421_PartUniversity Physics Solution

# 421_PartUniversity Physics Solution - Periodic Motion 13.54...

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Periodic Motion 13-13 13.54. IDENTIFY: The ornament is a physical pendulum: 2/ TI m g d π = (Eq.13.39). T is the target variable. SET UP: 2 5/ 3 , IM R = the moment of inertia about an axis at the edge of the sphere. d is the distance from the axis to the center of gravity, which is at the center of the sphere, so . dR = EXECUTE: 2 2 5/3 / 2 5/3 0.050 m/(9.80m/s ) 0.58 s. TR g ππ == = EVALUATE: A simple pendulum of length 0.050 m R = has period 0.45 s; the period of the physical pendulum is longer. 13.55. IDENTIFY: Pendulum A can be treated as a simple pendulum. Pendulum B is a physical pendulum. Use the parallel- axis theorem to find the moment of inertia of the ball in B for an axis at the top of the string. SET UP: For pendulum B the center of gravity is at the center of the ball, so dL = . For a solid sphere with an axis through its center, 2 2 cm 5 R = . /2 R L = and 2 1 cm 10 L = . EXECUTE: Pendulum A : 2 A L T g = . Pendulum B : The parallel-axis theorem says 22 11 cm 10 I I ML ML =+ = . 2 11 11 11 2 1 . 0 5 10 10 10 AA L L TT T mgd MgL g ⎛⎞ = = = ⎜⎟ ⎝⎠ . It takes pendulum B longer to complete a swing. EVALUATE: The center of the ball is the same distance from the top of the string for both pendulums, but the mass is distributed differently and I is larger for pendulum B , even though the masses are the same. 13.56. IDENTIFY: If the system is critically damped or overdamped it doesn&t oscillate. With no damping, / mk ω = . With underdamping, the angular frequency has the smaller value 2 2 4 kb mm =− . SET UP: 2.20 kg m = , 250.0 N/m k = . 2 T ′ = and 10.22 rad/s 0.615 s T = . EXECUTE: (a) 250.0 N/m 10.66 rad/s 2.20 kg k m = . ′ < so the system is damped. 2 2 4 gives 250.0 N/m 2 2(2.20 kg) (10.22 rad/s) 13.3 kg/s 2.20 kg k bm m = = . (b) Since the motion has a period the system oscillates and is underdamped. EVALUATE: The critical value of the damping constant is 2 2 (250.0 N/m)(2.20 kg) 46.9 kg/s bk m = . In this problem b is much less than its critical value. 13.57. IDENTIFY and SET UP: Use Eq.(13.43) to calculate , and then /2 . f = (a) EXECUTE: 2 2.50 N/m (0.900 kg/s) (/ ) ( /4 ) 2 .47 r ad / s 0.300 kg 4(0.300 kg) km b m 2 = = (2.47 rad/s)/2 0.393 Hz f ′′ = (b) IDENTIFY and SET UP: The condition for critical damping is 2 m = (Eq.13.44) EXECUTE: 2 (2.50 N/m)(0.300 kg) 1.73 kg/s b EVALUATE: The value of b in part (a) is less than the critical damping value found in part (b). With no damping, the frequency is 0.459 Hz; f = the damping reduces the oscillation frequency. 13.58. IDENTIFY: From Eq.(13.42) ( ) 21 exp 2 b t m . SET UP: ln( ) x ex EXECUTE: 1 2 2 2(0.050 kg) 0.300 m ln ln 0.0220 kg/s. (5.00 s) 0.100 m mA b tA = EVALUATE: As a check, note that the oscillation frequency is the same as the undamped frequency to 3 4.8 10 %, so Eq. (13.42) is valid. × 13.59. IDENTIFY: () x t is given by Eq.(13.42). / x vd x d t = and / xx v d t = . SET UP: (cos )/ sin dt d t t ωω . (sin )/ cos d t t = . ( )/ tt de d t e α . EXECUTE: (a) With 0 φ = , (0) x A = .

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13-14 Chapter 13 (b) (2) cos sin , 2 bm t x dx b vA e ω t ωω t dt m ⎡⎤ ′′ == ⎢⎥ ⎣⎦ and at 0, /2 ; x tvA b m the graph of x versus t near 0 t = slopes down.
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421_PartUniversity Physics Solution - Periodic Motion 13.54...

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