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Unformatted text preview: 14.2. Model: The airtrack 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.
 Fall '09
 geller
 Simple Harmonic Motion

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