ass us to estiate, we let /. That is, we let 49000 N / 500 g 9.9 rad / s,
78 CHAPTER 5 so that T 0.63 s. Taing the (natural) log of both sides of the above equation, and rearranging, we find 500 g ln 0.69. 0 3 b g/s. T 0.63 s Note: if one worries about the approxiation, it is quite possible (though essy) to use Eq. 5-43 in its full for and solve for b. The result would be (quoting ore figures than are significant) = ln b = 086 g / s ( ln ) + 4 which is in good agreeent with the value gotten the easy way above. 6. (a) We set = d and find that the given expression reduces to x = F /b at resonance. (b) In the discussion iediately after Eq. 5-6, the boo introduces the velocity aplitude v = x. Thus, at resonance, we have v = F /b = F /b. 6. With = /T then Eq. 5-8 can be used to calculate the angular frequencies for the given pendulus. For the given range of.00 < < 4.00 (in rad/s), we find only two of the given pendulus have appropriate values of : pendulu (d) with length of 0.80 (for which = 3.5 rad/s) and pendulu (e) with length of. (for which =.86 rad/s). 63. With M = 000 g and = 8 g, we adapt Eq. 5- to this situation by writing T M 4. If d = 4.0 is the distance traveled (at constant car speed v) between ipulses, then we ay write T = v/d, in which case the above equation ay be solved for the spring constant: v v = M 4. d M 4 d Before the people got out, the equilibriu copression is x i = (M + 4)g/, and afterward it is x f = Mg/. Therefore, with v = 6000/3600 = 4.44 /s, we find the rise of the car body on its suspension is = 4 = 4 g g d xi x f = 0.050. M + 4 v F H I K
79 64. Since = f where f =. Hz, we find that the angular frequency is = 3.8 rad/s. Thus, with x = 0.00, the acceleration aplitude is a = x =.9 /s. We set up a ratio: F a = =.9 I a = 0.9. g g 9.8 g g HG KJ F H I K 65. (a) The proble gives the frequency f = 440 Hz, where the SI unit abbreviation Hz stands for Hertz, which eans a cycle-per-second. The angular frequency is siilar to frequency except that is in radians-per-second. Recalling that radians are equivalent to a cycle, we have = f.8 0 3 rad/s. (b) In the discussion iediately after Eq. 5-6, the boo introduces the velocity aplitude v = x. With x = 0.00075 and the above value for, this expression yields v =. /s. (c) In the discussion iediately after Eq. 5-7, the boo introduces the acceleration aplitude a = x, which (if the ore precise value = 765 rad/s is used) yields a = 5.7 /s. 66. (a) First consider a single spring with spring constant and unstretched length L. One end is attached to a wall and the other is attached to an object. If it is elongated by x the agnitude of the force it exerts on the object is F = x. Now consider it to be two springs, with spring constants and, arranged so spring is attached to the object. If spring is elongated by x then the agnitude of the force exerted on the object is F = x. This ust be the sae as the force of the single spring, so x = x. We ust deterine the relationship between x and x. The springs are unifor so equal unstretched lengths are elongated by the sae aount and the elongation of any portion of the spring is proportional to its unstretched length. This eans spring is elongated by x = CL and spring is elongated by x = CL, where C is a constant of proportionality. The total elongation is x = x + x = C(L + L ) = CL (n + ), where L = nl was used to obtain the last for. Since L = L /n, this can also be written x = CL (n + )/n. We substitute x = CL and x = CL (n + )/n into x = x and solve for. With = 8600 N/ and n = L /L = 0.70, we obtain n n 0.70 0.70.0 (8600 N/) 0886 N/. 0 4 N/ (b) Now suppose the object is placed at the other end of the coposite spring, so spring exerts a force on it. Now x = x. We use x = CL and x = CL (n + ), then solve for. The result is = (n + )..
730 CHAPTER 5 n 4 ( ) (0.70.0)(8600 N/) 460 N/.5 0 N/ (c) To find the frequency when spring is attached to ass, we replace in a / f / with (n + )/n. With f a / f /, we obtain, for f 00 Hz and n = 0.70, f ( n) n 0.70.0 n n 0.70 = f (00 Hz) 3. 0 Hz. (d) To find the frequency when spring is attached to the ass, we replace with (n + ) to obtain ( n) f n f = 0.70.0(00 Hz).6 0 Hz. 67. The agnitude of the downhill coponent of the gravitational force acting on each ore car is w x = 0000 g 9.8 / s sin b gc h where = 30 (and it is iportant to have the calculator in degrees ode during this proble). We are told that a downhill pull of 3 x causes the cable to stretch x = 0.5. Since the cable is expected to obey Hooe s law, its spring constant is = 3 wx 5 = 9.8 0 N /. x (a) Noting that the oscillating ass is that of two of the cars, we apply Eq. 5- (divided by ). f 0000 g 5 9.80 N /. Hz. (b) The difference between the equilibriu positions of the end of the cable when supporting two as opposed to three cars is = 3 w x wx x = 0.050. 68. (a) Hooe s law readily yields (0.300 g)(9.8 /s )/(0.000 ) = 47 N/. (b) With =.00 g, the period is T = 0. 733 s. 69. THINK The piston undergoes siple haronic otion. Given the aplitude and frequency of oscillation, its axiu speed can be readily calculated.
73 EXPRESS Let the aplitude be x. The axiu speed v is related to the aplitude by v = x, where is the angular frequency. ANALYZE We use v =x = fx, where the frequency is f = (80 rev)/(60 s) = 3.0 Hz and the aplitude is half the stroe, or x = 0.38. Thus, v = (3.0 Hz)(0.38 ) = 7. /s. LEARN In a siilar anner, the axiu acceleration is a x f x 3.0 Hz 0.38 35 /s. Acceleration is proportional to the displaceent x in SHM. 70. (a) The rotational inertia of a hoop is I = R, and the energy of the syste becoes E I x and is in radians. We note that r = v (where v = dx/dt). Thus, the energy becoes F R E v x H G I r K J which loos lie the energy of the siple haronic oscillator discussed in Section5-4 if we identify the ass in that section with the ter R /r appearing in this proble. Maing this identification, Eq. 5- yields R / r r R. (b) If r = R the result of part (a) reduces to /. (c) And if r = 0 then = 0 (the spring exerts no restoring torque on the wheel so that it is not brought bac toward its equilibriu position). 7. Since T = 0.500 s, we note that = /T = 4 rad/s. We wor with SI units, so = 0.0500 g and v = 0.50 /s. (a) Since /, the spring constant is 4 rad/s 0.0500 g 7.90 N/.
73 CHAPTER 5 (b) We use the relation v = x and obtain x v = = 0.50 = 0.09. 4 (c) The frequency is f = / =.00 Hz (which is equivalent to f = /T). 7. (a) We use Eq. 5-9 and the parallel-axis theore I = I c + h where h = R = 0.6. For a solid dis of ass, the rotational inertia about its center of ass is I c = R /. Therefore, (b) We see a value of r R such that R / R 3R T 0.873s. gr g R r gr 3R g and are led to the quadratic forula: Thus, our result is r = 0.6/ = 0.0630. a f or. 3R 3R 8R R r R 4 73. THINK A ass attached to the end of a vertical spring undergoes siple haronic otion. Energy is conserved in the process. EXPRESS The spring stretches until the agnitude of its upward force on the bloc equals the agnitude of the downward force of gravity: y 0 = g, where y 0 = 0.096 is the elongation of the spring at equilibriu, is the spring constant, and =.3 g is the ass of the bloc. As the bloc oscillate, its speed is a axiu as it passes the equilibriu point, and zero at the endpoints. ANALYZE (a) The spring constant is (b) The period is given by = g/ y 0 = (.3 g)(9.8 /s )/(0.096 ) =.33 0 N/. T.3 g 0.6 s. f 33 N / (c) The frequency is f = /T = /0.6 s =.6 Hz.
733 (d) The bloc oscillates in siple haronic otion about the equilibriu point deterined by the forces of the spring and gravity. It is started fro rest y = 5.0 c below the equilibriu point so the aplitude is 5.0 c. (e) At the initial position, yi y0 y 9.6 c 5.0 c 4.6 c 0.46, the bloc is not oving but it has potential energy U.3 g9.8 /s 0.46 33 N / 0.46 i gyi yi 0.44 J. When the bloc is at the equilibriu point, the elongation of the spring is y 0 = 9.6 c and the potential energy is U f gy0 y0 0.6 J..3 g9.8 /s 0.096 33 N / 0.096 We write the equation for conservation of energy as U U v and solve for v: Ui U f 0.44 J 0.6J v 0.5 /s..3g LEARN Both the gravitational force and the spring force are conservative, so the wor done by the forces is independent of path. By energy conservation, the inetic energy of the bloc is equal to the negative of the change in potential energy of the syste: where the relation y0 K U ( U f Ui ) Ui U f g( yi y0) ( yi y0) gy ( y0 y) y0 g y ( y) y0 y y( g y0) ( y) ( ) y g was used. 74. The distance fro the relaxed position of the botto end of the spring to its equilibriu position when the body is attached is given by Hooe s law: i f
734 CHAPTER 5 x = F/ = (0.0 g)(9.8 /s )/(9 N/) = 0.03. (a) The body, once released, will not only fall through the x distance but continue through the equilibriu position to a turning point equally far on the other side. Thus, the total descent of the body is x = 0.. (b) Since f = /, Eq. 5- leads to f = 6. z. (c) The axiu distance fro the equilibriu position gives the aplitude: x = x = 0.0. 75. (a) Assue the bullet becoes ebedded and oves with the bloc before the bloc oves a significant distance. Then the oentu of the bulletbloc syste is conserved during the collision. Let be the ass of the bullet, M be the ass of the bloc, v 0 be the initial speed of the bullet, and v be the final speed of the bloc and bullet. Conservation of oentu yields v 0 = ( + M)v, so v0 v = = + M a0.050 g fa50 / sf =.85 / s. 0.050 g + 4.0 g When the bloc is in its initial position the spring and gravitational forces balance, so the spring is elongated by Mg/. After the collision, however, the bloc oscillates with siple haronic otion about the point where the spring and gravitational forces balance with the bullet ebedded. At this point the spring is elongated a distance am f g /, soewhat different fro the initial elongation. Mechanical energy is conserved during the oscillation. At the initial position, just after the bullet is ebedded, the inetic energy is ( M ) v and the elastic potential energy is ( Mg / ). We tae the gravitational potential energy to be zero at this point. When the bloc and bullet reach the highest point in their otion the inetic energy is zero. The bloc is then a distance y above the position where the spring and gravitational forces balance. Note that y is the aplitude of the otion. The spring is copressed by y, so the elastic potential energy is ( y ). The gravitational potential energy is (M + )gy. Conservation of echanical energy yields F I a f b g a f + + = M v Mg y + M + gy. H K