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Unformatted text preview: Chapter 9 Vibratory Motions 9.1 Undamped, Free Response for Single Degree of Freedom Sys tems 73 9.1.1 GOAL: Find coefficients for complex exponential representation of a sinusoid. GIVEN: Coefficients of the cosine and sine terms FORMULATE EQUATIONS: We’ll use the formulas cos x = 1 2 ( e ix + e ix ) and sin x = 1 2 i ( e ix e ix ) SOLVE: 2 cos ωt 3 sin ωt = 2 " e iωt + e iωt 2 # 3 " e iωt e iωt 2 i # = 1 + 3 i 2 e iωt + 1 3 i 2 e iωt b 1 = 1 + 3 i 2 cm , b 2 = 1 3 i 2 cm 74 9.1.2 GOAL: Find position, speed and acceleration at a particular time. GIVEN: Position versus time. FORMULATE EQUATIONS: We simply need to differentiate x ( t ) with respect to time. x ( t ) = 2 cos 3 t sin 3 t (1) v ( t ) = 6 sin 3 t 3 cos 3 t (2) a ( t ) = 18 cos 3 t + 9 sin 3 t (3) SOLVE: Evaluating (1), (2), and (3), at t = 1 . 5 s yields x = 0 . 556 m , v = 6 . 50 m/s , a = 5 . 00 m/s 2 75 9.1.3 GOAL: Find the maximum value of ˙ x . GIVEN: Position of x versus t . FORMULATE EQUATIONS: We need to differentiate x ( t ) with respect to time twice. The first gives us ˙ x . In order to find the maximum value of ˙ x we will differentiate again and set the resulting ¨ x equal to zero, which will allow us to solve for both the minima and maxima. x ( t ) = cos 4 t + 2 sin 4 t (1) ˙ x ( t ) = 4 sin 4 t + 8 cos 4 t (2) ¨ x ( t ) = 16 cos 4 t 32 sin 4 t (3) SOLVE: Set ¨ x equal to zero: 16 cos 4 t * 32 sin 4 t * = 0 32 sin 4 t * = 16 cos 4 t * tan 4 t * = 0 . 5 t * = 0 . 116 s CHECK: Evaluate ˙ x ( t * ): x (0 . 116) = 4 sin(0 . 464) + 8 cos(0 . 464) = 8 . 94 m/s The answer is positive and thus we’ve found the correct time for a maximum. 76 9.1.4 GOAL: Plot x , ˙ x , and ¨ x and comment on the effect of sensor noise. GIVEN: Position of x versus t . FORMULATE EQUATIONS: x ( t ) = 2 cos t + 0 . 05 sin 10 t (1) ˙ x ( t ) = 2 sin t + 0 . 5 cos 10 t (2) ¨ x ( t ) = 2 cos t 5 . 0 sin 10 t (3) SOLVE: The plots of x , ˙ x and ¨ x are shown below. The first plot looks like a slightly distorted cosine wave. The second plot ( ˙ x vs t ) has a much larger “wiggle” superimposed on the sine wave. In the third plot the “wiggle” distortion has grown so large that it dominates the response  being the most immediately noticed aspect of the plot. What this shows is the tendency of noise (an unavoidable consequence of imperfect sensors) to grow larger as the signal is differentiated. Because of this, experimentalists try to avoid differentiating experimentally obtained data. 77 9.1.5 GOAL: Determine if two spring arrangements affect the natural frequency of a springmass system. GIVEN: The two different spring arrangements. DRAW: FORMULATE EQUATIONS: For both cases, the FBD=IRD diagram yields m ¨ x = k 1 x k 2 x SOLVE: The equation of motion for both cases is the same, namely m ¨ x + ( k 1 + k 2 ) x = 0 Whether the spring acts on the mass from the left or from the right makes no difference in the governing equation and therefore no difference in the system’s natural frequency as well....
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 Spring '06
 KAM
 Force, Mass, Rotation, k2

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