PYHSICS:SIMPLE HARMONIC MOTION,

THE HARD WAY

INTRODUCTION

• In this Learning Objective presentation, we will discuss three different kinds of simple harmonic motion.

• We will focus the period of each of them.• These cases are different, but they are

somewhat related.• ENJOY AND HAVE FUN WITH PHYSICS!

FIRST CASE: THE SIMPLE ONE

This is probably the simplest model of simple harmonic motion, as what we have learned from our text book(PHYS 101), the period of this simple harmonic motion is :Given mass of m, constant of k, with no external forces.

T=2π√(m/k)

SECOND CASE: SOME VARIATION

In this case, the other side of the spring is attached to another object instead of the wall. So what would be the period for this motion?(m1 and m2 share the same mass of m, and the spring has Hooke’s constant of k. The system has no external forces.)

SOLUTION FOR THE SECOND

In this case, since there are no external forces, the center of mass of the whole system does not move during vibration. We can have a symmetrical line in the middle of the system, since it is the center of mass of the system and that would be our imaginary wall. From what we learned in the first case, the period was T=2π√(m/k) with the WHOLE spring, having a Hooke’s constant of k. For now, on both sides of the line, the Hooke’s constant for divided parts of the spring are both k/2, and so the period in this case would beT=2π√(m/2k)

THIRD CASE: THE HARD ONE

This is the HARD part! In addition to the 2 objects in the second case, what would happen if a third object was added? (All objects have the same mass of m, and the two springs have the same length L and Hooke’s constant k. The system has no external forces.)MAKE SURE YOU THINK ABOUT IT BEFORE GOING TO THE NEXT SLIDE!

SOLUTION FOR THE THIRD

Since the system has no external forces, the center of mass of the system does not move during vibration. And since all three objects have the same mass and the springs have the same length, the system is symmetrical. There will be 2 different scenarios in this system. One is that when m2 does not move, m1 and m3 compress and extend simultaneously. In this scenario, the period of the system would be the same as the FIRST CASE, T=2π√(m/k), because m2 is just like the wall in the first case, it is just that this “wall” has objects attached to it on both sides.

SOLUTION FOR THE THIRD

In the second scenario, m2 moves as well. In this case, we can divide m2 evenly in to two pieces and see the whole system as a combination of m1 - m2/2 system and m2/2 - m3 system. We only have to examine one of them to find the period of the entire system since they are symmetrical. If we solely look at m1-m2/2 system, the center of mass of this sub-system, would be in a distance of L/3 from m1 and 2L/3 from m2/2. Following the same approach in the second case, we have an imaginary wall at the point that is l/3 from m1. So the Hooke’s constant for the left side of the “wall” would be 3k. Therefore, the period in the second scenario would be T=2π√(m/3k)

THANK YOU !!!HAVE FUN WITH PHYSICS!!!