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13.1.1 Shm Part 2 Circular To Shm

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