Understanding Simple Harmonic Motion: Concepts, Energy, and Oscillation Types

1. Physics 211 12: Oscillatory Motion • Simple Harmonic Motion • Energy of a Simple Harmonic Oscillator • The Pendulum • Comparing Simple Harmonic Motion and Uniform Circular Motion • Damped Oscillations • Forced Oscillations

2. Simple Harmonic Motion Hookes Law F = -kx F = restoring force x = displacement from equilibrium position

3. x=-A x=0 X=A

4. 2 d x ( ) ( ) ( ) F t = - kx t = ma t = m dt 2 2 d x k Û = - x dt m 2 Differential Equation, need to find solution so that left hand side = right hand side From looking at graph of position versus time ( ) Guess : x t = A cos( w t ) ; A and w constants .

6. What is the meaning of the constant A at time t = 0 ( ) ( ) x 0 = A cos 0 = A which is the displacement from equilibrium at time t = 0 p at time t = w p p æ æ ö ö ( ) x = A cos w = A cos p = - A è ø è ø w w The max and min values of x are ± A Û A is the amplitude of the motion the maximum displacement from equilibrium position

7. cos(x) +1   cos(2n) = +1 -1 cos([2n+1]) = -1

8. Meaning of w 2 n p for t = Þ x = + A w ( ) 2 n + 1 p for t = Þ x = - A w motion repeats between ± A Harmonic Motion If it repeats itself exactly Simple Harmonic Motion (SHM) ( ) ( ) 2 n + 2 p 2 n p 2 p angle changes Time between repeats = - = ( 2 p ) w w w This is called the Period of the Oscillation, T Angular Frequency Rate of change of angle with time = 2 p w = = T

9. The number of times motion repeats in 1 second 1 w is the frequency f = = T 2 p rad [ ] [ ] f = Hz º cps; w = º s - 1 s 2 p m T = = 2 p w k When is the velocity the greatest/least When is the acceleration the greatest/least

10. ( ) 2 n p 2 n + 1 p x = ± A Û t = ; t = w w ( ) ( ) at these times the velocity v t = - A w sin w t Þ 2 n p 2 n p æ æ ö ö ( ) v = - A sin w = - A sin 2 n p = 0 è ø è ø w w ( ) ( ) æ 2 n + 1 p æ 2 n + 1 p ö ö ( ) ( ) v = - A sin w = - A sin 2 n + 1 p = 0 è ø è ø w w so velocity is zero at maximum displacement the acceleration on the other hand is a maximum ( ) 2 n p æ 2 n + 1 p æ ö ö a = - w A ; a = w A 2 2 è ø è ø w w The acceleration is zero when x = 0 The velocity is the greatest when x = 0 v = ± A w max

11. Energy of a Simple Harmonic Oscillator Mass experiences spring force, thus its P . E is 1 U ( x ) = kx 2 2 The spring force is a conservative force The total energy of the mass is 1 1 1 1 ( ) 2 ( ) 2 E = K + U = m v + kx = m v t + kx t = constant 2 2 2 2 2 2 tot \ The total energy when v = 0 is equal to 1 E = kA 2 2 tot which must be its value at ALL TIMES ! 1 1 1 ( ) Þ E t = m v + kx = kA 2 2 2 2 2 2 tot 1 1 k Þ when x = 0 , K = m v = kA Þ v = A 2 2 max max 2 2 m

12. The Pendulum  l T s mg cos mg sin W=mg

13. Restoring Force -mg when q F = - mg sin q » q is small s mg \ F = - mg q = - mg = - s l l º Hookes Law for the Pendulum 2 d s , This force provides the tangential acceleration 2 dt and we obtain a similar differential equation to before . Comparing to before we see mg k « & x « s l ( ) ( ) s t = A cos w t m ml l T = 2 p = 2 p = 2 p k mg g

14. The Physical Pendulum d d x W  W=mg Note that the pivot point could be inside the boundaries of the object

15. We formulate this for an objectsuspended so that its center of gravity is at a distance d from the pivot point, by using angular quantities: The restoring torque due to gravity t = - mgd sin q » - mgd q (small angle approx . ) using t I a = 2 d q mgd mgd 2 Þ a = = - q = - w q , where w = 2 dt I I ( ) ( ) Þ q t = A cos w t I Þ T = 2 p mgd

16. The Torsion Pendulum By suspending a mass at the end of a wire supported tightly at the other end, we make a torsion pendulum. By twisting the object through a small angle we produce a restoring torque d q 2 t = - kq = I a = I dt 2 2 d q k Þ = - q dt I 2 I T = 2 p k

17. Comparing Simple Harmonic Motion and Uniform Circular Motion ( ) ( ) x t = A cos w t is precisely the time variation of the x coordinate of a particle performing uniform circular motion about a fixed point at a fixed distance A . æ p ö ( ) ( ) y t = A sin w t = A cos w t - is the time variation è ø 2 of the y coordinate, which is a SHM variation . p Here though we have a phase shift of j = - 2 p The argument w t - is called the phase of the motion . 2

18. Damped Oscillations If there are frictional forces present D E = W < 0 tot nc Thus the total energy decreases and becomes a non constant function of time, E ( t ) ¹ constant. 1 ( ) 2 Þ E ( t ) = kA t 2 ( ) Þ A t decreases with time The differential equation describing the position of a particle undergoing damped SHM is of the form d x dx 2 m = - kx - b dt dt 2

19. Forced Oscillations If there is another external oscillating force acting on the object (in the direction of motion ) one says that the motion of the oscillator is forced by this external force. The differential equation describing such motion is d x dx 2 ( ) m = - kx - b + F cos w t dt dt 0 2

20. solutions to this equation give amplitudes of the form F 0 m A ( t ) = } { ( ) 2 ( ) b w 2 w - w + 2 2 m 0 w is the frequency of the SHM (i . e . no friction 0 and forcing oscillation ) If b (friction ) is small, then if w » w the amplitude 0 becomes larger and larger º RESONANCE