ch3sol-HW

# ch3sol-HW - 3 2 Derive Equation 3.3 x(t = Ce!n t cos d t...

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3 2. Derive Equation 3.3: x ( t ) = Ce n t cos( ! d t ) : Solution: This equation is derived by rewriting the general solution for the case ± < 1 (since this is an oscillatory solution). We write Equation 3.2 as x ( t ) = e n t A 1 e i p 1 2 ! n t + A 2 e i p 1 2 ! n t ± ; (3.1) where A 1 and A 2 must be complex conjugates to render x ( t ) e = cos ² + i sin ²; and letting A 1 ; 2 = ( a ± ib ) = 2 ; we obtain x ( t ) = e n t ( a cos ! d t + b sin ! d t ) : (3.2) We write this equation in the amplitude-phase form as x ( t ) = e n t C cos ( ! d t ) ; where the cosine of the argument is expanded C cos ( ! d t ) = C cos ! d t sin + C sin ! d t cos (3.3) Equating the equivalent expressions in Equations 3.2 and 3.3, we ±nd C cos = a C sin = b: The amplitude and the phase are then given by C = p a 2 + b 2 and = tan 1 b a : Care should be taken to ±nd the phase angle in the correct quadrant. For instance, if a < 0 and b > 0 ; the phase angle is in the second quadrant.

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4 CHAPTER 3 SDOF VIBRATION: WITH DAMPING 3. (a) The response of an oscillator with mass m = 1 kg is shown in Figure 3.40. What can you say about the properties of the oscillator and the response? Determine the equation of motion and the initial conditions. (b) Now one property is changed, with all others remaining the same. The resulting time history now looks like that in Figure 3.45. Again, what can you say about the properties of the oscillator and the response? What are the equation of motion and the initial conditions? Figure 3.40: Free vibration for Problem 3(a). Figure 3.45: Free vibration for Problem 3(b). Solution: (a) The oscillator is undamped and vibrates with an amplitude of 1 m with a period of a little more than 6 seconds. It takes about 18.4 s to complete three cycles. Therefore, the natural frequency is ! n = 2 2 (18 : 4 = 3) = 1 : 02 rad/s : Since there is no decay, the equation of motion is x + ! 2 n x = 0 : The initial displacement is zero. The initial velocity is found by approximating the slope at t = 0 s as 0 : 4 = 0 : 5 = 0 : 8 m/s : The free response is given by x ( t ) = x 0 cos ! n t + v 0 ! n sin ! n t: With x 0 = 0 ; we see that v 0 =! n is the amplitude of vibration. Therefore, the velocity is given by v ( t ) = _ x ( t ) = v 0 cos ! n t:
5 (b) Here, the one additional property is damping. We see the e/ect of damping in free vibration is the reduced peak amplitude of vibration with each cycle. The initial conditions are the same as above. The equation governing free vibration is x + 2 n _ x + ! 2 n x = 0 : From the equation of the logarithmic decrement, we can estimate and therefore c: We know that ± = ln x n x n +1 ± = 2 p 1 2 : From the graph we can estimate ± ln(0 : 98 = 0 : 89) = 0 : 096 amplitudes. Therefore we can solve for = s ± 2 4 ² 2 + ± 2 = 0 : 015

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## This note was uploaded on 02/26/2012 for the course 650 443 taught by Professor Benaroya during the Fall '11 term at Rutgers.

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ch3sol-HW - 3 2 Derive Equation 3.3 x(t = Ce!n t cos d t...

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