132_ch14 - 14.4. Model: The air-track glider attached to a...

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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 ω 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 = . (a) v max = A and = 2 π T = 2 2.0 s = A = v max = 0.40 m s = 0.127 m = 12.7 cm (b) The glider’s position at t = 0.25 s is x 0.25 s = 0.127 m ( ) cos ( ) 0.25 s ( ) [ ] = 0.090 m = 9.0 cm 14.5. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure Ex14.5. Solve: (a) The amplitude A = 10 cm. (b) The time to complete one cycle is the period, hence T = 2.0 s and f = 1 T = 1 2.0 s = 0.50 Hz (c) The position of an object undergoing simple harmonic motion is x t ( ) = A cos t + φ 0 ( ) . At t = 0 s, x 0 = 5 cm, thus 5 cm = 10 cm ( ) cos 0 s ( ) + 0 [ ] cos 0 = 5 cm 10 cm = 1 2 0 = cos 1 1 2 = 3 rad or 60 ° 14.9. Solve: The position of the object is given by the equation x t ( ) = A cos t + 0 ( ) = A cos 2 ft + 0 ( ) We can find the phase constant 0 from the initial condition: 0 cm = 4.0 cm ( ) cos 0 cos 0 = 0 0 = cos 1 0 ( ) = ± 1 2 rad Since the object is moving to the right, the object is in the lower half of the circular motion diagram. Hence, 0 = 1 2 rad. The final result, with f = 4.0 Hz, is x t ( ) = 4.0 cm ( ) cos 8.0 ( ) t 1 2 rad [ ] 14.13. Model: The air-track glider attached to a spring is in simple harmonic motion. Solve: Experimentally, the period is T = 12.0 s ( ) 10 oscillations ( ) = 1.20 s. Using the formula for the period, T = 2 m k k = 2 T 2 m = 2 1.20 s 2 0.200 kg ( ) = 14.15.
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132_ch14 - 14.4. Model: The air-track glider attached to a...

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