HW 5 Ch 14 - Oscillations Due: 11:59pm on Tuesday, October 13, 2015 To understand how points are awarded, read the Grading Policy for Policy for this assignment.
PhET Tutorial: Masses & Springs Learning Goal: To understand how the motion and energetics of a weight attached to a vertical spring depend on the mass, the spring constant, and initial conditions. For this tutorial, use the PhET simulation Masses & Springs. Springs. You can put a weight on the end of a hanging spring, stretch the spring, and watch the resulting motion.
Start the simulation. simulation. When you click the simulation link, you may be asked whether to run, open, or save the file. Choose to run or open it. You can drag a weight to the bottom of a spring and release it. You can put only one weight on any spring. With the weight on the spring, you can click and drag the weight up or down and release it. Adjusting the friction friction slider slider bar at top right increases or decreases the amount of thermal dissipation (due to air resistance and heating of the spring). You can adjust the spring constant of spring #3 using the softness spring 3 3 slider bar. The horizontal dashed line as well as the ruler can be dragged to any position, which is helpful for comparing positions of the springs. Feel free to play around with the simulation. When you are done and before starting Part A, set the friction slider bar to the middle and the gravitational acceleration back to "Earth".
Part A Place a 50 weight on spring #1, and and release release it. Ev entually, the weight will will come to rest at an equilibrium position, position, with the spring somewhat somewhat stretched compared to its original (unweighted) length. At this point, the upward force of the spring balances the force of gravity on the weight. With the weight in its equilibrium position, how does the amount the spring is stretched depend on the mass of the weight? ANSWER: The spring stretches less for a heavier weight. The spring stretches more for a heavier weight. The stretch does not depend on mass. Typesetting math: 100%
Correct Since the force of gravity on the weight increases as the mass increases, the upward force of the spring must increase for the two forces to balance (and the weight to therefore be in equilibrium). The force the spring exerts on the weight increases the more the spring is stretched from its unweighted length.
Part B Use the simulation to estimate the masses of the three colored, unlabeled weights. Then, place them into the appropriate mass bins.
Hint 1. How to approach the problem You learned in Part A that a heavier weight stretches the spring more at the equilibrium position. Compare the equilibrium positions for each of the colored weights to those for the weights with labeled masses. You might want to verify that the three springs all stretch the same amount for a particular weight, so you can compare the equilibrium positions simultaneously.
ANSWER:
Correct
Part C Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight. Use this concept to estimate the spring constant of spring #1. Express your answer in
to one significant figure.
Hint 1. How to approach the problem The spring constant is used in Hooke’s law for the force exerted by the spring: , wher where e is the amount the spring is stretched. When the spring is stretched to the equilibrium position, the force exerted by the spring pulling up on the weight is equal to the force of gravity pulling down down.. Use this relationship to solve for .
ANSWER: =
10
Correct
Part D Now, for parts D-F, you’ll investigate the energetics of the spring. Select 1 in the Show Energy of box, box, which shows an energy bar diagram. Select the g = 0 option (under the planet names), which simulates what happens without any gravitational forces (and consequently removes gravitational potential energy from the energetics). Adjust the friction friction slider to none (this none (this prevents any thermal energy from being generated). Place a weight on spring #1, stretch it, and release it. Watch how the kinetic energy and elastic potential energy vary with time. (You can slow down or stop time using the buttons next to the list of planets.) When is the elastic potential energy of the spring a maximum? ANSWER:
When the spring isn’t stretched or compressed When the spring is most stretched When the spring is most compressed Both when the spring is most compressed and when the spring is most stretched
Correct The elastic potential energy depends on the magnitude magnitude of the change in the length of the spring. Mathematically, the elastic potential energy is given by , where where is the spring constant and is the difference between between the length of the spring and its unweighted unweighted length. A compression of therefore therefore results in the same elastic potential energy energy as a stretch of .
Part E When is the kinetic energy of the mass a maximum? ANSWER: When the spring is most compressed Both when the spring is most compressed and when the spring is most stretched When the spring is most stretched When the spring is at its unweighted length (when it isn’t stretched or compressed)
Correct The total energy of the system is equal to the kinetic energy of the mass (since the spring has negligible mass) plus the elastic potential energy. Since this total energy is conserved, the kinetic energy is a maximum when the elastic potential energy is a minimum, which occurs when the spring is at its unweighted length.
Part F Select Earth Earth in the menu box so that there is now a force of gravity. Now the total energy of the mass/spring system is the sum of the kinetic energy, the elastic potential energy, and the gravitational potential energy. When is the kinetic energy a maximum? (It may help to watch the simulation in slow motion - 1/16 time.) ANSWER: When the spring is at its unweighted length (its equilibrium position without the weight attached) When the mass is at the equilibrium position When the spring is most stretched or most compressed
Correct The total potential energy (the gravitational potential energy plus elastic potential energy) is a minimum at the equilibrium position, even though there is some elastic potential energy when the mass is at this location. The kinetic energy is always a maximum when the total potential energy is a minimum (since the total energy is conserved).
Part G Now, for parts G-I, you’ll investigate what determines the frequency of oscillation. For these parts, turn off the friction using the slider bar. Select the stopwatch, and time how long it takes for a weight to oscillate back and forth 10 times. The period of oscillation is this time divided by 10. The frequency of oscillation is one divided by the period. How does the frequency of oscillation depend on the mass of the weight? ANSWER:
The frequency increases as the mass increases. The frequency is independent of the mass. The frequency decreases as the mass increases.
Correct A greater mass results in a lower frequency and a longer period of oscillation.
Part H The amplitude of oscillation is the maximum distance between the oscillating weight and the equilibrium position. Determine the frequency of oscillation for several different amplitudes by pulling the weight down different amounts. How does the frequency depend on the amplitude of oscillation? ANSWER: The frequency increases as the amplitude increases. The frequency decreases as the amplitude increases. The frequency is independent of the amplitude.
Correct Even though the weight has to travel farther each oscillation if the amplitude is greater, the spring on average exerts a stronger force, causing a greater acceleration and a greater average speed. The effects of the longer distance and faster speed cancel out so that the period of oscillation doesn’t change!
Part I The spring constant of spring #3 can be adjusted with the softness spring 3 slider 3 slider bar (harder means a greater spring constant, or stiffer spring). How does the frequency of oscillation depend on the spring constant?
Hint 1. How to approach the problem You can place a weight on spring #3 with the softness of spring 3 set 3 set to a very low (soft) value and place another weight with the same mass on spring #1. Then, release the two weights and compare their frequencies of oscillation.
ANSWER: The frequency increases as the spring constant increases. The frequency is independent of the spring constant. The frequency decreases as the spring constant increases.
Correct It turns out that the frequency frequency of oscillation depends depends on the square root of the ratio of the spring constant to mass:
, where where
is the frequency. A stiffer spring constant causes the frequency to increase. Sports cars use stiff springs, whereas large plush Cadillacs use soft springs for their suspension.
PhET Interactive Simulations University of Colorado http://phet.colorado.edu
PhET Tutorial: Pendulum Lab Learning Goal: To understand the relationships of the energetics, forces, acceleration, and velocity of an oscillating pendulum, and to determine how the motion of a pendulum depends on the mass, the length of the string, and the acceleration due to gravity.
For this tutorial, use the PhET simulation Pendulum Lab. Lab. This simulation mimics a real pendulum and allows you to adjust the initial position, the mass, and the length of the pendulum.
Start the simulation. simulation. You can drag the pendulum to an arbitrary initial angle and release it from rest. You can adjust the length and the mass of the pendulum using the slider bars at the top of the green panel. Velocity and acceleration vectors can be selected to be shown, as well as the forms of energy. Feel free to play around with the simulation. When you are done, click the Reset Reset button. button.
Part A Select to show the energy of pendulum 1. Be sure that friction friction is set to none none.. Drag the pendulum pendulum to an angle (with respect to the vertical) vertical) of , and and then then release release it. When Whe n the pen pendulu dulum m is at at , wha whatt form form(s) (s) of ene energy rgy doe does s itit h have? ave? Check all that apply. ANSWER: Kinetic energy Thermal energy Potential energy
Correct The pendulum starts off with no kinetic energy since it is released from rest, so it initially only has potential energy. When the pendulum is at , it is just as high above the ground as when it started, so it must have the same amount of potential energy as it initially had. Since the total energy energy is conserved, it can’t have have other other forms forms of energy energy at (if it did, it would have more energy energy there than it initially had) so it again again has only potential energy.
Part B Drag the pendulum pendulum to an angle (with respect to the vertical) vertical) of
, and and then then release release it.
Where is the pendulum swinging the fastest? ANSWER: at at at at
Correct The pendulum has the least potential energy at this location since it is at the lowest point in the arc (in fact, for this simulation, the potential energy reference location is here, so it has no potential energy). This means that the kinetic energy is greatest here, so the pendulum is moving the fastest.
Part C
Drag the pendulum pendulum to an angle (with respect to the vertical) vertical) of
, and then release release it. Select to show the acceleration acceleration vector.
With the pendulum swinging back and forth, at which locations is the acceleration equal to zero? ANSWER: The Th e ac accel celer erat atio ion n is ze zero ro wh when en th the e an angl gle e is ei eith ther er The acceleration is zero zero when the angle is
or
.
.
The acceleration is never equal to zero as it swings back and forth.
Correct The pendulum is moving in a circular path so its velocity is never constant. In fact, for most locations, the acceleration has both a radial component (the centripetal acceleration, which is directed along the rope) and a tangential component (due to the speed changing, directed along the path; path; the only place the tangential tangential acceleration acceleration is zero is when the angle angle is ).
Part D With the pendulum pendulum swinging swinging back and forth, how does the tension tension of the rope rope compare compare to the force force of gravity when the angle is
?
Hint 1. How to approach the problem Look at the direction direction of the acceleration when when the angle is (you can slow down down or pause the simulation to see this more clearly), and think about the relationship between the net force acting on the pendulum and the acceleration (Newton’s 2nd law of motion). It would probably help to draw a free-body diagram for the mass.
ANSWER: The tension is equal to the force of gravity. The tension is greater than the force of gravity only if it is swinging really fast. The tension is less than the force of gravity. The tension is greater than the force of gravity.
Correct Since the acceleration of the pendulum pendulum is directed directed up when when the angle is , the net force must must be directed up up (Newton’s (Newton’s 2nd law). This means means that the upward force of tension must be stronger than the downward force of gravity.
Part E Drag the pendulum pendulum to an angle (with respect to the vertical) vertical) of
, and and then then release release it.
With the pendulum swinging back and forth, where is the tension equal to zero? ANSWER: The Th e te ten nsi sio on is ze zerro at th the e angle les s
and
The tension is zero when the angle is The tension is zero zero when the angle is
. and
.
.
The tension is never zero.
Correct At these locations, the acceleration is solely due to gravity and directed downward. downward. Thus, the net force acting on t he pendulum i s also directed downward, meaning there are no horizontal forces. This requires the tension to be zero.
Part F Now, for parts F-I, you will investigate how the period of oscillation depends on the properties of the pendulum. The period of oscillation is the amount of time it takes for the pendulum to take a full swing, going from the original angle to the other side, and
returning to the original angle. You can determine the period by selecting other tools, tools, which gives you a stopwatch. With the pendulum swinging, you can start the stopwatch when the pendulum is at its original angle and time how long it takes to complete 10 swings. The period will be this time interval divided by 10 (this method is more accurate than trying to time one swing). Set the length of the pendu pendulum lum to 1.0 and the mass to 1.0 . Click Reset Reset,, and then drag the pendulum to an angle (with respect to the vertical) of and release it. What is the period of oscillation? ANSWER: 4.0 1.5 20.0 2.0 1.0 0.5
Correct A
-
-
pendulum pendulum completes one osc illation in 2.0 .
Part G How does the period of oscillation depend on the initial angle of the pendulum when released? (Be sure to measure the period for initial angles much greater than .) ANSWER: The period is longer when the initial angle is greater. The period is shorter when the initial angle is greater. The period is independent of the initial angle.
Correct Unlike Unlike a harm harmonic onic oscillator oscillator such as a mass on a spring, spring, the perio period d actually actually depe depends nds on on the initial angle. angle. For For small small angles angles (e.g., (e.g., pretty good approximation that the period doesn’t change, but for larger angles the period does in fact increase.
), it is a
Part H Keeping the length of the pendulum fixed, determine the period for a few different masses. (Alternatively, you can set up two pendulums by selecting Show 2nd pendulum. pendulum. Adjust the lengths to be the same, and have one pendulum with a higher mass. You can release one and then release the other, with the same angle, when the first one is back at that angle.) How does the period of the pendulum depend on mass? ANSWER: A heavier pendulum has a shorter period. The period is independent of the pendulum’s mass. A heavier pendulum has a longer period.
Correct The period of a pendulum does not depend on mass. The reason for this result is very similar to the reason that, without air resistance, all objects fall to the ground at the same rate (changing the mass changes both the inertia and the force of gravity by the same amount).
Part I Now, keep the mass fixed to any value you choose and measure the period for several different pendulum lengths. How does the period of the pendulum depend on the length? ANSWER:
A longer pendulum pendulum has a shorter period. A longer pendulum pendulum has a longer period. The period is independent of the length of the pendulum.
Correct A longer length causes the period to increase. In fact, as you can verify with the stopwatch, the period of oscillation is proportional proportional to the square root of the length; if you increase the length from 0.5 to 2.0 , the period doubles.
Part J Now, change change the planet planet where the experiment experiment takes place to see how the period period of oscillation depends on the acceleration due due to gravity, ; is larger than this value on Jupiter and smaller than this value on the Moon).
(on Earth, Earth,
How does the period period of osci llati on depend depend on the value of ? ANSWER: The period period of of oscillation is independent independent of the value of . The period period of oscillati oscillati on is longer longer on planets planets with a higher higher value of . The period period of oscillati oscillati on is shorter on planets planets with a higher higher value of .
Correct A higher value of gives rise to greater accelerations, more energy, and thus greater average speeds. In fact, for s mall angles of oscillation, the period is inversely proportional proportional to the square root of ; , where where is the length and is the period. period.
PhET Interactive Simulations University of Colorado http://phet.colorado.edu
Problem 14.50 It has recently become possible to "weigh" DNA molecules by measuring the influence of their mass on a nano-oscillator. Figure shows a thin rectangular cantilever etched out of silic on (den (density sity 2300 ) with a small gold gold dot at the end. If pulled pulled down and released released,, the end of the cantilever vibrates with simple harmonic motion, moving up and down like a diving board after a jump. When bathed with DNA molecules whose ends have been modified to bind with gold, one or more molecules may attach to the gold dot. The addition of their mass causes a very slight-but measurable-decrease in the oscillation frequency. frequen cy. A vibrating cantilever of mass can be modeled as a block of mass attached to a spring. (The factor of arises from the moment of inertia of a bar pivoted at one end.) Neither the mass nor the spring constant can be determined very accurately-perhaps to only two significant figures-but the oscillation frequency can be measured with very high precision simply by counting the oscillations. In one experiment, the cantilever was initially vibrating at exactly 13 . Attachment of a DNA molecule caused the freque frequency ncy to decrease by 57 .
Part A What was the mass of the DNA?
Express your answer to two significant figures and include the appropriate units. ANSWER: =
4.4*10^-21
All attempts used; correct answer withheld by instructor
Problem 14.80 The figure shows shows a 240 uniform rod rod pivoted at one end. end. The other other end is attached to a horizontal horizontal spring. The spring spring is neither stretched nor nor compressed when the rod hangs straight down.
Part A What is the rod's oscillation period? You can assume that the rod's angle from vertical is always small. Express your answer to two significant figures and include the appropriate units. ANSWER: =
0.60
Correct
Problem 14.58 A geologist needs to determine the local value of . Unfortun Unfortunately, ately, his only tools are a meter s tick, a saw, and a stopwatch. He s tarts by hanging the meter stic k from one one end end and measuring its frequency as it swings. swings. He then saws off 20 —using the centimeter markings—and markings—and measures measures the frequen frequency cy again. After two more cuts, these are his data: Length,
Frequency,
100
0. 61
80
0. 67
60
0. 79
40
0. 96
Part A You want to determine determine the local value of of by graphing graphing the data. Select the appropr appropriate iate variables to graph graph on each each axis that will produce a straight-line graph with either a slope or intercept that is related related to . Sort
variables all variables
into the appropriate bins.
Hint 1. How to approach the problem
A physical pendu pendulum lum has moment of inertia
and dist ance
between betwee n the pivot and the center of mass. Its oscillation freque frequency ncy is
ANSWER:
ANSWER:
Correct
Part B
Graphing either Graphing versus or versus gives a straight line. In the graph shown, we chose to plot horizontal axis. From the equation for the line of best fit given, determine the local value of . Express your answer to three significant figures and include the appropriate units. ANSWER:
on the vertical axis and
on the
= 9.76
Correct
Conceptual Question 14.10 Suppose the damping constant
of an an oscillator increases. increases.
Part A Is the medium more resistive or less resistive? ANSWER: The medium is more resistive. The medium is less resistive.
Correct
Part B Do the oscillations damp out more quickly or less quickly? ANSWER: The oscillations will damp out more quickly. The oscillations will damp out less quickly.
Correct
Part C Is the time constant
increased or decreased?
ANSWER: is increased. is decreased.
Correct
Problem 14.32 The two graphs in the figure are for two different vertical mass-spring systems.
Part A If both syst ems have the same mass, what is the ratio
of their spring constants?
ANSWER: =
2.25
Correct
Problem 14.31 The figure is the position-versus-time graph of a particle in simple harmonic motion.
Part A What is the phase constant? ANSWER:
Correct
Part B Whatt is the Wha the velocity at
=0 ?
Express your answer with the appropriate units. ANSWER: =
-13.6
Correct
Part C What is
?
Express your answer with the appropriate units. ANSWER: = 15.7
Correct
Problem 14.37
Part A When the displacement of a mass on a spring is
, what what percentag percentage e of the energ energy y is kinetic energy? energy?
ANSWER: 75.0 75 .0
Correct
Part B At what displacement, as a fraction of
, is the ener energy gy half kinetic and half potential?
ANSWER: 0.707
Correct Score Summary: Your score on this assignment is 86.7%. You received 7.81 out of a possible total of 9 points.
