The distance of the object from the equilibrium position is m.

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Abel Clarke
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1 Answers, Even-Numbered Problems, Chapter (a) A = 0.0 m (b).60 s (c) 0.65 Hz Whenever the object is released from rest, its initial displacement equals the amplitude of its SHM. (a) so Hz. (b) A = 0.0 cm. (c) T = 6.0 s (d) 0.9 rad/s k = 0.9 N m. (a),..7 Hz. (b)..58 Hz (a) = Acos( ωt + φ) (b) 8. m/s (maimum magnitude of velocity) 4. 0 m/s (maimum magnitude of acceleration) (c) a = ω Acos ωt da dt A ωt / =+ ω sin Maimum magnitude of the jerk is ω A = 6. 0 m/s 7 (a) A = 0.8 m (b) φ = 58.5 (or.0 rad) (c) = Acos( ωt + φ) gives = (0.8 m)cos([.rad/s] t +.0 rad) The distance of the object from the equilibrium position is 0.5 m..09 s (a). s (b) 7.64 mm

2 (c) 0.69 m/s (d). N/m s. (a) =± A/ ; magnitude is A / v =± ωa/ ; magnitude is ω A/. (b) the occurrences of K = U are equally spaced in time, with a time interval between them of π/ ω. K (c) E = 4 and U E = 4 (a) a = 5. m/s. v = 0.96 m/s ma ma (b) The speed is 0.8 m/s. (c) 0.7 s (d) The conservation of energy equation relates v and and F = ma relates a and. So the speed and acceleration can be found by energy methods but the time cannot. Specifying uniquely determines a but determines only the magnitude of v ; at a given the object could be moving either in the + or direction. (a) J (b) 0.04 m (c) 0.65 m/s (a) s (b) m. m (c) T = π. The force constant remains the same. m decreases, so T decreases. k (a) 5. 0 N/m. (b) s. (c) 0.45 m/s

3 (a) At the top of the motion, the spring has no potential energy, the cat has no kinetic energy, and the gravitational potential energy relative to the bottom is.9 J. This is the total energy, and is the same total for each part. U = 0, K = 0, so U =.9 J. (b) grav spring (c) U = 0.98 J, U =.96 J, K = 0.98 J. spring grav (a) κ = 8.7 N m/rad (b) f =.7 Hz. T = / f = 0.46 s. (c) ω =.6 rad/s. θ ( t) = (.4 )cos([.6 rad/s] t) N m/rad. κ = (a) ω = dθ = ω Θ sin( ω t) and d θ = ω Θ cos( ω t). dt dt (b) When the angular displacement is Θ, Θ=Θ cos( ωt). This occurs at t = 0, so ω = 0. α = ω Θ. When the angular displacement is Θ, Θ = Θ cos( ωt), or = cos( ωt). ω Θ α =, since cos( ω t) =. 4 f =. 0 Hz. (a).8 s. (b).84 s..60 s. ω Θ ω = since sin( ω t) =. (a) The forces and acceleration are shown in Figure.46a. a = 0 and rad a = a = gsin θ. tan (b) The forces and acceleration are shown in Figure.46b.

4 (c) The forces and acceleration are shown in Figure.46c. U = K gives i f mgl( cos ) mv Θ = and v gl( cos ) = Θ. As the rod moves toward the vertical, v increases, a increases and a decreases. rad tan Figure (a).84 s (b).89 s (c) Eq.(.5) is more accurate. Eq.(.4) is in error by.84 s.89 s = %..89 s m. (a) kg m. (b).66 rad s s. (a). the system is damped. b =. kg/s. (b) Since the motion has a period the system oscillates and is underdamped. b = 0.00 kg/s. (a) A / (b) A 4

5 .6.64 The resonant frequency is 9 rad/s =. Hz, and this package does not meet the criterion..00 s..66. (a).68 s. (b) m (c) μ = 0.4 s increased μ gm ( + m) s A =. k The amplitude is T =.77 s. (a) The graph is given in Figure.7. The following answers canalso be found algebraically (b) A = 0.00 m. (c) J. (d). = 0.4 m. (e) rad. The dependence of U on is not linear and U = U does not occur at =. ma ma Figure.7 5

6 Hz. (a) Substitution gives = 0.0 m, or using t = T gives = A cos 0 = A. (b) Substitution gives ma =+ ( kg)(.06 m s ) = 4. 0 N, in the + -direction. A 4 (c) T t = arccos = s. π A (d) Using the time found in part (c), v = m/s. EVALUATE: We could also calculate the speed in part (d) from the conservation of energy epression, Eq.(.). (a) f = Hz. The new amplitude is m. (b) the new frequency is again Hz. the new amplitude is 0.46 m. (a) 4.05 kg. (b) t = (0.5) T, and so = Asin[ π(0.5)] = m. Since t > T / 4, the mass has already passed the lowest point of its motion, and is on the way up. (c) Taking upward forces to be positive, F mg = k, where is the spring displacement from equilibrium, sof = (60 N/m)( 0.00 m) + (4.05 kg)(9.80 m/s ) = 44.5 N. spring EVALUATE: When the object is below the equilibrium position the net force is upward and the upward spring force is larger in magnitude than the downward weight of the object s, or 84.5 min. c (a) U = Fd = c d = (b) From conservation of energy, A d 4 4 = c dt. m = c. v mv ( A ) d =, so dt Integrating from 0 to A with respect to and from 0 to T 4 with 6

7 respect to t, A d c T = To use the hint, let u =, so that d = a du A m 4 A and the upper limit of the u-integral is u =. Factoring A out of the square root, du. c T A = =, which may be epressed as T = 7.4 m. 0 4 u A m A c (c) The period does depend on amplitude, and the motion is not simple harmonic..86 (a) For the center of mass to be at rest, the total momentum must be zero, so the momentum vectors must be of equal magnitude but opposite directions, and the momenta can be represented as p and p. p p (b) K = =. tot m ( m/) (c) The argument of part (a) is valid for any masses. The kinetic energy is p p p m + m p K = + =. tot m m = mm ( mm /( m + m )).88 T = π M/ k s m. (a).97 m (b) take a slender rod of length 0.50 m and pivot it about an ais that is 0.5 cm above its center. (a) one spring stretches 0.50 m and the other stretches m, and so the equilibrium lengths are 0.50 m and 0.50 m. (b) 0.70 s. (a) T T L g g, g g T T Δ +Δ Δ = g π so ( )( ) Δ T = T g Δ g. 7

8 .00 (b) The clock runs slow; Δ T > 0, Δ g < 0 and 4.00 ( s) ΔT g +Δ g = g ( 9.80 m s ) = = m s. T ( 86, 400 s) I = ML and d = L in Eq.(.9 ), T = π L g. With the added 0 With ( ) mass, (( ) ) ( ) I = M L + y, m = M and d = L 4 + y. ( ) ( ( )) T L + y T = π L + y g L + y and r = =. The graph of the ratio r T L + yl 0 versus y is given in Figure.00. Figure.00 (b) From the epression found in part (a), T = T when y = L. At this point, a simple 0 pendulum with length y would have the same period as the meter stick without the added mass; the two bodies oscillate with the same period and do not affect the other s motion..0 kl0 (a) l =. k mω (b) The spring will tend to become unboundedly long. 8

CHAPTER 12 OSCILLATORY MOTION

CHAPTER 12 OSCILLATORY MOTION CHAPTER 1 OSCILLATORY MOTION Before starting the discussion of the chapter s concepts it is worth to define some terms we will use frequently in this chapter: 1. The period of the motion, T, is the time