Chapter 13 Solutions

Transcription

1 Chapter 3 Solutions 3. x = (4.00 m) cos (3.00πt + π) Compare this with x = A cos (ωt + φ) to find (a) ω = πf = 3.00π or f =.50 Hz T = f = s A = 4.00 m (c) φ = π rad (d) x(t = 0.50 s) = (4.00 m) cos (.75π) =.83 m 3. (a) Since the collision is perfectly elastic, the ball will rebound to the height of 4.00 m and then repeat the motion over and over again. Thus, the motion is periodic. To determine the period, we use: x = gt The time for the ball to hit the ground is t = x g = (4.00 m) 9.80 m/s = s This equals one-half the period, so T = (0.909 s) =.8 s (c) No The net force acting on the mass is a constant given by F = mg (except when it is in contact with the ground), which is not in the form of Hooe's law. 3.3 (a) 0.0 cm v max = ωa = π fa = 94. cm/s This occurs as the particle passes through equilibrium. (c) a max = ω A = (π f) A = 7.8 m/s This occurs at maximum excursion from equilibrium.

2 Chapter 3 Solutions *3.4 (a) x = (5.00 cm) cos t + π 6 At t = 0, x = (5.00 cm) cos π 6 = 4.33 cm v = dx = (0.0 cm/s) sin t + π 6 At t = 0, v = 5.00 cm/s (c) a = dv = (0.0 cm/s ) cos t + π 6 At t = 0, a = 7.3 cm/s (d) A = 5.00 cm and T = π ω = π = 3.4 s 3.5 (a) At t = 0, x = 0 and v is positive (to the right). Therefore, this situation corresponds to x = A sin ω t and v = v i cos ω t Since f =.50 Hz, ω = π f = 3.00π Also, A =.00 cm, so that x = (.00 cm) sin 3.00πt v max = v i = Aω = (.00)(3.00π) = 6.00π cm/s The particle has this speed at t = 0 and next at t = T = 3 s (c) a max = Aω = (3.00π) = 8.0π cm/s The acceleration has this positive value for the first time at t = 3T 4 = s (d) Since T = 3 s and A =.00 cm, the particle will travel 8.00 cm in this time. Hence, in.00 s = 3T, the particle will travel 8.00 cm cm =.0 cm

3 Chapter 3 Solutions The proposed solution x(t) = x i cos ω t + v i ω sin ωt implies velocity and acceleration v = dx a = dv = x i ω sin ωt + v i cos ωt = x i ω cos ωt v i ω sin ωt = ω x i cos ωt + ω sin ωt = ω x v i (a) The acceleration being a negative constant times position means we do have SHM, and its angular frequency is ω. At t = 0 the equations reduce to x = x i and v = v i so they satisfy all the requirements. v ax = ( x i ω sin ωt + v i cos ωt) ( x i ω cos ωt v i ω sin ωt) x i cos ωt + sin ωt ω v i = x i ω sin ωt x i v i ω sin ω t cos ω t + v i cos ω t + x i ω cos ω t + x i v i ω cos ω t sin ω t + x i v i ω sin ω t cos ω t + v i sin ω t = x i ω + v i So this expression is constant in time. On one hand, it must eep its original value v i a ix i On the other hand, if we evaluate it at a turning point where v = 0 and x = A, it is A ω + 0 = A ω Thus it is proved. 3.7 = F x = ( g)(9.80 m/s ) m =.5 N/m and T = π m = π g.5 N/m = 0.67 s

4 4 Chapter 3 Solutions 3.8 (a) T =.0 s 5 =.40 s f = T =.40 = 0.47 Hz (c) ω = π f = π (0.47) =.6 rad/s 3.9 (a) ω = m = 8.00 N/m g = 4.00 s Therefore, position is given by x = 0.0 sin (4.00t) cm From this we find that v = 40.0 cos (4.00t) cm/s v max = 40.0 cm/s a = 60 sin (4.00t) cm/s a max = 60 cm/s t = 4.00 sin x 0.0 and when x = 6.00 cm, t = 0.6 s, and we find v = 40.0 cos [4.00(0.6)] = 3.0 cm/s a = 60 sin [4.00(0.6)] = 96.0 cm/s (c) Using t = 4.00 sin x 0.0 when x = 0, t = 0 and when x = 8.00 cm, t = 0.3 s Therefore, t = 0.3 s 3.0 m =.00 g, = 5.0 N/m, and A = 3.00 cm At t = 0, x = 3.00 cm (a) ω = m = = 5.00 rad/s so that, T = π ω = π 5.00 =.6 s

5 Chapter 3 Solutions 5 v max = Aω = ( m)(5.00 rad/s) = 0.50 m/s a max = Aω = ( m)(5.00 rad/s) = m/s (c) Because x = 3.00 cm and v = 0 at t = 0, the required solution is x = A cos ωt or x = 3.00 cos (5.00t) cm v = dx a = dv = 5.0 sin (5.00t) cm/s = 75.0 cos (5.00t) cm/s 3. f = ω π = π m or T = f = π m Solving for, = 4π m T = (4π) (7.00 g) (.60 s) = 40.9 N/m 3. (a) Energy is conserved between the maximum-displacement and the half-maximum points: (K + U) i = (K + U) f 0 + A = mv + mx (6.50 N/m)(0.00 m) = m (0.300 m/s) + (6.50 N/m)( m) 3.5 mj = m(0.300 m/s) + 8. mj (4.4 mj) m = m /s = 0.54 g ω = m = 6.50 N/m 0.54 g = 3.46 rad/s T = π ω = π 3.46/s =.8 s (c) a max = ω A = (3.46/s) (0.00 m) =.0 m/s

6 6 Chapter 3 Solutions 3.3 (a) v max = ωa A = v max ω =.50 m/s.00 rad/s = m x = (0.750 m) sin.00t 3.4 (a) v max = ωa A = v max ω = v ω x = A sin ω t = v ω sin ω t 3.5 The s must elapse between one turning point and the other. Thus the period is.00 s. ω = π T = 6.8/s and v max = ωa = (6.8/s)(0.00 m) = 0.68 m/s *3.6 m = 00 g, T = 0.50 s, E =.00 J; ω = π T = π 0.50 = 5. rad/s (a) = mω = (0.00 g)(5. rad/s) = 6 N/m E = A A = E = (.00) 6 = 0.78 m 3.7 By conservation of energy, mv = x v = m x = (3.6 0 m) =.3 m/s

7 Chapter 3 Solutions 7 Goal Solution G: If the bumper is only compressed 3 cm, the car is probably not permanently damaged, so v is most liely less than 0 mph (< 5 m/s). O: Assuming no energy is lost during impact with the wall, the initial energy (inetic) equals the final energy (elastic potential): A: K i = U f or mv = x v = x m = (3.6 0 m) N/m 000 g v =.3 m/s : The speed is less than 5 m/s as predicted, so the answer seems reasonable. If the speed of the car were sufficient to compress the bumper beyond its elastic limit, then some of the initial inetic energy would be lost to deforming the front of the car. In this case, some other procedure would have to be used to estimate the car s initial speed. 3.8 (a) E = A = (50 N/m)( m) = 0.53 J v max = Aω where ω = m = =.4 s v max = m/s (c) a max = Aω = ( m)(.4 s ) = 7.5 m/s 3.9 (a) E = A = (35.0 N/m)( m) = 8.0 mj v = ω A x = m A x v = ( ) (.00 0 ) =.0 m/s (c) (d) mv = A x = (35.0) [( ) ( ) ] =. mj x = E mv = 5.8 mj

8 8 Chapter 3 Solutions 3.0 (a) = F x = 0.0 N 0.00 m = 00 N/m ω = m = 50.0 rad/s f = ω =.3 Hz π (c) v max = ωa = 50.0 (0.00) =.4 m/s at x = 0 (d) a max = ω A = 50.0(0.00) = 0.0 m/s at x = ± A (e) E = A = (00)(0.00) =.00 J (f) v = ω A x = (0.00) =.33 m/s (g) a = ω x = (50.0) = 3.33 m/s 3. (a) In the presence of non-conservative forces, we use E = mv f mv i + mgy f mgy i + x f x i (0.0 N)(0.300 m) = (.50 g) v f (9.6 N/m)(0.300 m) 0 v f =.6 m/s f = µ n = 0.00(4.7 N) =.94 N (K + U) i + E = (K + U) f Fd cos 0 + fd cos 80 = mv f + x 6.00 J + (.94 N)(0.300 m)cos 80 = (.50 g) v f J v f = (6.00 J 0.88 J 0.88 J)/.50 g =.38 m/s

9 Chapter 3 Solutions 9 3. (a) E = A, so if A' = A, E' = (A') = (A) = 4E Therefore E increases by factor of 4. v max = m A, so if A is doubled, v max is doubled. (c) a max = m A, so if A is doubled, a max also doubles. (d) T = π m is independent of A, so the period is unchanged. 3.3 From energy considerations, v + ω x = ω A v max = ωa and v = ωa so wa + ω x = ω A From this we find x = 3A 4 and x = A 3 = ±.60 cm where A = 3.00 cm Goal Solution G: If we consider the speed of the particle along its path as shown in the setch, we can see that the particle is at rest momentarily at one endpoint while being accelerated toward the middle by an elastic force that decreases as the particle approaches the equilibrium position. When it reaches the midpoint, the direction of acceleration changes so that the particle slows down until it stops momentarily at the opposite endpoint. From this analysis, we can estimate that v = v max / somewhere in the outer half of the travel:.5 < x < 3. 3 cm 0 cm 3 cm v = 0 v = vmax v = 0 a a = 0 a O: We can analyze this problem in more detail by examining the energy of the system, which should be constant since we are told that the motion is SHM (no damping).

10 0 Chapter 3 Solutions A: From energy considerations (Eq. 3.3), v + ω x = ω A. The speed v will be maximum when x is zero. Thus, v max = ωa and v / = v max = ωa Substituting v / in for v, 4 ω A + ω x = ω A Solving for x we find that x = 3A 4 Given that A = 3.00 cm, x = ± A 3 (3.00 cm) 3 = ± = ±.60 cm : The calculated position is in the outer half of the travel as predicted, and is in fact very close to the endpoints. This means that the speed of the particle is mostly constant until it reaches the ends of its travel, where it experiences the maximum restoring force of the spring, which is proportional to x. *3.4 The potential energy is U s = x = A cos (ωt) The rate of change of potential energy is du s = A cos (ωt)[ ω sin (ωt)] = A ω sin ωt (a) This rate of change is maximal and negative at ωt = π, ωt = π + π, or in general, ωt = nπ + π for integer n. Then, t = π π(4n + ) (4n + ) = 4ω 4(3.60 s ) For n = 0, this gives t = 0.8 s while n = gives t =.09 s. All other values of n yield times outside the specified range. du s = A ω = (3.4 N/m)( m) (3.60 s ) = 4.6 mw max

11 Chapter 3 Solutions *3.5 (a) T = π g = gt 4π = (9.80 m/s )(.0 s) 4π = 35.7 m T moon = π g moon = π 35.7 m.67 m/s = 9. s *3.6 The period in Toyo is T t = π t g t and the period in Cambridge is T c = π c g c We now T t = T c =.00 s For which, we see t g t = c g c or g c g t = c t = =.005 *3.7 The swinging box is a physical pendulum with period T = π The moment of inertia is given approximately by I mgd. I = 3 m (treating the box as a rod suspended from one end). Then, with.0 m and d /, T π (/3)m mg(/) = π 3g = π (.0 m) 3(9.8 m/s ) =.6 s or T ~ 00 s 3.8 ω = π T T = π ω = π 4.43 =.4 s ω = g = g ω = 9.80 (4.43) = m 3.9 (a) mgh = mv h = ( cos θ) cos θ θ =.00 m θ = 5.0 v max = g ( cos θ) v max = 0.87 m/s h s m mg

12 Chapter 3 Solutions Iα = mg sin θ α max = mg sin θ m = g sin θ i =.54 rad/s (c) F max = mg sin θ i = (0.50)(9.80)(sin 5.0 ) = N 3.30 (a) The string tension must support the weight of the bob, accelerate it upward, and also provide the restoring force, just as if the elevator were at rest in a gravity field ( ) m/s T = π g = π 5.00 m 4.8 m/s T = 3.65 s T = π 5.00 m (9.80 m/s 5.00 m/s ) = 6.4 s (c) g eff = (9.80 m/s ) + (5.00 m/s ) =.0 m/s T = π = 5.00 m.0 m/s = 4.4 s 3.3 Referring to the setch we have F = mg sin θ and tan θ x R For small displacements, tan θ sin θ and F = mg R x = x and ω = m = g R x θ R mg sinθ mg

13 Chapter 3 Solutions (a) T = total measured time 50 The measured periods are: ength, (m) Period, T (s) T, s 4 3 T = π g so g = 4π T The calculated values for g are: Period, T (s) g (m/s ) , m Thus, g ave = 9.85 m/s this agrees with the accepted value of g = 9.80 m/s within 0.5%. (c) Slope of T versus graph = 4π /g = 4.0 s /m Thus, g = 4π slope = 9.85 m/s. You should find that % difference 0.5% f = Hz, d = m, and m =.0 g T = f ; T = π I mg d ; T = 4π I mg d I = T mg d 4π = f mg d 4π = (.0)(9.80)(0.350) (0.450 s ) (4π ) I = g m 3.34 (a) The parallel-axis theorem: I = I CM + Md = M + Md Pivot m = M(.00 m) + M(.00 m) = M 3 T = π I Mgd = π M(3 m ) Mg(.00 m) m CM.00 m T = π 3 m (9.80 m/s ) =.09 s

14 4 Chapter 3 Solutions For the simple pendulum T = π.00 m 9.80 m/s =.0 s difference =.09 s.0 s.0 s = 4.08% 3.35 (a) The parallel axis theorem says directly I = I CM + md so T = π I (I CM + md ) mg d = π mg d When d is very large T π d g gets large. When d is very small T π I CM mg d gets large. So there must be a minimum, found by dt dd = 0 = d dd π (I CM + md ) / (mgd) / = π (I CM + md ) / (mgd) 3/ mg + π(mgd) / (I CM + md ) / md = Error! mg,(i CM + md ) / (mgd) 3/ ) + Error! = 0 This requires I CM md + md = 0 or I CM = md 3.36 We suppose the stic moves in a horizontal plane. Then, I = m = (.00 g)(.00 m) = 0.67 g m T = π I/κ κ = 4π I T = 4π (0.67 g m ) (80 s) = 03 µn m

15 Chapter 3 Solutions T = 0.50 s; I = mr = ( g)( m) (a) I = g m I d θ = κ θ; κ I = ω = π T θ κ = Iω = ( ) π 0.50 = N m rad 3.38 (a) The motion is simple harmonic because the tire is rotating with constant velocity and you are looing at the motion of the boss projected in a plane perpendicular to the tire. Since the car is moving with speed v = 3.00 m/s, and its radius is m, we have: ω = 3.00 m/s m = 0.0 rad/s Therefore, the period of the motion is: T = π ω = π (0.0 rad/s) = 0.68 s 3.39 The angle of the cran pin is θ = ωt. Its x-coordinate is x = A cos θ = A cos ωt where A is the distance from the center of the wheel to the cran pin. This is of the form x = A cos(ωt + φ), so the yoe and piston rod move with simple harmonic motion. ω Piston A x = A x(t ) 3.40 E = mv + x de = mv d x + xv Use Equation 3.3: md x = x bv de = v( x bv) + vx de = bv < 0

16 6 Chapter 3 Solutions 3.4 θ i = 5.0 θ (t = 000) = 5.50 x = Ae bt/m x 000 x i = Ae bt/m A = = e b(000)/m e b m = s *3.4 Show that x = Ae bt/m cos (ωt + φ) is a solution of x b dx = m d x () and ω = m b m () x = Ae bt/m cos (ωt + φ) (3) dx = Ae bt/m b m cos (ωt + φ) Ae bt/m ω (sin(ωt + φ)) (4) d x = b Ae m bt/m b m cos (ωt + φ) Ae bt/m ω sin (ωt + φ) Ae bt/m b m ω sin (ωt + φ) + Ae bt /m ω cos (ωt + φ) (5) Substitute (3), (4) into the left side of () and (5) into the right side of (); Ae bt/m cos (ωt + φ) + b m Ae bt/m cos (ωt + φ) + bωae bt /m sin (ωt + φ) = b Ae bt/m b m cos (ωt + φ) Ae bt/m ω sin (ωt + φ) + b Ae bt/m ω sin (ωt + φ) mω Ae bt/m cos (ωt + φ)

17 Chapter 3 Solutions 7 Compare the coefficients of Ae bt/m cos (ωt + φ) and Ae bt/m sin (ωt + φ): cosine-term: + b m = b b m mω = b 4m (m) m b 4m = + b m sine-term: bω = + b (ω) + b ω = bω Since the coefficients are equal, x = Ae bt/m cos (ωt + φ) is a solution of the equation. *3.43 ( a) For resonance, her frequency must match f0 = ω 0 π = π m = π N/m.5 g =.95 Hz From x = A cos ωt, v = dx/ = Aω sin ωt, and a = dv/ = Aω cos ωt, the maximum acceleration is Aω. When this becomes equal to the acceleration of gravity, the normal force exerted on her by the mattress will drop to zero at one point in the cycle: Aω = g or A = g ω = g /m = gm A = (9.80 m/s )(.5 g) N/m 3.44 F = 3.00 cos (π t) N and = 0.0 N/m =.85 cm (a) ω = π T = π rad/s so T =.00 s In this case, ω 0 = m = = 3.6 rad/s Taing b = 0 in Equation 3.37 gives A = F 0 m (ω ω 0) = 3 [4π (3.6) ] A = m = 5.09 cm

18 8 Chapter 3 Solutions *3.45 F 0 cos (ωt) x = m d x ω 0 = x = A cos (ωt + φ) () m () dx = Aω sin (ωt + φ) (3) d x = Aω cos (ωt + φ) (4) Substitute () and (4) into (): F 0 cos (ωt) A cos (ωt + φ) = m( Aω ) cos (ωt + φ) Solve for the amplitude: (A maω ) cos (ωt + φ) = F 0 cos ωt These will be equal, provided only that φ must be zero and (A maω ) = F 0 Thus, A = F 0 /m m ω 3.46 From Equation 3.37 with no damping, A = F 0 /m (ω ω 0) ω = πf = (0.0π s ) ω 0 = m = 00 (40.0/9.80) F 0 = ma(ω ω 0 ) F 0 = (.00 0 )( ) = 38 N = 49.0 s *3.47 A = F ext /m (ω ω 0) + (bω/m) With b = 0, A = F ext /m (ω ω 0) = F ext /m ±(ω ω0) = ± F ext/m ω ω 0 Thus, ω = ω 0 ± F ext/m A = m ± F ext ma This yields ω = 8.3 rad/s or ω = 4.03 rad/s = 6.30 N/m 0.50 g ±.70 N (0.50 g)(0.440 m) Then, f = ω gives either f =.3 Hz or f = 0.64 Hz π

19 Chapter 3 Solutions 9 *3.48 The beeper must resonate at the frequency of a simple pendulum of length 8. cm: f = π g = π 9.80 m/s 0.08 m =.74 Hz 3.49 Assume that each spring supports an equal portion of the car's mass, i.e. m 4. Then T = π m 4 and = 4π m 4T = 4π 500 (4)(.50) = 6580 N/m 3.50 T T 0 = π/ω π/ω 0 = π/ /m π/ /m 0 T T 0 = m m 0 = =.0 T =.0.50 =.57 s 3.5 et F represent the tension in the rod. pivot (a) At the pivot, F = Mg + Mg = Mg P A fraction of the rod s weight Mg y as well as the y weight of the ball pulls down on point P. Thus, the tension in the rod at point P is M F = Mg y + Mg = Mg + y Relative to the pivot, I = I rod + I ball = 3 M + M = 4 3 M I For the physical pendulum, T = π where m = M and d is the distance from the mgd pivot to the center of mass of the rod and ball combination. Therefore, d = M(/) + M M + M = 3 4 and T = π (4/3)M (M)g(3/4) = 4π 3 g For =.00 m, T = 4π 3 (.00 m) 9.80 m/s =.68 s

20 0 Chapter 3 Solutions Goal Solution G: The tension in the rod at the pivot = weight of rod + weight of M = Mg. The tension at point P should be slightly less since the portion of the rod between P and the pivot does not contribute to the tension. The period should be slightly less than for a simple pendulum since the mass of the rod effectively shortens the length of the simple pendulum (massless rod) by moving the center of mass closer to the pivot, so that T < π O: The tension can be found from applying Newton s Second aw to the static case. The period of oscillation can be found by analyzing the components of this physical pendulum and using Equation 3.8. A: (a) At the pivot, the net downward force is: T = Mg + Mg = Mg At P, a fraction of the rod's mass (y/) pulls down along with the ball. g Therefore, T = Mg y + Mg = Mg + y Relative to the pivot, I total = I rod + I ball = 3 M + M = 4 3 M For a physical pendulum, T = π I mgd In this case, m = M and d is the distance from the pivot to the center of mass. ( M + M) d = (M + M) = 3 4 so we have, T = π I mgd = π (4M )4 3(M)g(3) = 4π 3 g For =.00 m, T = 4π 3 (.00 m) 9.80 m/s =.68 s : In part (a), the tensions agree with the initial predictions. In part we found that the period is indeed slightly less (by about 6%) than a simple pendulum of length. It is interesting to note that we were able to calculate a value for the period despite not nowing the mass value. This is because the period of any pendulum depends on the location of the center of mass and not on the size of the mass.

21 Chapter 3 Solutions 3.5 (a) Total energy = A = (00 N/m)(0.00 m) =.00 J At equilibrium, the total energy is: (m + m ) v = (6.0 g) v = (8.00 g)v Therefore, (8.00 g)v =.00 J, and v = m/s This is the speed of m and m at the equilibrium point. Beyond this point, the mass m moves with the constant speed of m/s while mass m starts to slow down due to the restoring force of the spring. The energy of the m -spring at equilibrium is: m v = (9.00 g)(0.500 m/s) =.5 J This is also equal to (A'), where A' is the amplitude of the m -spring system. Therefore, (00)(A') =.5 or A' = 0.50 m The period of the m -spring system is: T = π m =.885 s and it taes T = 0.47 s after it passes the equilibrium point for the spring to become fully 4 stretched the first time. The distance separating m and m at this time is: D = v T 4 A' = (0.500 m/s)(0.47 s) 0.50 m = = 8.56 cm 3.53 d x max = Aω f max = µ s n = µ s mg = maω n f A = µ s g ω = 6.6 cm mg

22 Chapter 3 Solutions 3.54 The maximum acceleration of the oscillating system is a max = Aω = 4π Af. The friction force exerted between the two blocs must be capable of accelerating bloc B at this rate. Thus, if Bloc B is about to slip, f = f max = µ s n = µ s mg = m(4π Af ) or A = µ s g 4π f 3.55 M D = M H ω D ω H = /M D /M H = M H M D = f D = f H = Hz 3.56 The inetic energy of the ball is K = mv + IΩ, where Ω is the rotation rate of the ball about its center of mass. Since the center of the ball moves along a circle of radius 4R, its displacement from equilibrium is s = (4R)θ and its speed is v = ds = 4R dθ. Also, since the ball rolls without slipping, v = ds = RΩ so Ω = v R = 4 dθ 5R h t θ s R The inetic energy is then K = m 4R dθ + 5 mr 4 dθ mr = dθ 0 When the ball has an angular displacement θ, its center is distance h = 4R( cos θ) higher than when at the equilibrium position. Thus, the potential energy is U g = mgh = 4mgR( cos θ). For small angles, ( cos θ) θ (see Appendix B). Hence, U g mgrθ, and the total energy is E = K + U g = mr dθ 0 + mgrθ Since E = constant in time, de = 0 = mr dθ 5 d θ + 4mgRθ dθ This reduces to 8R 5 d θ + gθ = 0, or d θ = 5g 8R θ

23 Chapter 3 Solutions 3 This is in the classical form of a simple harmonic motion equation with ω = The period of the simple harmonic motion is then T = π ω = π 8R 5g 5g 8R (a) i a a h T = π g dt = π g d () We need to find (t) and d = i + a h ; d. From the diagram in (a), = dh But dm = ρ dv = ρa dh. Therefore, dh = dm ρa ; d = ρa dm () Also, d = i ρa dm t = i (3) Substituting Equation () and Equation (3) into Equation (): dt = π g ρa dm i + ρa dm t

24 4 Chapter 3 Solutions (c) Substitute Equation (3) into the equation for the period. T = π g i + ρa dm t Or one can obtain T by integrating : T idt π = T g dm ρa t 0 i + ρa dm t T T i = π g dm ρa ρa dm i + ρa dm t i But T i = π i g, so T = π g i + ρa dm t 3.58 ω 0 = m = π T (a) = ω 0 m = 4π m T m' = (T') 4π = m T' T 3.59 For the pendulum (see setch) we have τ = Iα and d θ = α τ = Mg sin θ + xh cos θ = I d θ θ h For small amplitude vibrations, use the approximations: sin θ θ cosθ and x s = hθ M

25 Chapter 3 Solutions 5 Therefore, d θ = Mg + h I θ = ω θ ω = Mg + h M = π f f = π Mg + h M Goal Solution G: The frequency of vibration should be greater than that of a simple pendulum since the spring adds an additional restoring force: f > π g O: We can find the frequency of oscillation from the angular frequency, which is found in the equation for angular SHM: d θ = ω θ. The angular acceleration can be found from analyzing the torques acting on the pendulum. A: For the pendulum (see setch), we have τ = Iα and d θ = α The negative sign appears because positive θ is measured clocwise in the picture. We tae torque around the point of suspension: τ = Mg sin θ + xh cos θ = Iα For small amplitude vibrations, use the approximations: sin θ θ, cos θ, and x s = hθ Therefore, with I = m, d θ = Mg + h I θ = Mg + h M θ

26 6 Chapter 3 Solutions This is of the form d θ = ω θ required for SHM, with angular frequency, ω = Mg + h M = πf The ordinary frequency is f = ω π = π Mg + h M : The frequency is greater than for a simple pendulum as we expected. In fact, the additional portion resembles the frequency of a mass on a spring scaled by h/ since the spring is connected to the rod and not directly to the mass. So we can thin of the solution as: f = Mg + h 4π M = g 4π + h 4π M = f pendulum + h f spring *3.60 (a) At equilibrium, we have τ = 0 = mg + x 0 where x 0 is the equilibrium compression. After displacement by a small angle, Pivot θ τ = mg + x = mg + (x 0 θ) = θ = Iα = 3 m d θ So d θ = 3 m θ The angular acceleration is opposite in direction and proportional to the displacement, so we have simple harmonic motion with ω = 3 m. f = ω π = π 3 m = π 3(00 N/m) 5.00 g =.3 Hz

27 Chapter 3 Solutions 7 *3.6 One can write the following equations of motion: mg T = ma = m d x T' x = 0 (for the mass) (describes the spring) R(T T') = I d θ = I d x R (for the pulley) T R M T with I = MR. m Combining these equations gives the equation of motion m + M d x + x = mg The solution is x(t) = A sin ωt + Error!, with frequency f = ω π = π m + M = π 00 N/m 0.00 g + M ( a ) For M = 0, f = 3.56 Hz For M = 0.50 g, f =.79 Hz (c) For M = g, f =.0 Hz 3.6 (a) ω 0 = m = 5.8 rad/s F s mg = ma = m 3 g F s = 4 3 mg = 6. N x s = F s = 5.3 cm (c) When the acceleration of the car is zero, the new equilibrium position can be found as follows: F ' s = mg = 9.6 N = x ' s x ' s = 3.9 cm Thus, A = x' s x s =.3 cm The phase constant is π rad

28 8 Chapter 3 Solutions 3.63 (a) T = π ω = π g = 3.00 s E = mv = (6.74)(.06) = 4.3 J (c) At maximum angular displacement, mgh = mv h = v g = 0.7 m h = cos θ = ( cos θ) cos θ = h θ = 5.5 *3.64 Suppose a 00-g bier compresses the suspension.00 cm. Then, = F x = 980 N.00 0 m = N/m If total mass of motorcycle and bier is 500 g, the frequency of free vibration is f = π m = π N/m 500 g =.58 Hz If he encounters washboard bumps at the same frequency, resonance will mae the motorcycle bounce a lot. Assuming a speed of 0.0 m/s, these ridges are separated by (0.0 m/s).58/s =.7 m ~ 0 m In addition to this vibration mode of bouncing up and down as one unit, the motorcycle can also vibrate at higher frequencies by rocing bac and forth between front and rear wheels, by having just the front wheel bounce inside its for, or by doing other things. Other spacings of bumps will excite all of these other resonances. *3.65 y = (0.0 cm) [ cos (0.60 m x)] y (0.0 cm) + (0.60 m x) or y (0.0 cm)(0.60 m x) x

29 Chapter 3 Solutions 9 The geometric slope of the wire is dy dx = (0.00 m)(0.60 m ) (x) = ( m )x If m is the mass of the bead, the component of the bead s weight that acts as a restoring force is F = mg dy dx = m(9.80 m/s )( m )x = ma Thus, a = (0.050 s )x = ω x. Since the acceleration of the bead is opposite the displacement from equilibrium and is proportional to the displacement, the motion is simple harmonic with ω = s, or ω = 0.4 rad/s. *3.66 (a) For each segment of the spring dk = (dm) v x x dx v Also, v x = x l v M and dm = m l dx Therefore, the total inetic energy is K = Mv + l x v m 0 l l dx = M + m 3 v ω = m eff and m effv = M + m 3 v m Therefore, T = π M + ω = π 3

30 30 Chapter 3 Solutions 3.67 (a) F = T sin θ j where θ = tan y Therefore, for a small displacement y sin θ tan θ = y and F = Ty j For a spring system, F = x becomes F = T Therefore, ω = m = T m y 3.68 (a) Assuming a Hooe's aw type spring, Static stretching of a spring F = Mg = x y =.7386x 0.8 and empirically Mg =.74x 0.3 so.74 N/m ± 6% M, g x, m Mg, N Suspended weight, N Extension, m We may write the equation as theoretically Squared period as a function of the mass of an object bouncing on a spring T = 4π M + 4π 3 m s and empirically T =.7 M so = 4π.7.8 N/m ± 3% Period squared, seconds squares Time, s T, s M, g T, s Mass, g The values.74 N/m ± 6% and.8 N/m ± 3% differ by 4%, so they agree.

31 Chapter 3 Solutions 3 (c) Utilizing the axis-crossing point, m s = g 8 grams ± % in agreement with 7.4 grams (a) K + U = 0 Thus, K top + U top = K bot + U bot where K top = U bot = 0 M R Therefore, mgh = Iω, but θ θ h = R R cos θ = R( cos θ) v m ω = v R and I = MR + mr + mr Substituting we find mgr( cos θ) = MR + mr + v mr R mgr( cos θ) = M 4 + mr 4R + m v and v ( cos θ) = 4gR M m + r R + so v = Rg( cos θ) M m + r R + T = π I m T gd CM m T = m + M d CM = mr + M(0) m + M T = π MR + mr + mr mgr

32 3 Chapter 3 Solutions 3.70 (a) We require Ae b t/m = A e +b t/m = or bt m = ln or (0.00 g/s) (0.375 g) t = t = 5.0 s The spring constant is irrelevant. We can evaluate the energy at successive turning points, where cos (ω t + φ) = ± l and the energy is x = A e b t /m We require A e b t /m = A or e +b t /m = t = m ln b = g(0.693) 0.00 g/s =.60 s (c) From E = A, the fractional rate of change of energy over time is de d da da E = A A = = A A A two times faster than the fractional rate of change in amplitude. 3.7 (a) When the mass is displaced a distance x from equilibrium, spring is stretched a distance x and spring is stretched a distance x. By Newton's third law, we expect x = x. When this is combined with the requirement that x = x + x, we find x = ( + ) x The force on either spring is given by F = x = ma ( + ) where a is the acceleration of the mass m. This is in the form and T = π m eff = π F = eff x = ma m( + ) (a) m m

33 Chapter 3 Solutions 33 In this case each spring is stretched by the distance x which the mass is displaced. Therefore, the restoring force is F = ( + )x and eff = ( + ) so that T = π m ( + ) 3.7 For θ max = 5.00, the motion calculated by the Euler method agrees quite precisely with the prediction of θ max cos ωt. The period is T =.0 s. Time, t (s) Angle, θ ( ) Ang. speed ( /s) Ang. Accel. ( /s ) θ max cos ωt

34 34 Chapter 3 Solutions Angle ( ) 6 Simple Pendulum, small amplitude Time (s) 4 6

35 Chapter 3 Solutions 35 For θ max = 00, the simple harmonic motion approximation θ max cos ωt diverges greatly from the Euler calculation. The period is T =.7 s, larger than the small angle period by 3%. Time, t (s) Angle, θ ( ) Ang. speed ( /s) Ang. Accel. ( /s ) θ max cos ωt Angle ( ), A Cos ωt Simple Pendulum, large amplitude Time (s) 50 00