<ul><li> 1. SOLUTION MANUAL FOR</li></ul><p> 2. Problems and Solutions Section 1.1 (1.1 through 1.19)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. Notethat the data contain some error. Also calculate the standard deviation.m(kg) 10 11 12 13 14 15 16x(m) 1.14 1.25 1.37 1.48 1.59 1.71 1.82Solution:Free-body diagram:mkkxmg0 1 22015Plot of mass in kg versus displacement in mComputation of slope from mg/xm(kg) x(m) k(N/m)10 1.14 86.0511 1.25 86.3312 1.37 85.9313 1.48 86.1714 1.59 86.3815 1.71 86.0516 1.82 86.2410mxFrom the free-body diagram and staticequilibrium:kx = mg (g = 9.81m/ s2)k = mg/ x =!kin= 86.164The sample standard deviation incomputed stiffness is:! =(ki' ) 2n#i=1n '1= 0.164 3. 1.2 Derive the solution of mx + kx = 0 and plot the result for at least two periods for the casewith n = 2 rad/s, x0 = 1 mm, and v0 = 5 mm/s.Solution:Given:m!x!+ kx = 0 (1)Assume: x(t) = aert . Then: rt x! = are and rt x ar e2 !! = . Substitute into equation (1) toget:mar2ert + kaert = 0mr2 + k = 0r = kmiThus there are two solutions:x1= c1ekmi!' #$% &amp;t, and x2= c2e' kmi!' #$% &amp;twhere (n=km= 2 rad/sThe sum of x1 and x2 is also a solution so that the total solution is:2! = + = +it it x x x c e c e221 2 1Substitute initial conditions: x0 = 1 mm, v0 = 5 mm/sx(0) = c1+ c2= x0= 1!c2= 1' c1, and v(0) = x! (0) = 2ic1' 2ic2= v0= 5 mm/s!'2c1+ 2c2= 5 i. Combining the two underlined expressions (2 eqs in 2 unkowns):'2c1+ 2 ' 2c1= 5 i!c1=12'54i, and c2=12+54iTherefore the solution is:x =12!54i'# $%&amp; 'e2it +12+54i'%'&amp; $# e!2itUsing the Euler formula to evaluate the exponential terms yields:x =12!54i'# $%&amp; '(cos2t + i sin2t ) +12+54i'# $%&amp; '(cos2t ! i sin 2t )( x(t ) = cos2t +52sin2t =32sin(2t + 0.7297) 4. Using Mathcad the plot is:5x t cos 2. t .2sin 2. t0 5 1022x tt 5. 1.3 Solve mx + kx = 0 for k = 4 N/m, m = 1 kg, x0 = 1 mm, and v0 = 0. Plot the solution.Solution:This is identical to problem 2, except v0= 0. !n=km= 2 rad/s'# $%&amp; '. Calculating theinitial conditions:x(0) = c1+ c2= x0= 1! c2=1 ' c1v(0) = x (0) = 2ic1' 2ic2= v0= 0! c2= c1c2= c1= 0.5x(t) =12e2it +12e'2it =12(cos2t + isin 2t) +12(cos2t ' i sin2t)x(t)= cos (2t )The following plot is from Mathcad:1Alternately students may use equation (1.10) directly to getx(t ) =22 (1)2 + 022sin(2t + tan!1[2 '10])= 1sin(2t +#2) = cos2tx t cos 2. t0 5 101x tt 6. 1.4 The amplitude of vibration of an undamped system is measured to be 1 mm. The phaseshift from t = 0 is measured to be 2 rad and the frequency is found to be 5 rad/s.Calculate the initial conditions that caused this vibration to occur. Assume the responseis of the form x(t ) = Asin(!nt +').Solution:Given: A = 1mm, ' = 2 rad, ! = 5rad/s . For an undamped system:x(t ) = Asin !n( t +' ) = 1sin(5t + 2) andv(t ) = x! (t ) = A!ncos !n( t +' ) = 5 cos(5t + 2)Setting t = 0 in these expressions yields:x(0) = 1sin(2) = 0.9093 mmv(0) = 5 cos(2) = - 2.081 mm/s1.5 Find the equation of motion for the hanging spring-mass system of Figure P1.5, andcompute the natural frequency. In particular, using static equilibrium along withNewtons law, determine what effect gravity has on the equation of motion and thesystems natural frequency.Figure P1.5Solution:The free-body diagram of problem system in (a) for the static case and in (b) for thedynamic case, where x is now measured from the static equilibrium position.(a) (b)From a force balance in the static case (a): mg = kxs, where xs is the static deflection ofthe spring. Next let the spring experience a dynamic deflection x(t) governed bysumming the forces in (b) to get 7. m!x!(t ) = mg ! k(x(t ) + xs)'m!x!(t ) + kx(t ) = mg ! kxs'm!x!(t ) + kx(t ) = 0'#n=kmsince mg = kxsfrom static equilibrium.1.6 Find the equation of motion for the system of Figure P1.6, and find the natural frequency.In particular, using static equilibrium along with Newtons law, determine what effectgravity has on the equation of motion and the systems natural frequency. Assume theblock slides without friction.Figure P1.6Solution:Choosing a coordinate system along the plane with positive down the plane, the free-bodydiagram of the system for the static case is given and (a) and for the dynamic casein (b):In the figures, N is the normal force and the components of gravity are determined by theangle as indicated. From the static equilibrium: !kx+ mgsin' = 0 . Summing forcessin (b) yields: 8. !Fi = m!x!(t )'m!x!(t ) = #k(x + xs) + mgsin$'m!x!(t ) + kx = #kxs+ mgsin$ = 0'm!x!(t ) + kx = 0'%n=kmrad/s1.7 An undamped system vibrates with a frequency of 10 Hz and amplitude 1 mm. Calculatethe maximum amplitude of the system's velocity and acceleration.Solution:Given: First convert Hertz to rad/s: !n= 2'fn= 2'(10) = 20' rad/s. We also have thatA= 1 mm.For an undamped system:x(t)= A sin(' t +! ) nand differentiating yields the velocity: v(t) = A!ncos !n ( t + ') . Realizing that both thesin and cos functions have maximum values of 1 yields:= ' = 1(20! )= 62.8mm/smax n v ALikewise for the acceleration: a(t)= #' 2Asin (' t +! ) n n2= 2 = ( ) = 3948mm/s 2max' 1 20!n a A 9. 1.8 Show by calculation that A sin (nt + ) can be represented as Bsinnt + Ccosnt andcalculate C and B in terms of A and .Solution:This trig identity is useful: sin(a + b) = sin acosb + cosasinbGiven: Asin(' t + ! ) = Asin(' t)cos(! ) +Acos(' t)sin(! ) n n n= Bsin!nt + Ccos!ntwhere B = Acos' and C = Asin'1.9 Using the solution of equation (1.2) in the form x(t) = Bsin!nt + Ccos!ntcalculate the values of B and C in terms of the initial conditions x0 and v0.Solution:Using the solution of equation (1.2) in the formx(t) = Bsin!nt + Ccos!ntand differentiate to get:x (t) = !nBcos(!nt) ' !nCsin(!nt)Now substitute the initial conditions into these expressions for the position and velocityto get:x0= x(0) = Bsin(0) + Ccos(0) = Cv0= x (0) = !nBcos(0) ' !nCsin(0)nB(1) '!nC(0) =!nB= !Solving for B and C yields:B =v0!n, and C = x0Thus x(t) =v0!nsin!nt + x0cos!nt 10. 1.10 Using Figure 1.6, verify that equation (1.10) satisfies the initial velocity condition.Solution: Following the lead given in Example 1.1.2, write down the general expressionof the velocity by differentiating equation (1.10):x(t) = Asin(!nt + ')# x (t) = A!ncos(!nt + ')# v(0) = A!ncos(!n0 + ') = A!ncos(')From the figure:Figure 1.62 +A = x0v0!n'# $%&amp; '2, cos( =v0!n2 +x0v0!n'# $%&amp; '2Substitution of these values into the expression for v(0) yieldsv(0) = A!ncos' = x02 +v0!n#$ %' ( 2&amp;(!n)v0!n2 +x0v0!n#$ %&amp;' (2= v0verifying the agreement between the figure and the initial velocity condition. 11. 1.11 (a)A 0.5 kg mass is attached to a linear spring of stiffness 0.1 N/m. Determine the naturalfrequency of the system in hertz. b) Repeat this calculation for a mass of 50 kg and astiffness of 10 N/m. Compare your result to that of part a.Solution: From the definition of frequency and equation (1.12)(a) !n=km=.5.1= 0.447 rad/sfn=!n2'=2.2362'= 0.071 Hz(b) !n=5010= 0.447rad/s, fn=!n2'= 0.071 HzPart (b) is the same as part (a) thus very different systems can have same naturalfrequencies. 12. 1.12 Derive the solution of the single degree of freedom system of Figure 1.4 by writingNewtons law, ma = -kx, in differential form using adx = vdv and integrating twice.Solution: Substitute a = vdv/dx into the equation of motion ma = -kx, to get mvdv = -kxdx. Integrating yields:v222 x2= !'n2+ c2 , where c is a constantor v2 = !'n2x2 + c2 #v =dxdt= !'n2x2 + c2 #dt =dx!'n2x2 + c2, write u ='nx to get:t ! 0 =1'nduc2 ! u2 $ =1'nsin!1 uc%&amp; '() *+ c2Here c2 is a second constant of integration that is convenient to write as c2 = -/n.Rearranging yields!nt +' = sin#1 !nxc$% &amp;'( )*!nxc= sin(!nt +' )*x(t ) = Asin(!nt +' ), A =c!nin agreement with equation (1.19). 13. 1.13 Determine the natural frequency of the two systems illustrated.(a) (b)Figure P1.13Solution:(a) Summing forces from the free-body diagram in the x direction yields:-k1x+x-k2xFree-body diagram for part amx = !k1 x ! k2 x'mx + k1x + k2 x = 0mx + x k1+ k2 ( ) = 0, dividing by m yields :x +k1+ k2m#$%&amp;x = 0Examining the coefficient of xyields:!n=k1+ k2m(b) Summing forces from the free-body diagram in the x direction yields:-k1x-k2x+x-k3xFree-body diagram for part bm!x! = !k1x ! k2x ! k3x,'m!x!+ k1x + k2x + k3x = 0'm!x!+ (k1+ k2+ k3)x = 0' !x!+(k1+ k2+ k3)mx = 0'#n=k1+ k2+ k3m 14. 1.14* Plot the solution given by equation (1.10) for the case k = 1000 N/m and m = 10 kg fortwo complete periods for each of the following sets of initial conditions: a) x0 = 0 m, v0 =1 m/s, b) x0 = 0.01 m, v0 = 0 m/s, and c) x0 = 0.01 m, v0 = 1 m/s.Solution: Here we use Mathcad:a) all units in m, kg, sm 10k 1000 !x0 0.0T2. !!nfn 'n2. 'x t A. sin . !n t 'parts b and c are plotted in the above by simply changing the initial conditions asappropriate1A .!n2x0 .0.22!nv02! atan. 'n x0v00 0.5 1 1.50.10.10.2x txb txc ttn:=kmv0 1 15. 1.15* Make a three dimensional surface plot of the amplitude A of an undamped oscillatorgiven by equation (1.9) versus x0 and v0 for the range of initial conditions given by 0.1 vo0.001=0.010.001= 10So!d=km1'# ( 2 ) &gt;10$km1' # 2 ( ) &gt;100,$ k = m1001 '# 2(1) Choose ! = 0.01'km&gt; 100.01(2) Choose m = 1 kg ! k &gt; 100.01(3) Choose k = 144 N/m &gt;100.01!'n= 144rads=12rads!'d= 11.99rads! c = 2m#'n= 0.24kgs 82. 1.84 Repeat problem 1.83 if the mass is restricted to lie between 10 kg &lt; m &lt; 15 kg.Solution: Referring to the above problem, the relationship between m and k isk &gt;1.01x10-4 mafter converting to meters from mm. Choose m =10 kg and repeat the calculationat the end of Problem 1.82 to get n (again taking = 0.01). Then k = 1000 N/mand:!'n=31.0 # 1010rads=10rads!'d= 9.998rads! c = 2m$'n= 2.000kgs 83. 1.85 Use the formula for the torsional stiffness of a shaft from Table 1.1 to design a 1-m shaft with torsional stiffness of 105 Nm/rad.Solution: Referring to equation (1.64) the torsional stiffness iskt=GJp!Assuming a solid shaft, the value of the shaft polar moment is given byJp=!d 432Substituting this last expression into the stiffness yields:kt=G!d432!Solving for the diameter d yieldsd =kt(32)!G!'#$%14Thus we are left with the design variable of the material modulus (G). Choosesteel, then solve for d. For steel G = 8 1010 N/m2. From the last expression thenumerical answer isd =105 Nmrad(32)(1m)8 ! 1010 Nm2'# $%&amp; '(( ))++++*,..-14= 0.0597 m1.86 Repeat Example 1.7.2 using aluminum. What difference do you note?Solution:For aluminum G = 25 109 N/m2From example 1.7.2, the stiffness is k = 103 =34Gd64nRand d = .01 mSo, 103=25 !109 ( )(.01)4364nRSolving for nR3 yields: nR3 = 3.906 10-3m3Choose R = 10 cm = 0.1 m, so that 84. n =3.906 ! 10'3(0.1)3= 4 turnsThus, aluminum requires 1/3 fewer turns than steel.1.87 Try to design a bar (see Figure 1.21) that has the same stiffness as the spring ofExample 1.7.2. Note that the bar must remain at least 10 times as long as it iswide in order to be modeled by the formula of Figure 1.21.Solution:From Figure 1.21,EAlk =9N/mFor steel, E = 210 ! 102From Example 1.7.2, k = 103 N/mSo, 103=210 !109 ( )Al8 ( )Al = 2.1 !10If A = 0.0001 m2 (1 cm2), then8 ( ) 10l = 2.1 !10'4 ( ) = 21,000 m (21km or 13 miles)Not very practical at all. 85. 1.88 Repeat Problem 1.87 using plastic (E = 1.40 109 N/m2) and rubber (E = 7 106N/m2). Are any of these feasible?Solution:From problem 1.53,EAlk 10 N/m 3 = =9N/mFor plastic, E = 1.40 ! 102So, l = 140mFor rubber, E = 7 !106N/m2So, l = 0.7 mRubber may be feasible, plastic would not.1.89 Consider the diving board of Figure P1.89. For divers, a certain level of staticdeflection is desirable, denoted by . Compute a design formula for the dimensionsof the board (b, h and ! ) in terms of the static deflection, the average divers mass, m,and the modulus of the board.Figure P1.89Solution: From Figure 1.15 (b), !k = mg holds for the static deflection. Theperiod is:T =2!'n= 2!mk= 2!mmg / #= 2!#g(1)From Figure 1.24, we also have thatT =2!'n= 2!m!33EI(2)Equating (1) and (2) and replacing I with the value from the figure yields: 86. 2!m!33EI= 2!12m!33Ebh3= 2!'g#!3bh3='E4mgAlternately just use the static deflection expression and the expression for thestiffness of the beam from Figure 1.24 to get!k = mg '!3EI!3= mg '!3bh3=!E4mg 87. Problems and Solutions Section 1.8 (1.90 through 1.93)1.90 Consider the system of Figure 1.90 and (a) write the equations of motion in termsof the angle, , the bar makes with the vertical. Assume linear deflections of thesprings and linearize the equations of motion. Then (b) discuss the stability of thelinear systems solutions in terms of the physical constants, m, k, and ! . Assumethe mass of the rod acts at the center as indicated in the figure.Figure P1.90Solution: Note that from the geometry, the springs deflect a distancekx = k(!sin!) and the cg moves a distance!2cos! . Thus the total potentialenergy isU = 2 !12k(!sin')2 #mg!2cos'and the total kinetic energy isT =12JO!!2 =12m'23!!2The Lagrange equation (1.64) becomesddt!T!'!#$ %&amp;' (+!U!'=ddtm'23'!#$ %&amp;' (+ 2k'sin' cos' )12mg'sin' = 0Using the linear, small angle approximations sin! ' ! and cos! ' 1 yieldsa)m!23!'+ 2k!2 ' mg!2#$ %&amp;' (! = 0Since the leading coefficient is positive the sign of the coefficient of determinesthe stability.b)if 2k! !mg2&gt; 0'4k &gt;mg!' the system is stableif 4k = mg '#(t) = at + b' the system is unstableif 2k! !mg2&lt; 0'4k mg and c &gt; 0.The result of adding a dashpot is to make the system asymptotically stable. 89. 1.92 Replace the massless rod of the inverted pendulum of Figure 1.37 with a solidobject compound pendulum of Figure 1.20(b). Calculate the equations ofvibration and discuss values of the parameter relations for which the system isstable.Solution:m2m1k k0!m2g2Fsp +Moment about O: ' =!!!M I om1gl2sin! + m2gl sin! ' 2kl2cos! #$sin! l2%&amp;=122 ml+ ml123#$%&amp; !When is small, sin and cos 1.m13+ m2!'#$l2% +kl22&amp; m12gl &amp; m2 gl!' '#$% = 0m13+ m2!'#$l % +kl2&amp; m12+ m2!'#$g() *+% = 0, -For stability,kl2&gt;m12+ m2!'#$g.1.93 A simple model of a control tab for an airplane is sketched in Figure P1.93. Theequation of motion for the tab about the hinge point is written in terms of theangle from the centerline to beJ !!!+ (c ' fd) !!+ k! = 0 .Here J is the moment of inertia of the tab, k is the rotational stiffness of the hinge,c is the rotational damping in the hinge andfd!!is the negative damping provided 90. by the aerodynamic forces (indicated by arrows in the figure). Discuss thestability of the solution in terms of the parameters c and fd.Figure P1.93 A simple model of an airplane control tabSolution: The stability of the system is determined by the coefficient of!! sincethe inertia and stiffness terms are both positive. There are three casesCase 1 c - fd&gt; 0 and the systems solution is of the form !(t) = e'at sin(#nt +$)and the solution is asymptotically stable.Case 2 c - fd&lt; 0 and the systems solution is of the form !(t) = eat sin('nt +#)and the solution is oscillates and grows without bound, a..</p>