Chapter 14 - 14.4. Model: The air-track glider attached to...

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Unformatted text preview: 14.4. Model: The air-track glider attached to a spring is in simple harmonic motion. Visualize: The position of the glider can be represented as x(t) = A cos w t. Solve: The glider is pulled to the right and released from rest at t = 0 s . It then oscillates with a period T = 2.0 s and a maximum speed v max = 40 cm s = 0.40 m s . (a) v max = w A and w = 2p T = 2p 2.0 s = p rad s fi A = vmax w = 0.40 m s p rad s = 0.127 m = 12.7 cm (b) The glider's position at t = 0.25 s is x0.25 s = (0.127 m ) cos (p rad s )(0.25 s ) = 0.090 m = 9.0 cm [ 14.6. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure Ex14.6. Solve: (a) The amplitude A = 20 cm. (b) The period T = 4.0 s, thus 1 = = 0.25 Hz T 4.0 s (c) The position of an object undergoing simple harmonic motion is x( t ) = A cos(w t + f 0 ) . At t = 0 s, x0 = -10 cm. Thus, 10 cm ^ 1^ 2p -1 -1 - 10 cm = (20 cm ) cos f 0 fi f 0 = cos = cos = rad = 120 20 cm 2 3 f = 1 Because the object is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram and thus must have a phase constant between and 2 radians. Therefore, f 0 = - 2 p rad = -120 . 3 14.10. Solve: The position of the object is given by the equation x( t ) = Acos (wt + f 0 ) = Acos (2pft + f 0 ) We can find the phase constant f 0 from the initial condition: - A = A cos f 0 fi f 0 = cos -1 ( -1) = p rad The final result, with f = 0.50 Hz, is x( t ) = (10.0 cm ) cos (p rad s )t + p [ 14.15. Model: The mass attached to the spring oscillates in simple harmonic motion. Solve: (a) The period T = 1 f = 1 2.0 Hz = 0.50 s . (b) The angular frequency w = 2pf = 2 p (2 Hz) = 4p rad /s . (c) Using energy conservation 1 2 kA = 2 1 2 kx 0 + 2 1 2 mv 0 x 2 2 Using x0 = 5.0 cm, v0x = -30 cm/s and k = mw2 = (0.2 kg) (4 p rad s) , we get A = 5.54 cm. (d) To calculate the phase constant f 0 , Acos f 0 = x 0 = 5. 0 cm fi f 0 = cos -1 5.0 cm ^ = 0.445 rad 5.54 cm (e) The maximum speed is v max = w A = (4p rad s )(5.54 cm ) = 69.6 cm s . (f) The maximum acceleration is a max = w A = w (wA ) = (4 p rad s)( 69.6 cm s ) = 875 cm s 2 2 (g) The total energy is E = 1 mv max = 2 (h) The position at t = 0.40 s is 2 1 2 ( 0.200 kg )(0.696 m s ) = 0.0484 J . 2 x0.4 s = (5.54 cm ) cos (4p rad s )( 0.40 s ) + 0.445 rad = + 3.81 cm [ 14.17. Model: The block attached to the spring is in simple harmonic motion. Visualize: Solve: (a) The conservation of mechanical energy equation Kf + Usf = Ki + Usi is 1 2 mv 1 + 1 k ( Dx ) = 2 m v0 = k 2 2 1 2 mv 0 + 0 J fi 0 J + 2 1 2 kA = 2 1 2 mv 0 + 0 J 2 fi A= 1.0 kg (0.40 m s ) = 0.10 m = 10.0 cm 16 N m (b) We have to find the velocity at a point where x = A/2. The conservation of mechanical energy equation K2 + U s2 = K i + U si is 1 2 mv 2 + 2 1 A^ 1 1 1 1 1 2^ 1 1 1 3 1 2 2 2 2 2^ 2^ k ~ = mv 0 + 0 J fi mv 2 = mv 0 - kA ~ = mv 0 - mv 0 ~ = mv 0 ~ 2 2 2 2 2 2 4 2 4 2 4 2 fi v2 = 3 v0 = 4 3 ( 0.40 m s ) = 0.346 m s = 34 .6 cm s 4 2 14.22. Model: Assume a small angle of oscillation so there is simple harmonic motion. Solve: The period of the pendulum is T0 = 2p L0 g = 4.0 s (a) The period is independent of the mass and depends only on the length. Thus T = T0 = 4.0 s. (b) For a new length L = 2L0, 2 L0 T = 2p = 2T0 = 5.66 s g (c) For a new length L = L0/2, T = 2p L0 2 g = 1 2 T0 = 2.83 s (d) The period is independent of the amplitude as long as there is simple harmonic motion. Thus T = 4.0 s. 14.25. Model: Assume the pendulum to have small-angle oscillations. In this case, the pendulum undergoes simple harmonic motion. 2 Solve: Using the formula g = GM R , the periods of the pendulums on the moon and on the earth are Tearth = 2 p L g = 2p 2 L earth Rearth GM earth and Tmoon = 2 p L moon R 2 moon GM moon 2 Because Tearth = Tmoon, 2p 2 Learth Rearth GM earth = 2p 2 Lmoon Rmoon GM moon M ^ R ^ fi Lmoon = moon ~ earth ~ L earth M earth Rmoon 2 7.36 10 22 kg ^ 6.37 10 6 m ^ = ~ ~ (2.0 m ) = 33 cm 24 6 5.98 10 kg 1.74 10 m 14.29. Model: The motion is a damped oscillation. Solve: The position of the air-track glider is x( t ) = Ae w = - ( t 2t ) cos(w t + f 0 ) , where t = m b and b2 4m 2 k m - Using A = 0.20 m, f 0 = 0 rad , and b = 0.015 kg/s, w = 4.0 N m 0.250 kg - (0.015 kg s ) 2 2 4 (0.250 kg ) T = 2p = = 16 - 9 10 -4 rad s = 4.00 rad s Thus the period is = 1.57 s w 4.0 rad s -1 The amplitude at t = 0 s is x0 = A and the amplitude will be equal to e A at a time given by 1 m - ( t 2t ) A = Ae fi t = 2 t = 2 = 33.3 s e b The number of oscillations in a time of 33.3 s is (33.3 s)/(1.57 s) = 21. 2 p rad 14.33. Visualize: Please refer to Figure P14.33. Solve: The position and the velocity of a particle in simple harmonic motion are x( t ) = Acos (wt + f 0 ) and vx (t ) = - A w sin (wt + f 0 ) = -v max sin (wt + f 0 ) (a) At t = 0 s, the equation for x yields (5.0 cm ) = (10.0 cm ) cos(f 0 ) fi f 0 = cos -1 (0.5 ) = 1 p rad 3 Because the particle is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram, and the phase constant is between and 2 radians. Thus, f 0 = - 1 p rad. 3 (b) At t = 0 s, 2p ^ p ^ v 0 x = - A w sinf 0 = - (10.0 cm ) ~ sin - ~ = +6. 80 cm T 3 (c) The maximum speed is v max = w A = 2p ^ (10.0 cm ) = 7.85 cm s 8.0 s Assess: The positive velocity at t = 0 s is consistent with the position-versus-time graph and the negative sign of the phase constant. 14.40. Model: The spring undergoes simple harmonic motion. Solve: (a) Total energy is E = 1 2 kA2. When the displacement is x = 1 2 2 1 2 1 2 2 1 4 1 2 A, the potential energy is U = kx = k ( A) = ( 1 2 kA 2 )= 1 4 E fi K = E-U = 3E 4 One quarter of the energy is potential and three-quarters is kinetic. (b) To have U = 1 E requires 2 U= 1 2 kx = 2 1 2 E= 1 2 ( 1 2 kA 2 )fi x = A 2 14.43. Model: The ball attached to a spring is in simple harmonic motion. Solve: (a) Let t = 0 s be the instant when x0 = -5 cm and v0 = 20 cm/s. The oscillation frequency is w = k m = 2.5 N m 0.10 kg = 5.0 rad / s Using Equation 14.27, the amplitude of the oscillation is A= x0 + 2 v0 ^ w 2 2 = (-5 cm ) + 2 2 20 cm / s ^ ~ = 6.40 cm 5 rad / s 2 (b) The maximum acceleration is a max = w A = 160 cm s . (c) For an oscillator, the acceleration is most positive (a = a max ) when the displacement is most negative ( x = - x max = - A ) . So the acceleration is maximum when x = -6.40 cm . (d) We can use the conservation of energy between x0 = -5 cm and x1 = 3 cm: 1 2 mv 0 + 1 kx 0 = 2 2 2 1 2 mv 1 + 1 kx 1 fi v1 = 2 2 2 v0 + 2 k m (x 2 0 - x1 = 0.283 m s = 28 .3 cm s 2 ) Because k is known in SI units of N/m, the energy calculation must be done using SI units of m, m/s, and kg. 14.52. Model: The blocks undergo simple harmonic motion. Visualize: The length of the stretched spring due to a block of mass m is DL 1 . In the case of the two block system, the spring is further stretched by an amount DL 2 . Solve: The equilibrium equations from Newton's second law for the single block and double block systems are ( DL )k = mg and ( DL 1 1 + DL 2 )k = (2m ) g Using DL 2 = 5.0 cm , and subtracting these two equations, gives us (DL fi 1 + DL2 )k - DL1 k = (2m ) g - mg fi ( 0.05 m )k = mg k 9.8 m s 2 w = = w 2 fi w = 14.0 rad s fi f = = 2.23 Hz m 0.05 m 2p 14.54. Model: The two blocks are in simple harmonic motion, without the upper block slipping. We will also apply the model of static friction between the two blocks. Visualize: Solve: The net force acting on the upper block m1 is the force of friction due to the lower block m2. The model of static friction gives the maximum force of static friction as f s max = m s n = m s ( m1 g ) = m1 a max fi a max = m s g Using ms = 0.5 , a max = mS g = (0. 5) 9.8 m s = 4.9 m s . That is, the two blocks will ride together if the maximum acceleration of the system is equal to or less than amax. We can calculate the maximum value of A as follows: a max = w Amax = 2 ( 2 ) 2 k m1 + m2 A max fi A max = a max ( m1 + m 2 ) k = (4.9 m s 2 (1.0 kg + 5.0 kg ) 50 N m ) = 0.588 m 14.67. Model: Assume that rmarble << R hoop and that q is a small angle. Visualize: Solve: The marble is like an object on an inclined plane. The net force on the marble in the tangential direction is - w sin q = ma = mR a = mR d 2q dt 2 fi - mg sin q = mR d 2q dt 2 where a is the angular acceleration. With the small-angle approximation sin q q , this becomes d 2q dt 2 =- g R q = -w q 2 This is the equation of motion of an object in simple harmonic motion with a period of T = 2p w = 2p R g 14.69. Model: The vertical oscillations constitute simple harmonic motion. Visualize: Solve: At the equilibrium position, the net force on mass m on Planet X is: k g Fnet = kDL - mg X = 0 N fi = X m DL 2 For simple harmonic motion k m = w , thus w = 2 gX DL fiw = gX DL = 2p T fi gX = 2p ^ 2p ^ DL = ~ T 14.5 s 10 2 2 ( 0.312 m ) = 5.86 m s 2 14.71. Model: The oscillator is in simple harmonic motion. Solve: (a) The maximum displacement at time t of a damped oscillator is xmax (t ) = Ae - t 2t fi- t 2t x ( t) ^ = ln max ~ A Using xmax = 0.98A at t = 0.5 s, we can find the time constant t to be 0.5 s t == 12.375 s 2ln ( 0.98) 25 oscillations will be completed at t = 25T = 12.5 s. At that time, the amplitude will be xmax, 12.5 s = (10.0 cm )e -12. 5 s ( 2 ) (12 .375 s ) = 6.03 cm (b) The energy of a damped oscillator decays more rapidly than the amplitude: E(t) = E0et/t. When the energy is 60% of its initial value, E(t)/E0 = 0.60. We can find the time this occurs as follows: t t E (t ) ^ E( t) ^ = ln ~ fi t = -t ln ~ = -(12.375 s ) ln( 0.60 ) = 6.32 s E0 E0 14.79. Model: The vertical movement of the car is simple harmonic motion. Visualize: The fact that the car has a maximum oscillation amplitude at 5 m/s implies a resonance. The bumps in the road provide a periodic external force to the car's suspension system, and a resonance will occur when the "bump frequency" fext matches the car's natural oscillation frequency f0. Solve: Now the 5.0 m/s is not a frequency, but we can convert it to a frequency because we know the bumps are spaced every 3.0 meters. The time to drive 3.0 m at 5.0 m/s is the period: Dx 3.0 m T = = = 0.60 s v 5.0 m s The external frequency due to the bumps is thus f ext = 1 T = 1.667 Hz . This matches the car's natural frequency f0, which is the frequency the car oscillates up and down with if you push the car down and release it. This is enough information to deduce the spring constant of the car's suspension: 2p m m where we used m = mtotal = mcar + 2mpassenger = 1200 kg. When at rest, the car is in static equilibrium with Fnet = 0 N. The downward weight mtotalg of the car and passengers is balanced by the upward spring force kDy of the suspension. Thus the compression Dy of the suspension is m g Dy = total k Initially mtotal = mcar + 2mpassenger = 1200 kg, causing an initial compression Dyi = 0.0894 m = 8.94 cm. When three additional passengers get in, the mass increases to mtotal = mcar + 5mpassenger = 1500 kg. The final compression is Dyf = 0.1117 m = 11.17 cm. Thus the three new passengers cause the suspension to "sag" by 11.17 cm 8.94 cm = 2.23 cm. 1.667 Hz = 1 k fi k = m (2pf ext ) = 131,600 2 N ...
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