Like this presentation? Why not share!

# 13.1.1 Shm Part 2 Circular To Shm

## by Chris Staines, student at Warwick University on Mar 24, 2009

• 1,031 views

### Views

Total Views
1,031
Views on SlideShare
1,031
Embed Views
0

Likes
0
38
1

No embeds

### Categories

Uploaded via SlideShare as Microsoft PowerPoint

### Report content

11 of 1 previous next

• anthonyarulappan hello this power point is so good pls forward it to my email id
anthonygsv@gmail.com
4 years ago
Are you sure you want to

## 13.1.1 Shm Part 2 Circular To ShmPresentation Transcript

• 13.1.1 simple harmonic motion Part 2: translating circular motion to simple harmonic
• 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
• Linking circular motion and SHM
• The arrangement shown below can be used to demonstrate the link between circular motion and SHM
• 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
• 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
• 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
• 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
• 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:
•
• 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
•
• 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
• 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!
• 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 
• 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 )
• 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
• 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
• 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
• 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
• Time traces of SHM
• Displacement
T/4 T/2 3T/4 T Note: the gradient = velocity
• Velocity
T/4 T/2 3T/4 T Note: when  = 0, a = max
• Acceleration
T/4 T/2 3T/4 T a = 0 when  = max
• 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
• 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