HW10_Sol_F10

HW10_Sol_F10 - Nov 1 2010 PHYS 2101 HW#10 WileyPlus Problem...

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Unformatted text preview: Nov. 1, 2010 PHYS 2101 HW#10 WileyPlus Problem Solutions 1 HW #10 Solutions – or at least most of them… (USING BOOK VALUES 9 th Edition in problems; problem # indicated by ) 1) [15-CQ3] Given that acceleration a(t) of a particle undergoing SHM is graphed as follows, we know that the position, velocity and acceleration are given by : x ( t ) = x max cos ω t + ϕ ( ) , v ( t ) = dx dt = − x max ω sin ω t + ϕ ( ) , and a ( t ) = dv dt = − x max ω 2 cos ω t + ϕ ( ) . Thus the velocity vs time is phase shifted by + π 2 from the acceleration and the position vs time is phase shifted by + π (completely out of phase) from the acceleration. Thus (a) the point that corresponds to the particle at − x m is 2 ; (b) at point 4 the velocity of the particle is POSITIVE ; and at point 5 , the particle is at a position between 0 and +x m . 2) [15-5] (a) The amplitude is half the range of the displacement, or x m = 1.0 mm. (b) The maximum speed v m is related to the amplitude x m by v m = ω x m , where ω is the angular frequency. Since ω = 2 π f , where f is the frequency, (c) The maximum acceleration is 3) [15-9] (a) Making sure our calculator is in radians mode, we find x = 6.0cos 3 π 2.0 ( ) + π 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 3.0 m. (b) Differentiating with respect to time and evaluating at t = 2.0 s, we find v = dx dt = − 3 π 6.0 ( ) sin 3 π 2.0 ( ) + π 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 49 m/s. (c) Differentiating again, we obtain a = dv dt = − 3 π ( ) 2 6.0 ( ) cos 3 π 2.0 ( ) + π 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 2.7 × 10 2 m/s 2 . (d) In the second paragraph after Eq. 15-3, the textbook defines the phase of the motion. In this case (with t = 2.0 s) the phase is 3 π (2.0) + π /3 ≈ 20 rad. (e) Comparing with Eq. 15-3, we see that ω = 3 π rad/s. Therefore, f = ω /2 π = 1.5 Hz....
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This note was uploaded on 01/26/2011 for the course PHYS 2101 taught by Professor Grouptest during the Spring '07 term at LSU.

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HW10_Sol_F10 - Nov 1 2010 PHYS 2101 HW#10 WileyPlus Problem...

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