AP B Physics: Springs & Pendulum like to do. Unit 5. (chap.10, sections 1-4)

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1 AP B Physics: Springs & Pendulum like to do. Unit 5 (chap.10, sections 1-4) 1

2 Objectives: Pendulums, Springs, Simple Harmonic Motion 1. Simple harmonic motion (dynamics and energy relationships) a) Sketch or identify a graph of displacement as a function of time, and determine from such a graph the amplitude, period, and frequency of the motion. b) Write displacement in the form Atsinw or Atcosw to describe the motion. d) State the relations between acceleration, velocity, and displacement, and identify locations where these quantities are zero or achieve their greatest positive and negative values. e) State and apply the relation between frequency and period. g) State how the total energy of an oscillating system depends on the amplitude of the motion, sketch a graph of kinetic or potential energy as a function of time for all points of motion h) Calculate the kinetic and potential energies of an oscillating system as functions of time, and prove that the sum of kinetic and potential energy is constant. 2. Mass on a spring b) Apply the expression for the period of oscillation of a mass on a spring. c) Analyze problems in which a mass hangs from a spring and oscillates vertically. d) Analyze problems in which a mass attached to a spring oscillates horizontally. 3. Pendulum and other oscillations Students should be able to apply their knowledge of simple harmonic motion to the case of a pendulum, so they can: b) Apply the expression for the period of a simple pendulum. c) State what approximation must be made in deriving the period. Misconceptions The period of oscillation depends on the amplitude. The restoring force is constant at all points in the oscillation. The heavier a pendulum bob, the shorter its period. All pendulum motion is perfect simple harmonic motion, for any initial angle. A pendulum accelerates through lowest point of its swing. Amplitude of oscillations is measured peak-to-peak. The acceleration is zero at the end points of the motion of a pendulum. 2

3 Cliff notes 1. Oscillating motion describes pendulums, springs, and sound waves. A restoring force pulls back an object so that it repeatedly moves up and down or side to side. 2. SHM or simple harmonic motion is when the restoring force depends on displacement: a. F=-kx where the constant is k for springs or mg/l for pendulums b. the force varies so the acceleration is not constant (use energy not tneoms!) c. at maximum stretch: max position, max acceleration, minimum velocity d. period of oscillations depends only on mass of object or length of pendulum spring: T= 2π (m/k) pendulum: T= 2π (l/g) e. natural frequency of oscillation: ω = 2 π/t = 2π π f (ω= radians/sec) f. pendulums: restoring force = mg sin() = constant*x so SHM if < 15 o 3. All springs: x= stretch, F= -k x, ω= (k/m), Elastic energy= ½ k x 2 spring constant= k (but will be higher for shorter springs- stiffer) Elastic energy= area under F vs. x curve (but F depends on x) 4. Horizontal springs: x measure from equilibrium point before add mass 5. Vertical springs: x measure from equilibrium point after add mass 3

4 What does the period of a simple pendulum depend on? Pendulum Class Data Fnet = T- mg (but note it s not constant! depends on angle) a = Fnet/m = also not constant so can t use TNEOM s (energy ok) Only mg provides torque since tension attached to pivot point (lever arm =0) mg*sin() is the restoring force which as small angles is proportional to angle so is a simple harmonic motion (force varies with angle distance directly) Which graph shows: Period (T) vs. length? _B_ T vs. mass _A A B C more pendulums? See these web sites: 4

5 Pendulum: SHM at small angles L θ θ L Max acceleration Max force Max position zero velocity r h L x L* θ Max acceleration Max force Max position zero velocity Min acceleration Min force Min position Max velocity Bottom Line: Period depends only on the length of the string at small angles! Torque due to gravity (tension always perpendicular to arc motion) Period: T= 2π (l/g) gives the period at small angles Frequency: f= 1/T (frequency is measured in 1/sec or Hertz) y : h= L Lcos(θ ) radius: r= L*sin(θ) velocity: ½ m V 2 = mgh where h = y (only vertical drop matters) x : x= L* θ (in radians) (approximately horizontal for small angles) centripetal acceleration: horizontal acceleration: a = mv 2 /r = F net /m = T- mgcos(θ ) / m a = g sin(θ) θ T total energy: E= mgh + ½ mv 2 θ net force: F net = T mg (use vectors and free body diagrams!) mg Max values (derive from energy and free body diagrams!) acceleration: a max = gsin(θ) at highest point velocity: 1/2m(V max ) 2 = mg y max at lowest point displacement: y max = L- Lcos(θ max ) and X max = L*(θ max ) at highest point 5

6 Angle Sin() Angle Sin() Angle Sin() Angle Sin() SamplePendulum Problems 1. A pendulum follows simple harmonic motion only at small enough angles so that sin(θ ) θ. Make a chart and estimate the largest angle that meets this criteria *** if you use excel, use the formula: =SIN(RADIANS(A2)) A. close to A grandfather clock swings back and forth in two second. (mistake in original packet- period is 2 seconds- tic- toc) a. What is the frequency in Hz of this papa clock? 0.5 Hz f= 1/T = 1/2sec = 0.5 Hz b. How long is the clock s arm assuming it follows simple harmonic motion? 0.248m T= 2π (l/g) so l = g*t 2 /4π 2 = 9.8*4/4*pi*pi = 0.99 m c. What would be the period if the mass of the clock arm was doubled? No change d. How long should the arm be if we want it to work with the same period while on Mars? (g= 4m/s 2 ). l = g*t 2 /4π 2 = 4/9.8 of 0.99m =.40 m 6

7 3. A one meter long pendulum with a mass of 1kg is let go at θ= 7 0. a. Draw a picture and determine the change in height. y= 0.007m y= l-lcos(7) = 1-1cos(7) =.007 b. How much energy does it have when it is let go? 0.073J PE= mgh = mgy = 1*9.8*.007 =.073 c. How fast is it traveling when the pendulum is at θ= 0 0? 0.38 m/s 1/2mv 2 = mgh so v= (2*g*h)= (2*9.8*.007) = 0.38 m/s d. what it its period? T= 2 sec T= 2π (l/g) = T= 2π (1/9.8) = sec e. How much time did it take to go from θ= 7 0 to θ= 0 0? T= 0.5 sec this is ¼ of a period so just ¼ * 2 = 0.5 sec 4. A heavy pendulum called a charpy impact pendulum is used to test the strength of m = 10,0000 kg materials by breaking them! How much energy was needed to break the steel pieces above if a 2m long pendulum had to be dropped from at least at θ = 10 0 to fracture and break the steel? E= 2978J PE= mgh = 10,000 * 9.8 * (2-2cos(10) =

8 Angular Motion (see chap.8 for refresher) (2πradians=360 o ) θ = s/r (radians= arc length/radius) s or x ω = θ/t (radians/sec of angular velocity) r θ α = ω/t (radians/sec 2 of angular acceleration) * angular frequency: ω = 2π/T= 2 π f arc shown is about ½ the radius so it s 0.5 radians!) If compare to tangential motion: x= r* θ v= r*ω a= r* α a c = r *ω 2 ( Centripetal) Simple Harmonic Motion: *Harmonic force depends linearly on displacement: F= constant * x *Harmonic forces create sinusoidal motion such as in pendulums, springs, and elastic materials. *velocity is 180 out of phase with displacement (at max velocity, displacement is min : x=0) *acceleration is in phase with displacement (at max acceleration, displacement is max, x= A ) * displacement in angle : θ = ωt x = A*cos(ωt) where x max = Α V = ωα * sin(ωt ) where v max = ω* Α a = -ω 2 Α 2 * cos(ωt) where a max = ω 2 Α 2 Term Springs ( x = stretch) Pendulum (small angles only) Amplitude A= max stretch from equilibrium A L* θ max Force F= -k* x F= T-mg gsin(θ)/m Period T= 2π (m/k) T= 2π (l/g) Energy Emax = 1/2k( x) 2 Emax = mg* (L-Lcos(θ) ) 8

9 Sinusoidal Motion due to harmonic forces A B C Fill out the plots below using the teacher demo or lame web site B C B A B C B A B C B A B C B A B force time acceleration time position time velocity time B C B A B C B A B C B A B C B A B Summary: At max position, what is else is maximum: _at C or A, max force and acceleration? At minimum position, only at B, velocity is maximum. If the force is minimum, then the acceleration is _min (min or max?) If the stretch is the most, the force is max (the most or least?) Work=Force times displacement, but force depends on displacement (depends or doesn t depend) 9

10 Horizontal Spring motion! A B C Position Displacement Force Total Energy Position #1 : (no stretch, max speed, no force) x =0 (equilibrium) F=0 E= 0 + 1/2m(V max ) 2 Position #2: (little stretch, some elastic and kinetic) x = X 2 -X 1 F=-k(X 2 -X 1 ) E= 1/2k(X 2 -X 1 ) 2 + 1/2mV 2 2 Position #3: (max stretch, most force, no velocity) x = X 3 -X 1 F=-k(X 3 -X 1 ) E= 1/2k(X 3 -X 1 ) Measure all distances from the equilibrium point where there is no stretching No gravity involved, Friction is optional (motion similar, just won t last as long) Max stretch: most force, acceleration, displacement No stretch: most velocity as mass oscillates back and forth Find max speed using energy: 1/2m(V max ) 2 = ½ k (X 3 -X 1 ) 2 Find spring constant using frequency: : ω = 2 π/t π = 2π f Find max acceleration using a max = F max / m = -k x max / m 10

11 Vertical Spring Motion! (JUST LIKE HORIZONTAL IF MEASURE X FROM b ) Force diagrams (g= +10m/s 2 ) A. No weight- spring not stretched x= 0, F= 0 A B. Add weight but just sits there Fs F= -k(x b -X a ) + mg=0 k= mg/(x b -X a ) b B C mg c C. Pull weight down F= -k (X c -X a ) + mg = -k(x c -X b ) + -k (X b -X a ) + mg but note the second term is mg from part B. = -k(x c -X b ) + -mg + mg == -k(x c -X b ) so just measure from B where it hangs naturally! * TREAT LIKE A HORIZONTAL SPRING BUT MEASURE STRETCH FROM THE EQUILIBRIUM POINT FOUND AFTER HANG ON MASS AT POINT B SO. F(x) = -k * (X - X b ) Now Vertical Spring acts like a Horizontal spring! ( Fun by any stretch of the imagination) 11

12 Sample Spring Problems 5. A moving cart hits a spring, traveling at 5.0 m/s at the time of contact. At the instant the cart is motionless, by how much is the spring compressed? Initial E k E g E e Energy Flow Diagram Final E k E g E e E int Initial Final 0 0 1/2mv 2 = 1/2kx 2 so x= v*sqrt (m/k)= 5*sqrt(8/50)= 2.0m 6.. A block is placed on a spring, compressing it 0.30m. m = 500 g v = 0 Initial E k E g E e Energy Flow Diagram Final E k E g E e E int k = 100 N m ²x = 0.30 m Initial Final 0 What height does the block reach when launched by the spring? 0 Mgh = 1/2k x 2 so h= 1/2kx 2 / mg = ½*100*(0.30) 2 / ( 0.5kg *9.8) =.92m 7a... A spring whose spring constant is 850 N/m is compressed 0.40 m. What is the maximum speed it can give to a 500. g ball? 1/2k x 2 = 1/2mv 2 so v= x* sqrt(k/m) = 0.40* sqrt(850/.5kg) = 16.5m/s b. If spring were compressed twice as much, what would the velocity of the ball be? Twice x means twice velocity so both are squared terms of energy 12

13 8.. Suppose in the lab one group found that F = N x Construct a graphical m representation of force vs. stretch. (Hint: make the maximum displacement 0.25 m. ) a. Graphically determine the amount of energy stored while stretching the spring described above from x = 0 to x = 10. cm. E=area = 2 boxes= 2* 50*.05= 5J b. Graphically determine the amount of energy stored while stretching the spring described above from x = 15 to x = 25 cm. area = 8boxes= 8* 2.5= 20J Force (N) Data Set 1 Data Set 2 The graph at left was made from data collected during an investigation of the relationship between the amount two different springs stretched, when different forces were applied. c.for each spring determine the spring constant. K= slope = 36/4.5 = 8.0 N/m K= slope = 36/ 7 = 5.1 N/m d. For just the stiffest spring, find: force to stretch the spring to 3.0 m. F= kx = 8 N/m * 3m = 24 N ²x (m) energy stored when stretched to 3.0 m. E=1/2 kx 2 = ½ * 8 N/m * (3m) 2 = 36 Nm 13

14 AP Problems 9. A pendulum consists of a small object of mass m fastened to the end of an inextensible cord of length L. Initially, the pendulum is drawn aside through an angle of 60 with the vertical and held by a horizontal string as shown in the diagram above. This string is burned so that the pendulum is released to swing to and fro. a. draw a force diagram identifying forces acting on the object while it is held by the string. b. Determine the tension in the cord before the string is burned. Wt = Tc*cos(60) so Tc = Wt /cos(60)= 2Wt.= 2mg c. Show that the cord, strong enough to support the object before the string is burned, is also strong enough to support the object as it passes through the bottom of its swing. T mg = mv2/r where r= L, v = sqrt (2gh) and h = L- Lcos(60) = L/2 V= sqrt(2gl/2) = sqrt(g/l) or v2 = g/l plug in: T = mg + m (g/l)/l = 2mg but this is equal to Tc (tension of cord so ok, won t break) d. The motion of the pendulum after the string is burned is periodic. Is it also simple harmonic? Why, or why not? Not simple harmonic since the force = 2mg which is not kx or ksin() 14

15 10. Suppose the object undergoes this motion because it is the bob of a simple pendulum as shown above. If the object breaks loose from the string at the instant the pendulum reaches its lowest point and hits the ground at point P shown, what is the horizontal distance d it travels? X= V x * t t= time to fall, get from tneom s or energy since y= 0.5m= ½ *g*t 2 or t=.319 sec find Vx at bottom of swing using energy of pendulum (PE = KE) Angle and length are not given (from another problem) so pick L= 1m and angle=5 degrees: 1/2mV x 2 = mgh where h = L- L*cos(θ) = 1-1cos(5) =.0038 V x = sqrt (2gh) = sqrt (2*9.8*.0038) = 0.27 m/s Then x= V x *t = 0.27 m/s *.319 sec =.087 m (yipeee) 15

16 11. A heavy ball swings at the end of a string as shown with negligible air resistance. Point P is the lowest point reached by the ball in its motion. point Q is one of the two highest points. a. Draw and label the vectors that could represent the velocity and acceleration of the ball at points P & Q. If a vector is zero, state so. horizontal and vertical directions are dashed. no horizontal acceleration, max speed max acceleration, no speed i. Point P ii. Point Q b. After several swings, the string breaks. The mass of the string and air resistance are negligible. On the following diagrams, sketch the path of the ball if the break occurs when the ball is at point P or Q. In each case, describe the motion of the ball after the break. horizontal motion free falls from test since v=0 i. Point P ii. Point Q 16

17 12. A spring that can be assumed to be ideal hangs from a stand, as shown above. a. You wish to determine experimentally the spring constant k of the spring. i. What additional, commonly available equipment would you need? force meter & meter stick (F=kx) or known mass, stopwatch (T= 2π (m/k) ii. What measurements would you make? force, stretching distance or time for a period iii. How would k be determined from these measurements? K= slope of F vs. x or calculate from T= 2π (m/k) b. Assume that the spring constant is 500 N/m. A 2.0-kg mass is attached to the lower end of the spring and released from rest. Determine the frequency of oscillation of the mass. f= 1/T but T=2π (m/k) = 2π (2/500) = 0.4 sec..so T = 2.5 Hz c. Suppose that the spring is now used in a spring scale that is limited to a maximum value of 25 N, but you would like to weigh an object of mass M that weighs more than 25 N. You must use commonly available equipment and the spring scale to determine the weight of the object without breaking the scale. i. Draw a clear diagram that shows one way that the equipment you choose could be used with the spring scale to determine the weight of the object, lots of ideas: levers, water (buoyance), counter weights, etc and equipment for these ii. Explain how you would make the determination. Include how avoid friction or unknowns 17

18 13. A 3.0 kg object subject to a restoring force F is undergoing simple harmonic motion with a small amplitude. The potential energy U of the object as a function of distance x from its equilibrium position is shown above. This particular object has a total energy E: of 0.4 J. (a) What is the object's potential energy when its displacement is +4 cm from its equilibrium position? read off graph, U or elastic potenial energy is about 0.05 J (or use PE = 1/2kx 2 a so 0.4J= 1/2k(0.1) 2 to gt k= 80 then do again U= ½*80*(.042) 2 =.06J (almost same) (b) What is the farthest the object moves along the x axis in the positive direction? Explain your reasoning. 10 cm since max energy is 0.4 J and will be all potential at ends (KE=0) (c) Determine the object's kinetic energy when its displacement is 7 cm. total is always 0.4J = potenial + kinetic potential is about 0.18 from graph so kinetic = = 0.22J (d) What is the object's speed at x = 0? all kinetic energy = total energy = 0.4 J = 1/2mv 2 = ½* 3kg * v 2 solve and get v= 0.5 m/s 18

19 14. One end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance h above the floor. A block of mass M is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance X, as shown above. The block is released, follows the trajectory shown, and strikes the floor a horizontal distance D from the edge of the table. Air resistance is negligible. Determine expressions for the following quantities in terms of M, X, D, h, and g. Note that these symbols do not include the spring constant. a. The time elapsed from the instant the block leaves the table to the instant it strikes the floor h= ½*g*t 2 so t= (2*h/g) b. The horizontal component of the velocity of the block just before it hits the floor V x = x/t = D/t = D/ (2*h/g) c. The work which is done on the block by the spring Work = kinetic energy gained = 1/2mv 2 = ½*m *D 2 /(2hg)= md 2 g/4h d. The value of the spring constant k. elastic = potential ½*k*x 2 = ½ * m*v 2 ½*k*x 2 = md 2 g/4h k= md 2 g/(2h*x 2 ) 19

20 15a. Label the graphs as either force, velocity, position, or acceleration b. Which point (A,B,C) matches with equilibrium, max extension, and max compression? A B C B C B A B C B A B C B A B C B A B? force time acceleration? time position? time velocity? time B C B A B C B A B C B A B C B A B 20

21 16. waves Unit I: Worksheet 2 The diagram to the right shows a kg block attached to a Hookean spring. The block experiences no net force when it is at position C. When the block is to the left of point C the spring pulls it to the right. When the block is to the right of point C, the spring pushes it to the left. The mass is pulled to the left from point C to point A and released. The block then oscillates between positions A and E. Assume that the system consists of the block, spring, and the horizontal surface on which the spring oscillates. 1. When there is no friction between the block and the surface, the force required to hold the block at rest at position A is 5.00 N. Calculate the spring constant for the spring. K= F/x = 5 N/ 0.1m = 50 N/m 2. Calculate the elastic energy at the following positions: a. Position A Ee= ½ k*x 2 = ½ * 50 *.10 2 = 0.25 J 0 A B C D E 10.0 cm 5.0 cm 5.0 cm cm b. Position B. Ee= ½ k*x 2 = ½ * 50 *.05 2 =.0625 J c. Position C. Ee= ½ k*x 2 = ½ * 50 * 0 2 = 0 d. Position D. Ee= ½ k*x 2 = ½ * 50 *.05 2 =.0625 J e. Position E. Ee= ½ k*x 2 = ½ * 50 *.1 2 =.25 J 3. Calculate the kinetic energy of the system at each of the following positions: 1. Position A KE = 0 b. Position B KE = total Ee = =.1875J = ½ mv 2 if want velocity c. Position C KE=.25 J d. Position D KE = J e. Position E KE = 0 4. Create a set of qualitative energy pie charts for this system starting from the instant of release at point A and ending after the system has completed one full cycle and returns to position A Position A Position B Position C Position D Position E Position D Position C Position B Position A 21

22 17. Waves Unit I: Worksheet 3 1. An ideal Hookean spring of spring constant 20.0 N/m is connected to a kg block in the arrangement shown to the right. The (*) represents the position of the center of the block when the spring is unstretched. (Positions are not to scale.) From this position the experimenter slowly lowers the block from (*) until it reaches point B where the system is at rest. a. How far does the spring stretch when the kg block is slowly lowered to position B? * b. The block is then pulled to position A, 10.0 cm below position B. It is released and allowed to oscillate between positions A and C. In the space below calculate the elastic energy, gravitational energy, and kinetic energy of the system at positions A, B, and C. C B 10.0 cm 10.0 cm A c. How fast is the block moving at the instant the center is even with position B? E k d. Create quantitative energy bar graphs for the system at the positions indicated for one complete cycle of the system starting when it is released at position A until it returns to position A the first time. E g E e E diss E k E g E e E diss E k E g E e E diss E k E g E e E diss E k E g E e E diss Position A Position B Position C Position B Position A 22