426_PartUniversity Physics Solution

426_PartUniversity Physics Solution - 13-18 Chapter 13 The...

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13-18 Chapter 13 EVALUATE: The dependence of U on x is not linear and 1 max 2 UU = does not occur at 1 max 2 x x = . Figure 13.72 13.73. IDENTIFY: 2 m T k π = so the period changes because the mass changes. SET UP: 3 2.00 10 kg/s dm dt =− × . The rate of change of the period is dT dt . EXECUTE: (a) When the bucket is half full, 7.00 kg m = . 7.00 kg 21 . 4 9 s 125 N/m T == . (b) 1/2 1 2 22 () dT d dm dm mm dt dt dt dt kk m k ππ = . 34 ( 2.00 10 kg/s) 2.12 10 s per s (7.00 kg)(125 N/m) dT dt −− × = × . dT dt is negative; the period is getting shorter. (c) The shortest period is when all the water has leaked out and 2.00 kg m = . Then 0.795 s T = . EVALUATE: The rate at which the period changes is not constant but instead increases in time, even though the rate at which the water flows out is constant. 13.74. IDENTIFY: Use x Fk x to determine k for the wire. Then 1 2 k f m = . SET UP: Fm g = moves the end of the wire a distance l Δ . EXECUTE: The force constant for this wire is mg k l = Δ , so 2 3 11 1 9 . 8 0 m s 11.1 Hz. 2 2 . 0 0 1 0 m kg f π m π l π = = Δ× EVALUATE: The frequency is independent of the additional distance the ball is pulled downward, so long as that distance is small. 13.75. IDENTIFY and SET UP: Measure x from the equilibrium position of the object, where the gravity and spring forces balance. Let x + be downward. (a) Use conservation of energy (Eq.13.21) to relate x v and x . Use Eq. (13.12) to relate T to k / m . EXECUTE: 2 1 2 x mv kx kA += For 0, x x mv kA and /, vAk m = just as for horizontal SHM. We can use the period to calculate /: 2 / km T mk = implies /2 / . km T = Thus ( ) 2 / 2 0.100 m /4.20 s 0.150 m/s. vA T = (b) IDENTIFY and SET UP: Use Eq.(13.4) to relate x a and x . EXECUTE: so ( / ) xx ma kx a k m x x + - direction is downward, so here 0.050 m x = − 2 (2 / ) ( 0.050 m) (2 /4.20 s) (0.050 m) 0.112 m/s x aT =+ = (positive, so direction is downward) (c) IDENTIFY and SET UP: Use Eq.(13.13) to relate x and t . The time asked for is twice the time it takes to go from 0 x = to 0.050 m. x
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Periodic Motion 13-19 EXECUTE: () cos( ) xt A t ωφ =+ Let /2, so 0 at 0. φ π =− = = Then cos( /2) sin sin(2 / ). x At A t A t T ω πω = −= = Find the time t that gives 0.050 m: x 0.050 m (0.100 m) sin(2 / ) tT = 2 / arcsin(0.50) /6 and /12 4.20 s/12 0.350 s t T == = = = The time asked for in the problem is twice this, 0.700 s. (d) IDENTIFY: The problem is asking for the distance d that the spring stretches when the object hangs at rest from it. Apply Newton’s 2nd law to the object. SET UP: The free-body diagram for the object is given in Figure 13.75. EXECUTE: x x Fm a = 0 mg kd = (/) dm k g = Figure 13.75 But /2 / km T = (part (a)) and m / k 2 (/2) T = 22 2 4.20 s (9.80 m/s ) 4.38 m. T dg ⎛⎞ ⎛ = ⎜⎟ ⎜ ⎝⎠ ⎝ EVALUATE: When the displacement is upward (part (b)), the acceleration is downward. The mass of the partridge is never entered into the calculation. We used just the ratio k / m , that is determined from T . 13.76. IDENTIFY: cos ( ) A t , sin( ) x vA t = −+ and 2 x ax . 2 / T = . SET UP: x A = when 0 t = gives 0 = . EXECUTE: 2 0.240 m cos 1.50 s t x ⎛⎞ = ⎜⎟ ⎝⎠ .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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426_PartUniversity Physics Solution - 13-18 Chapter 13 The...

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