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

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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:
Background image of page 2
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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 .
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 15

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online