1. Oscillations x(t)=x m cos(  t+  ) v(t)=-  x m sin (  t+  ) v m =  x m ‘amplitude’ shifted by T/4 (90 0 ) a(t)=-  2 x m cos(  t+  ) a m =  2 x m ‘amplitude’ shifted by 2T/4 (180 0 ) d 2 x/dt 2 =-  2 x x(t) = x m cos(  t +  )

2. Example A body oscillates with SHM according to x(t)= (6.0m) cos( 3  t +  /3) At t=2.0 s, what are (a) the displacement, (b) the velocity,(c) the acceleration, (d) the phase of the motion,(e) the frequency, (f) the period ?

3. Solution x(t)= (6.0m) cos( 3  t +  /3) x m =6.0m,  =3  rads/s,  =  /3 rads (constants!) (a) x(t=2) = 6cos(6  +  /3)=6cos(60 0 )=3.0m (b) v(t)= -(3  )(6)sin(3  t +  /3) v(t=2)=-18  sin(6  +  /3)=- 18  3 1/2 /2 m/s (c) a(t) = - (3  ) 2 (6) cos( 3  t +  /3) a(t=2)= -54  2 cos(60 0 )= -27  2 m/s 2

4. Solution cont’d x(t)= (6.0m) cos( 3  t +  /3) (d) phase =  t+  = 3  t +  /3 = (19/3)  rads (e)  = 2  /T = 2  f ==> f =  / 2  = 1.5 s - 1 (f) T=1/f = 2/3 s

5. Force Law for SHM Newton’s second law F = m a ‘a’ non-zero ==> there is a force F = ma = m(-  2 x) = -m  2 x = - k x force  -displacement (in oppositedirection) Hooke’s law for springs with k = m  2 SHM d 2 x/dt 2 = -  2 x or F = - k x

6. a= -  2 x plus F=ma ==> F= -k x where k=m  2  =(k/m) 1/2 T= 2  /  = 2  (m/k) 1/2

7. Example A small body of mass 0.12 kg is undergoing SHM of amplitude 8.5 cm and period 0.20 s (a) what is maximum force?(b) if the motion is due to a spring, what is k? What do we have to know? (1) F=ma (2) F=-kx (3) a= -  2 x or in other words x(t)=x m cos(  t+  )

8. Solution (a) F max = m a max = m|  2 x m | (1)+(3)  =2  /T = 2  /.2 = 10  rads/s F max = (.12 kg)(.085m)(10  s -1 ) 2 =10. N (b) k= m  2 =(.12kg)(10  s -1 ) 2 =1.2x10 2 N/m (2)+(3)

9. P is projection of P` x(t)= x m cos (  t+  ) x=rcos  Uniform Circular Motion

10. v=r  v x (t)= -  x m sin (  t+  ) v x =-vsin  Velocity

11. a=v 2 /r=r  2 =x m  2 a x (t)= -  2 x m cos (  t+  ) acceleration

12. Uniform Circular Motion Uniform circular motion r(t) = x(t) i + y(t) j x(t)= x m cos(  t+  ), y(t)= x m sin(  t+  ) r(t) is the sum of SHM along perpendicular directions

13. Problem Two particles execute SHM of the same amplitude and frequency along a straight line.They pass each other moving in opposite directions each time their displacement is half their amplitude. What is the phase difference between them? Where are the particles on the circle when they pass?

14. Solution x(t)=x m cos(  t+  ) x(t) = x m /2 when phase angle = ± 60 0   phase difference= 120 0

15. Energy in SHM As particle oscillates, its speed varies and hence so does its kinetic energy K Where does the energy go when speed is zero? If the oscillation is produced by a spring, the spring is compressed or stretched to some maximum amount when speed is zero Energy is in the form of potential energy U

16. Energy in SHM U(t) = (1/2) k x 2 = (1/2)k x m 2 cos 2 (  t+  ) K(t) = (1/2) mv 2 = (1/2) m  2 x m 2 sin 2 (  t+  ) using k = m  2 and cos 2  + sin 2  =1 E(total) = U(t)+K(t) =(1/2) k x m 2 (constant!) =(1/2) m  2 x m 2