# SolSec1.1 - Problems and Solutions Section 1.1(1.1 through...

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Problems and Solutions Section 1.1 (1.1 through 1.17) 1.1 The spring of Figure 1.2 is successively loaded with mass and the corresponding (static) displacement is recorded below. Plot the data and calculate the spring's stiffness. Note that the data contain some error. Also calculate the standard deviation. m( kg ) 10 11 12 13 14 15 16 x( m ) 1.14 1.25 1.37 1.48 1.59 1.71 1.82 Solution: Free-body diagram: m k kx mg Plot of mass in kg versus displacement in m Computation of slope from mg / x m (kg) x (m) k (N/m) 10 1.14 86.05 11 1.25 86.33 12 1.37 85.93 13 1.48 86.17 14 1.59 86.38 15 1.71 86.05 16 1.82 86.24 0 1 2 10 15 20 m x From the free-body diagram and static equilibrium: kx mg g m s km g x == = (. / ) / 981 2 µ Σ k n i 86 164 . Standard deviation in computed stiffness: σ = = = () k n i i n 2 1 1 0.164

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1.2 Derive the solution of mx kx ˙˙ += 0 and plot the result for at least two periods for the case with ω n = 2 rad/s, x 0 = 1 mm, and v 0 = 5 mm/s. Solution: Given: 0 = + kx x m & & (1) Assume: xt ae rt () = : Then: rt are x = & and rt e ar x 2 = & & Substitute into equation (1): mar e kae mr k r k m i rt rt 2 2 0 0 Thus there are two solutions: xc e e k m k m it k m n 11 2 2 2 == , and where rad/s ω The sum of x 1 and x 2 is also a solution so that the total solution is: it it e c e c x x x 2 2 2 1 2 1 + = + = Substitute initial conditions: x 0 = 1 mm, v 0 = 5 mm/s c x c c v x ic ic v cc i i ci 01 1 00 2 2 5 22 5 2 5 1 2 5 4 1 2 5 4 12 0 2 1 0 1 2 =+= =⇒=− () = () =−= = −+ = = =− =+ ˙ Therefore the solution is:
xi e i e t i t i t i t it it =− ++ + () =+ = + 1 2 5 4 1 2 5 4 1 2 5 4 22 1 2 5 4 Using the Euler formula to evaluate the exponential terms yields: cos sin cos sin x cos2t 5 2 sin2t 3 2 sin 2t 0.7297 Using Mathcad the plot is: x t cos . 2t . 5 2 sin . 0 5 10 2 2 x t t

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1.3 Solve mx kx ˙˙ += 0 for k = 4 N/m, m = 1 kg, x 0 = 1 mm, and v 0 = 0. Plot the solution. Solution: This is identical to problem 2, except v k m n 0 02 == = . ω rad/s . Calculating the initial conditions: xc c x c c v x ic ic v c c cc xt e e t i t t it it 01 1 00 2 2 0 05 1 2 1 2 1 2 22 1 2 12 0 2 1 0 2 1 21 () =+= =⇒=− () = () =−= = = =+ = + +− ˙ . cos sin cos sin x ( t )= cos (2 t ) The following plot is from Mathcad: Alternately students may use equation (1.10) directly to get t tt sin( tan [ ]) sin( ) cos = + + = 0 2 2 0 2 2 22 2 1 π x t cos .
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SolSec1.1 - Problems and Solutions Section 1.1(1.1 through...

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