14_P02InstructorSolution

# 14_P02InstructorSolution - 14.2. Model: The air-track...

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Unformatted text preview: 14.2. Model: The air-track glider oscillating on a spring is in simple harmonic motion. Solve: The glider completes 10 oscillations in 33 s, and it oscillates between the 10 cm mark and the 60 cm mark. 33 s (a) T= = 3.3 s oscillation = 3.3 s 10 oscillations 1 1 (b) f= = = 0.303 Hz ≈ 0.30 Hz T 3.3 s ω = 2π f = 2π ( 0.303 Hz ) = 1.90 rad s (c) (d) The oscillation from one side to the other is equal to 60 cm − 10 cm = 50 cm = 0.50 m. Thus, the amplitude is A = 1 ( 0.50 m ) = 0.25 m. 2 (e) The maximum speed is ⎛ 2π ⎞ vmax = ω A = ⎜ ⎟ A = (1.90 rad s )( 0.25 m ) = 0.48 m s ⎝T ⎠ ...
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## This note was uploaded on 12/10/2011 for the course PHYS 1211 taught by Professor Geller during the Fall '09 term at University of Georgia Athens.

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