Harmonic+Motion

=__HARMONIC MOTION__=

Wednesday



Thursday



Tuesday





Wednesday



Thursday



Friday



PRACTICE TEST

1) Draw a pendulum in motion and use force vectors (arrows) to explain why the pendulum moves in a circular fashion, and why it slows down and changes direction. 2) Why does Hooke's Law have a negative sign in it? 3) The variable omega is "angular velocity", but for a spring system there is no circular motion at all. Why do we still use this variable? 4) What causes the oscillations to dampen when a ball is bouncing on a table? 5) Which variables affect the period of a pendulum and which ones affect the period of a spring-mass system? 6) A spring is at rest and then a mass is attached, causing it to stretch 20 cm. As soon as it reaches the new equilibrium, you pull it down 30 cm and release it. Which value is ΔX and which value is amplitude? 7) Does the spring constant for a certain spring change if you go to another planet? Why or why not? ( Miley was wrong, by the way) 8) Someone hands you a spring, a 100 gram mass, and a meter stick and asks you to find the spring constant. How would you do it? 9) You have indentical pendulums at the equator and at the North Pole. Which one oscillates faster and why? 10) Start with the formula for angular velocity and show how you get to the correct formulas for the period of a spring-mass and a pendulum.

11) The length of a pendulum is 2.5 meters. You pull it back 30 degrees and release it. Graph the motion of the pendulum over 10 seconds below and label the period, amplitude, and find the frequency. (you’ll need to label the graph as well)

12) A 567 gram mass is attached to a spring and is allowed to reach a new equilibrium. If the spring stretches 40 cm, what is the value of the spring constant? if you pull the mass down 25 cm and release it, what will the period and frequency of the oscillations be? How long will it take to complete 3000 oscillations?

13) A pendulum with a length of 5 meters is pulled back at an angle of 35 degrees and released. However, due to friction, each successive swing (one oscillation is two swings) loses 20 % of it’s previous amplitude. Graph the motion of the pendulum over 15 seconds—including the damping. How high is the pendulum off the ground at the end of the 15 seconds? (You’ll have to estimate using your graph and some geometry).

14) Imagine you were on the surface of a strange planet with a spring, a mass, a meter stick and a stopwatch. How could you use those things to determine the g of this new planet? You will need to find an equation for g using only the variables you can measure with your given equipment.

15) A pendulum with length of 50 cm is pulled back to an angle of 50 degrees and released. Draw a graph showing this harmonic motion over the course of ten seconds. At what times is it at the top of its motion? Where is it in its phase at 7.5 seconds?

__ANSWER KEY FOR 1-10__

1) see your notes 2) Because the spring force is always in the opposite direction as the force applied. 3) Harmonic motion works similarly to circular motion in the way the motion is repeated. Like an object on a wheel going up and down, if we want to increase the oscillation rate we spun the wheel faster. Angular velocity describes the rate of this spinning which is analogous to the rate at which things oscillate. 4) Energy lost to vibrations in the table, the creation of sound waves, energy lost to deformation of the ball, and air resistance. 5) Length affects the period of a pendulum and mass and the spring constant affect the period of a spring-mass system. 6) 20 cm is the x value and 30 cm is the amplitude. 7) No, the spring constant won't change. K is affected by gravity but that will also change the x value, so g/x will be a constant. Basically, imagine holding the spring sideways so that gravity isn't a factor. The spring should resist the force of you compressing it the same on earth as anywhere else. (Kim D) 8) See practice test question 14 9) The one at the North Pole will oscillate faster because the value of g is greater there due to less centrifugal force. At the equator there is more centrifugal force which means objects experience a slightly lower free-fall acceleration. 10) See notes (start with the formula for omega, get to frequency, switch to period, then simplify the expression