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# 13.1.1 Shm Part 2 Circular To Shm

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### 13.1.1 Shm Part 2 Circular To Shm

1. 1. 13.1.1 simple harmonic motion Part 2: translating circular motion to simple harmonic
2. 2. Simple harmonic motion <ul><li>Linear motion: a - constant in size + direction </li></ul><ul><li>Circular motion: a - constant in size only </li></ul><ul><li>Oscillatory motion: a changes periodically in size + direction (like x and  ) </li></ul><ul><li>SHM is a special form of oscillating motion </li></ul><ul><li>Pendulums and masses on springs exhibit SHM </li></ul><ul><li>A body oscillates with SHM if the displacement changes sinusodially </li></ul>
3. 3. Linking circular motion and SHM <ul><li>The arrangement shown below can be used to demonstrate the link between circular motion and SHM </li></ul>
4. 4. <ul><li>Adjusting the speed of the turntable will allow both shadows to have the same motion (i.e. move in phase) </li></ul><ul><li>The shadows are the components of each motion parallel to the screen and are sinusoidal </li></ul><ul><li>Let N represent the sinusoidal motion of one shadow </li></ul><ul><li>It oscillates about O (equilibrium point) in a straight line between A and B </li></ul>O N A B
5. 5. Displacement and velocity <ul><li>When N is left of O: </li></ul><ul><li>- x is left </li></ul><ul><li>-  is left when moving away from O and right when moving towards O </li></ul><ul><li>When N is right of O: </li></ul><ul><li>- x is right </li></ul><ul><li>-  is right when moving away from O and left when moving towards O </li></ul>
6. 6. <ul><li>The size of the restoring force increases with x BUT always acts towards equilibrium point (O) </li></ul><ul><li>F  - x </li></ul><ul><li> resulting acceleration must behave likewise, since F = ma and m is constant </li></ul><ul><li>i.e. a increases with x but acts towards O </li></ul><ul><li>a  - x </li></ul><ul><li>In oscillations a and x always have opposite signs </li></ul>
7. 7. Definition <ul><li>“ If the acceleration of a body is directly proportional to the distance from a fixed point, and is always directed toward that point, then the motion is simple harmonic” </li></ul><ul><li>a  - x or </li></ul><ul><li>a = -(+ve constant) x </li></ul><ul><li>Many mechanical oscillations are nearly simple harmonic, especially at small amplitudes </li></ul><ul><li>Any system obeying Hooke’s law will exhibit SHM when vibrating </li></ul>
8. 8. Equations of SHM <ul><li>Consider the ball rotating on the turntable </li></ul><ul><li>The ball moves in a circle of radius r </li></ul><ul><li>It has uniform angular velocity  </li></ul><ul><li>The speed, v around the circumference will be constant and equal to  r ( v =  r) </li></ul><ul><li>At time t the ball (and hence the bob of the pendulum) are in the positions shown: </li></ul>
9. 10. Displacement <ul><li>Angle  =  t (since  =  /t) </li></ul><ul><li>The displacement of the ball along OF from O is given by: x = r cos  = r cos  t </li></ul><ul><li>For the pendulum (and masses on springs), the radius of the circle is equal to the amplitude of its oscillation i.e. r = A </li></ul><ul><li>Hence x = A cos  t </li></ul><ul><li>But  = 2  f </li></ul><ul><li>x = A cos 2  f t </li></ul> r x
10. 12. Velocity <ul><li>The velocity of the pendulum bob is equal to the component of the ball’s velocity parallel to the screen (i.e. along y-axis) </li></ul>Bob Ball  v =  r O  r  velocity = - v sin  Ball Bob
11. 13. <ul><li>Velocity of bob = - v sin  = -  r sin  </li></ul><ul><li> =  /t   =  t </li></ul><ul><li>Velocity of pendulum bob,  = -  r sin  t </li></ul><ul><li>Sin  is +ve when 0 °    180° (i.e. bob or ball moving down) </li></ul><ul><li>Sin  is –ve when 180 °    360° (bob or ball moving up) </li></ul><ul><li>Negative sign ensures velocity is negative when moving down and positive when moving up! </li></ul>
12. 14. Variation of velocity with displacement <ul><li>sin 2  + cos 2  = 1 </li></ul><ul><li> sin  =  (1- cos 2  ) </li></ul><ul><li> = ±  r sin  </li></ul><ul><li>  = ±  r  (1- cos 2  ) </li></ul><ul><li>From earlier x = r cos  , so x /r = cos  </li></ul><ul><li> ( x /r) 2 = cos 2  </li></ul>
13. 15. <ul><li>By substituting: </li></ul><ul><li>Velocity = ±  r  (1- cos 2  ) </li></ul><ul><li>= ±  r  (1 - ( x /r) 2 ) </li></ul><ul><li>= ±   (r 2 – x 2 ) </li></ul><ul><li>Recall  = 2  f </li></ul><ul><li>Hence velocity of pendulum in SHM of amplitude A is given by: </li></ul><ul><li> = ± 2  f  (A 2 – x 2 ) </li></ul>
14. 16. Acceleration <ul><li>The acceleration of the bob is equal to the component of the acceleration of the ball parallel to the screen </li></ul><ul><li>The acceleration of the ball, a =  2 r towards O </li></ul><ul><li>So the component of a along OF =  2 r cos  </li></ul> Bob Ball a =  2 /r a =  2 r cos  O  O
15. 17. <ul><li>Hence, the acceleration of the bob is given by: </li></ul><ul><li>a = -  2 r cos  (-ve since moving down) </li></ul><ul><li>Since x = r cos  and  = 2  f </li></ul><ul><li>a = -(2  f ) 2 x </li></ul><ul><li>Since (2  f ) 2 or  2 is a +ve constant, equation states that acceleration of the bob towards the equilibrium point O is proportional to the displacement x from O </li></ul><ul><li>The acceleration is zero at O and maximum when the bob is at the limits of its motion when the direction and motion changes </li></ul>
16. 18. Time period <ul><li>Period T is time taken for the bob to complete one oscillation </li></ul><ul><li>In the same time the ball has made one revolution of the turntable </li></ul><ul><li> T = circumference of circle </li></ul><ul><li> speed of ball </li></ul><ul><li>T = 2  r </li></ul><ul><li>  </li></ul>
17. 19. <ul><li>Since  = r  </li></ul><ul><li>T = 2  </li></ul><ul><li> </li></ul><ul><li>For a particular SHM  is constant and independent of the amplitude (or radius) of the oscillation </li></ul><ul><li>If the amplitude increases, the body travels faster  T is unchanged </li></ul><ul><li>A motion with constant T, whatever the amplitude, is isochronous - this is an important characteristic of SHM </li></ul>
18. 20. Time traces of SHM <ul><li>Displacement </li></ul>T/4 T/2 3T/4 T Note: the gradient = velocity
19. 21. <ul><li>Velocity </li></ul>T/4 T/2 3T/4 T Note: when  = 0, a = max
20. 22. <ul><li>Acceleration </li></ul>T/4 T/2 3T/4 T a = 0 when  = max
21. 23. <ul><li>All graphs are sinusoidal </li></ul><ul><li>When the velocity is zero, the acceleration is a maximum and vice versa </li></ul><ul><li>There is a phase difference between them </li></ul><ul><li>Between  and a phase difference = T/4 </li></ul><ul><li>Between x and a phase difference = T/2 </li></ul>
22. 24. Summary: equations of SHM <ul><li>Frequency f =  /2  </li></ul><ul><li>Period T = 2  /  </li></ul><ul><li>Displacement x = A cos  t </li></ul><ul><li> = A cos 2  f t </li></ul><ul><li>Velocity  =    A 2 – x 2 </li></ul><ul><li>=  2  f  A 2 – x 2 </li></ul><ul><li>Acceleration a = -(2  f ) 2 x </li></ul>