Section 5.1-2 Mass Spring Systems

1 Asst. Prof. HottovySM212- Section Mass Spring SystemsName:Purpose:To investigate the mass Spring Systems in Chapter :Work on the following activity with 2-3 other students during class (but be sureto complete your own copy) and finish the exploration outside of in 2/07 :In this worksheet we will be exploring the Spring /mass system modeledby homogeneous, linear, second order differential equations with constant coefficients of theform,md2xdt2+ dxdt+kx= Review:(a) Find the general solution to the differential + 2x= 0.(b) If the units ofx areLengthTime2, what must be the units of in the DE above?(c) If sin(t) has a period of 2 , then what must be the period of sin( t)?

2 12. Spring mass Systems withfree motion: Consider a massmattached to a Spring inthe following pictureNote that the position (x) is positive when the Spring isbelowthe equilibriumposition. This means that velocity is positive when the mass is moving downward(falling in the direction of gravity).In the first picture, nothing is happening. In the second picture, the mass hangsstilland is unmoving. In the third picture, the mass is stretched and is currently moving(or about to be released and begin its motion).The equation of motion comes from Newton s second law. Mass times acceleration isequal to the sum of the forces. The forces acting on the mass are gravity (mg) and therestoration of the Spring (k(s+x)).

3 The restoration force comes from Hook s law. Thesis the distance the mass hangs from the natural length of the Spring . This is calledtheequilibrium position(middle picture above), andxis the distance of the massfrom equilibrium. We say thatx=at only these two forces acting on the mass, we can write the differential equationmx = k(s+x) +mg= kx+mg ks zero= kx.(a) Discuss whymg ks= 0 from equationmg ks= 0 is used to calculate the Spring constantk. To do soyou must be given the weight of the mass (Example: 2lbs = mg (remember lbsare a mass times gravity)) and the distance the Spring stretches under the weightof the mass.(b) Calculate the Spring constantkof the following Spring mass A mass weighing 4 pounds, attached to the end of a Spring , stretches it 3inches.

4 Calculate the Spring constantk. What are the units?Solution: We use the equationmg ks= 0, ormg=ks. Here the weightof the mass is given asmg=lbsFor thekspart of the equation, we must be careful. In the text they state that because we are using the engineering system of units, the measurements mustbe converted into feet. So blame the engineers here, not the 3 inches =ft,andks= solving forkgives,mg=ks= 4 =k4= k=.ii. A mass weight 24 pounds, attached to the end of a Spring , stretches it 4inches. Calculate the Spring constant. Watch the units!Solution:iii. A force of 400 Newtons stretches a Spring 2 meters. Find the Spring : Here the weight of the mass is replaced by 400 Newtons.

5 Somg= 400. Thus solving forkgives,3(c) A mass weighing 2 pounds stretches a Spring 6 inches. Att= 0 the mass isreleased from a point 8 inches below the equilibrium position with an upwardvelocity of43ft/s . Determine the equation of Write the : The Spring mass equation for free motion ismx = solve forkusing the same strategy above,k=.To write the mass of the Spring we convert weightWto mass using,m=Wg=2 lbs32ft/s2= initial conditions are given by the sentence Att= 0the mass is releasedfrom a point 8 inches below the equilibrium position with an upward velocityof43ft/s. From this we get that (WATCH THE SIGN! Refer to the pictureon page 2)x(0) =ftand the initial velocity isx (0) = 43ft/ssince the mass is traveling upward (refer to the picture on page 2).

6 So theIVP is116x = 4x, x(0) = 2/3, x (0) = 43ii. Solve this Now try another on your own: A mass weighing 8 pounds, attached to the end of aspring, stretches it 8 ft. Initially, the mass is released from a point 6 inches belowthe equilibrium position with a downward velocity of 3/2 ft/s. Find the equation You should have a final answer ofx(t) =12cos(2t) +34sin(2t).(1)It is difficult to answer the very physical question: What is the maximum distancefrom equilibrium that the mass obtains? To do so we writex(t) in a different form forx(t): For a general solutionx(t) =c1cos( t) +c2sin( t)wherec1,c2are not zero,x(t) can be written asx(t) =Asin( t+ )whereA= c21+c2,tan = called theamplitudeand thephase the extra motivatedstudent:To see that this formula works, use the sine angle addition formulasin( + ) = sin( ) cos( ) + cos( ) sin( )and the picture below to rewrite the new form into the general solution in equation(1).

7 (a) Find the amplitude and phase shift of the general solution in equation (1).65. Now consider the first example we did. The general solution wasx(t) =23cos(8t) 16sin(8t).in the new form it isx(t) =sin(8t+)6. Once we have this new form, the motion of the Spring is much easier to visualize. NotetheNote that the sine wave flips when plotted below because the position is positivebelow the equilibrium : Theperiodis the time it takes (in seconds) to execute a full cycle. For theexample above it isT=2 =2 8= 4seconds. Thefrequencyis the number of cycles that the mass makes in 1 second. For theexample above it isf=1T= 2 =cycles per second (inx = 2x) is called thecircular frequency.

8 For a mass Spring system , 2=kmThe circular frequency for the example above is, = km=cycles per second77. Solve the following problems.(a) A mass weighing 4 pounds is attached to a Spring whose Spring constant is 16lb/ft. What is the period of simple harmonic motion?(b) A 20-kilogram mass is attached to a Spring . If the frequency of simple harmonicmotion is 2/ cycles/s, what is the Spring constantk? What is the frequencyof simple harmonic motion if the original mass is replaced with an 80-kilogrammass? damped motion:Now we consider a mass on a Spring in which there is friction force slows the motion. The force is considered to be proportional theinstantaneous velocity.

9 So the differential equation for the mass Spring is nowmd2xdt2= kx dxdt friction rewrite this equation asd2xdt2+dxdt+x= make the analysis easier we rewrite this with new variablesx + 2 x + 2x= 0where 2 = /m,and 2=k/m.(a) Write the polynomial inmafter we substitutex=emtand simplifying. Solve thispolynomial for general and . (Please suppress the audible groan at having touse the quadratic equation.)(b) We saw from last Section that the type of general solution that results dependson how many real roots come from the above polynomial. So we must considerthe cases of 2 2. Why?99. Three cases:We consider the three cases for 2 2and investigate the type of motion.

10 (a) 2 2>0. For this case, the polynomialm2+ 2 m+ 2= 0 hasrealroots. The roots arem1=, m2=.Thus the general solution is:(b) Which of the three graphs below graphs this case? Describe in words case is calledover damped. It is because the mass gets damped to equilib-rium very quickly and does not have a chance to oscillate at its circular frequency(what it would do without friction).(c) 2 2<0. For this case, the polynomialm2+ 2 m+ 2= 0 hasrealroots. The roots arem1=, m2=.Thus the general solution is:10(d) Which of the three graphs is this case? Describe in words case is calledunder damped. It is called this because the friction doesnot damp enough to prevent oscillation.