2. 4.1.1Describe examples of oscillation. 4.1.2Define the terms displacement, amplitude, frequency, period, and phase difference. 4.1.3Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x. 4.1.4Solve problems using the defining equation a = -  2 x for SHM. 4.1.5Apply the equations as solutions to the defining equation a = -  2 x. 4.1.6Solve problems, both graphically and by calculation, for acceleration, velocity and displacement during SHM. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion

3. Describe examples of oscillation.  Oscillations are vibrations which repeat themselves. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion EXAMPLE: They can be driven externally, like a pendulum in a gravitational field. EXAMPLE: They can be driven internally, like a mass on a spring. x v = 0 v = v max v = 0 v = v max v = 0 FYI  In all oscillations v = 0 at the extremes and v = v max in the middle of the motion.

4. Describe examples of oscillation.  Oscillations are vibrations which repeat themselves. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion EXAMPLE: They can be very rapid vibrations such as in a plucked guitar string or a tuning fork.

5. Define the terms displacement, amplitude, frequency, period, and phase difference.  Consider a mass on a spring that is displaced 4 meters to the right and then released.  We call the maximum displacement x 0 the amplitude. In this example x 0 = 4 m.  We call the point of zero displacement the equilibrium position.  The period T (measured in s) is the time it takes for the mass to make one complete oscillation or cycle.  For this particular oscillation, the period T is about 24 seconds (per cycle). Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion x0x0 x equilibrium

6. Define the terms displacement, amplitude, frequency, period, and phase difference.  The frequency f (measured in Hz or cycles/s) is defined as how many cycles (oscillations, repetitions) occur each second.  Since period T is seconds per cycle, frequency must be 1/T. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion f = 1/T relation between T and f T = 1/f EXAMPLE: The cycle of the previous example repeated each 24 s. What are the period and the frequency of the oscillation? SOLUTION:  The period is T = 24 s.  The frequency is f = 1/T = 1/24 = 0.042 Hz

7. Define the terms displacement, amplitude, frequency, period, and phase difference.  We can pull the mass to the right and then release it to begin its motion:  Or we could push it to the left and release it:  The resulting motion would have the same values for T, f, and .  However, the resulting motion will have a phase difference of half a cycle. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion Start stretched Start compressed The two motions are half a cycle out of phase. x x

8. PRACTICE: Two identical mass-spring systems are started in two different ways. What is their phase difference?  The phase difference is a quarter of a cycle. Define the terms displacement, amplitude, frequency, period, and phase difference. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion Start stretched and then release Start unstretched with a push left x x

9. PRACTICE: Two identical mass-spring systems are started in two different ways. What is their phase difference?  The phase difference is three quarters of a cycle. Define the terms displacement, amplitude, frequency, period, and phase difference. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion Start stretched and then release Start unstretched with a push right x x

10. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x.  Before we define simple harmonic motion, which is a special kind of oscillation, we have to digress for a moment and revisit uniform circular motion.  Recall that UCM consists of the motion of an object at a constant speed v 0 in a circle of radius x 0.  Since the velocity is always changing direction, we saw that the object had a centripetal acceleration given by a = v 0 2 /x 0, pointing toward the center.  If we time one revolution we get the period T.  T is about 12 s.  And the frequency is f = 1/T = 1/12 = 0.083 s. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion x0x0 v0v0

11. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x.  We say that the angular speed of the object is 360 deg/12 s = 30 deg s -1.  In Topic 4 we must learn about an alternate and more natural method of measuring angles besides degrees.  They are called radians. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion  rad = 180° = 1/2 rev radian-degree-revolution conversions 2  rad = 360° = 1 rev EXAMPLE: Convert 30  into radians (rad) and convert 1.75 rad to degrees. SOLUTION:  30  (  rad / 180° ) = 0.52 rad.  1.75 rad (180° /  rad ) = 100°.

12. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x.  Angular speed will not be measured in degrees per second in this section. It will be measured in radians per second. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion  rad = 180° = 1/2 rev radian-degree-revolution conversions 2  rad = 360° = 1 rev EXAMPLE: Convert the angular speed of 30  s -1 from the previous example into radians per second. SOLUTION:  Since 30  (  rad / 180° ) = 0.52 rad,  then 30  s -1 = 0.52 rad s -1. FYI  Angular speed is also called angular frequency. x0x0 v0v0

13. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x.  Since 2  rad = 360° = 1 rev it should be clear that the angular speed  is just 2  /T.  And since f = 1/T it should also be clear that  = 2  f. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion  = 2  /T =  /t relation between , T and f  = 2f = 2f  rad = 180° = 1/2 rev radian-degree-revolution conversions 2  rad = 360° = 1 rev EXAMPLE: Find the angular frequency (angular speed) of the second hand on a clock. SOLUTION:  Since the second hand turns through one circle each 60 s, it has an angular speed given by  = 2  /T = 2  /60 = 0.105 rad s -1.

14. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion  = 2  /T =  /t relation between , T and f  = 2f = 2f  rad = 180° = 1/2 rev radian-degree-revolution conversions 2  rad = 360° = 1 rev EXAMPLE: A car rounds a 90° turn in 6.0 seconds. What was its angular speed during the turn? SOLUTION:  Since  needs radians we begin by converting  :  = 90°(  rad / 180° ) = 1.57 rad.  Now we use  =  /t = 1.57/6 = 0.26 rad s -1.

15. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion PRACTICE: Find the angular frequency of the minute hand of a clock, and the rotation of the earth in one day.  The minute hand takes 1 hour to go around one time. Thus  = 2  /T = 2  /3600 s = 0.0017 rad s -1.  The earth takes 24 h for each revolution so that  = 2  /T = ( 2  / 24 h )( 1 h / 3600 s ) = 0.000073 rad s -1.  This small angular speed is why we can’t really feel the earth as it spins.

16. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion PRACTICE: This is an old IB question. An object is traveling at speed v 0 in a circle of radius x 0. The period of the object’s motion is T. (a) Find the speed v 0 in terms of x 0 and T.  Since the object travels a distance of one circumference in one period v 0 = distance / time v 0 = circumference / period v 0 = 2  x 0 /T. (b) Show that v 0 = x 0 .  Since v 0 = 2  x 0 /T and  = 2  /T we have v 0 = 2  x 0 /T v 0 = x 0 2  /T v 0 = x 0 (2  /T) v 0 = x 0  x0x0 v0v0

17. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion PRACTICE: This is an old IB question. An object is traveling at speed v 0 in a circle of radius x 0. The period of the object’s motion is T. (c) Find the centripetal acceleration a C in terms of x 0 and .  Since the centripetal acceleration is a C = v 0 2 /x 0 and since v 0 = x 0 , v 0 2 = x 0 2  2 a C = v 0 2 /x 0 a C = x 0 2  2 /x 0 a C = x 0  2. x0x0 v0v0

18. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x.  You might be asking yourself how an oscillating mass- spring system might be related to uniform circular motion. The relationship is remarkable.  Consider a rotating disk that has a ball glued onto its edge. We project a strong light to produce a shadow of the ball’s motion on a screen.  Like the mass in the mass- spring system, the ball behaves the same at the arrows: Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion  x0x0 x  x0x0  x0x0  x0x0  x0x0  x0x0  x0x0  x0x0  x0x0  x0x0  x0x0  x0x0  x0x0 -x 0 x0x0 0 x

19. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x.  Note that the shadow is the x-coordinate of the ball.  Thus the equation of the shadow is x = x 0 cos .  Therefore the equation of the shadow’s x-coordinate is x = x 0 cos  t.  If we know , and if we know t, we can calculate x. x0x0 v0v0 x0x0 v0v0 Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion FYI  From  =  /t we get  =  t.  x -x 0 x0x0 0 x x = x 0 cos 

20. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x.  At precisely the same instant we can find the equation for the shadow of v.  Create a velocity triangle.  Working from the displacement triangle we can determine the angles in the velocity triangle.  The x-component of the velocity is opposite the theta, so we use sine: v = -v 0 sin .  But since  =  t we get our final equation: v = -v 0 sin  t.  Why is our v negative? Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion x0x0 v0v0   90-  

21. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x.  Remember the acceleration of the mass in UCM?  We found recently that a C = x 0  2.  And we know that it points to the center.  The x-component of the acceleration is adjacent to the theta, so we use cosine: a = -a C cos .  But since  =  t we get our final equation and a C = x 0  2 we have a = -x 0  2 cos  t.  Why is our a negative?  Since x = x 0 cos  t we can just write a = -  2 x. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion x0x0 v0v0  aCaC

22. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x.  Let’s put all of our equations in a box-there are quite a few!  We say a particle is undergoing simple harmonic motion (SHM) if it’s acceleration is of the form a = -  2 x.  Data Booklet has highlighted formulas. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion x = x 0 cos  t Set 1 - equations of simple harmonic motion v = -v 0 sin  t a = -x 0  2 cos  t a = -  2 x x 0 is the maximum displacement v 0 is the maximum speed  = 2  /T  = 2  f This equation set works only for a mass which begins at x = +x 0 and is released from rest at t = 0 s. v0 = x0v0 = x0 x0x0 v0v0 x

23. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x.  Without deriving the other set in the Physics Data Booklet, here they are:  If you are interested, this last set is derived from a clockwise-rotation beginning as shown:  Data Booklet has highlighted formulas. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion x0x0 v0v0 x x = x 0 sin  t Set 2 - equations of simple harmonic motion v = v 0 cos  t a = -x 0  2 sin  t a = -  2 x x 0 is the maximum displacement v 0 is the maximum speed  = 2  /T  = 2  f This equation set works only for a mass which begins at x = 0 and is given a positive velocity v 0 at t = 0 s. v0 = x0v0 = x0

24. Define simple harmonic motion (SHM) and state the defining equation as a = -  2 x.  We need one more formula before we can practice.  Set 1: x = x 0 cos  t, v = -v 0 sin  t and v 0 = x 0  :  Begin by squaring each equation from Set 1: v 2 = (-v 0 sin  t) 2 = v 0 2 sin 2  t. v 0 2 = x 0 2  2, and x 2 = x 0 2 cos 2  t.  Then v 2 = v 0 2 sin 2  t becomes v 2 = x 0 2  2 sin 2  t.  sin 2  t + cos 2  t = 1 yields sin 2  t = 1 - cos 2  t  so that v 2 = x 0 2  2 (1 - cos 2  t) or v 2 =  2 (x 0 2 - x 0 2 cos 2  t)  Then v 2 =  2 (x 0 2 - x 2 ), which becomes Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion v =  (x 0 2 - x 2 ) relation between x and v

25. Apply the equations as solutions to the defining equation a = -  2 x. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion EXAMPLE: A spring having a spring constant of 125 n m -1 is attached to a 5.0-kg mass, stretched +4.0 m as shown, and then released from rest. (a) Using Hooke’s law, show that the mass-spring system undergoes SHM with  2 = k/m. SOLUTION:  From Hooke’s law F = -kx.  From Newton’s second law F = ma.  Thus ma = -kx or a = -(k/m)x.  The result of a = -(k/m)x is of the form a = -  2 x where  2 = k/m.  Therefore, the mass-spring system is in SHM. x

26. Apply the equations as solutions to the defining equation a = -  2 x. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion EXAMPLE: A spring having a spring constant of 125 n m -1 is attached to a 5.0-kg mass, stretched +4.0 m as shown, and then released from rest. (b) Find the angular frequency, frequency and period of the oscillation. SOLUTION:  Since  2 = k/m = 125/5 = 25 then  = 5 rad s -1.  Since  = 2  f, then f =  /2  = 5/2  = 0.80 Hz.  Since  = 2  /T, then T = 2  /  = 2  /5 = 1.3 s. x

27. Apply the equations as solutions to the defining equation a = -  2 x. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion EXAMPLE: A spring having a spring constant of 125 n m -1 is attached to a 5.0-kg mass, stretched +4.0 m as shown, and then released from rest. (c) Show that the position and velocity of the mass at any time t is given by x = 4 cos 5t and that v = -20 sin 5t. SOLUTION:  Note:  2 = k/m = 125/5 = 25 so  = 5 rad s -1.  Note: x 0 = 4 m, and v 0 = x o  = 4(5) = 20 m s -1.  At t = 0 s x = +x 0 and v = 0, so use Set 1: x = x 0 cos  t and v = -v 0 sin  t x = 4 cos 5t and v = -20 sin 5t. x

28. Apply the equations as solutions to the defining equation a = -  2 x. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion EXAMPLE: A spring having a spring constant of 125 n m -1 is attached to a 5.0-kg mass, stretched +4.0 m as shown, and then released from rest. (d) Find the position, the velocity, and the acceleration of the mass at t = 0.75 s. Then find the maximum kinetic energy of the system. SOLUTION:  x = 4 cos 5t = 4 cos 5(.75) = 4 cos 3.75 = -3.3 m.  v = -20 sin 5t = -20 sin 5(.75) = +11 m s -1.  a = -  2 x = -5 2 (-3.3) = 82.5 m s -2.  E K,max = (1/2)mv max 2 = (1/2)mv 0 2 = (1/2)(5)(20 2 ) = 1000 J. x

29. Solve problems, both graphically and by calculation, for acceleration, velocity and displacement during SHM. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion EXAMPLE: The displace- ment x vs. time t for a 2.5-kg mass is shown in the sinusoidal graph. (a) Find the period, the angular velocity, and the frequency of the motion. SOLUTION:  The period is the time for one complete cycle. It is T = 6.0  10 -3 s.  = 2  /T = 2  /0.006 = 1000 rad s -1. [1047]  f = 1/T = 1/0.006 = 170 Hz.

30. Solve problems, both graphically and by calculation, for acceleration, velocity and displacement during SHM. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion EXAMPLE: The displace- ment x vs. time t for a 2.5-kg mass is shown in the sinusoidal graph. (b) Find the velocity and acceleration of the mass at t = 3.4 ms. SOLUTION:  From the graph x = -0.8  10 -3 m at t = 3.4 ms.  From the graph x 0 = 2.0  10 -3 m. Thus  v =  (x 0 2 - x 2 ) = -1047 (0.002 2 - 0.0008 2 ) = -1.9 m s -1.  Finally a = -  2 x = -(1047 2 )(-0.0008) = +880 m s -2. Why is v negative? Because the slope is!

31. Solve problems, both graphically and by calculation, for acceleration, velocity and displacement during SHM. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion PRACTICE: The displacement x vs. time t for a 2.5-kg mass is shown in the sinusoidal graph. (a) Find the period and the angular frequency of the mass. Then find the maximum velocity.  The period is the time for one complete cycle. It is T = 6.0 s.  = 2  /T = 2  /6 = 1.0 rad s-1. [1.047]  v 0 = x 0  = 1.4(1.047) = 1.5 m s -1.

32. Solve problems, both graphically and by calculation, for acceleration, velocity and displacement during SHM. Topic 4: Oscillations and waves 4.1 Kinematics of simple harmonic motion PRACTICE: The displacement x vs. time t for a 2.5-kg mass is shown in the sinusoidal graph. (b) Find the force acting on the mass at t = 3 s. Then find it’s velocity at that instant.  Use F = ma where m = 2.5 kg.  At t = 3 s we see that x = -1.4 m.  Then a = -  2 x = -(1.047 2 )(-1.4) = 1.5 m s -2.  Then F = ma = 2.5(1.5) = 3.8 n.  Because the slope is zero, so is the velocity.