1 / 17

Oscillations

Oscillations. x(t) = x m cos(  t +  ). 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. Example.

maire
Download Presentation

Oscillations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Oscillations x(t) = xm cos(t + ) • x(t)=xm cos(t+) • v(t)=- xm sin (t+ ) • vm= xm ‘amplitude’ • shifted by T/4 (900) • a(t)=- 2xm cos(t+ ) • am = 2xm ‘amplitude’ • shifted by 2T/4 (1800) • d2x/dt2 =- 2x

  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) • xm=6.0m, =3 rads/s,  =/3 rads (constants!) • (a) x(t=2) = 6cos(6+ /3)=6cos(600)=3.0m • (b) v(t)= -(3)(6)sin(3t + /3) v(t=2)=-18sin(6+/3)=-1831/2/2 m/s • (c) a(t) = - (3)2(6) cos( 3t + /3) a(t=2)= -54 2cos(600)= -27 2m/s2

  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(- 2x) = -m2x = - k x • force  -displacement (in oppositedirection) • Hooke’s law for springs with k = m 2 • SHM d2x/dt2 = - 2x • or F = - k x

  6. a= - 2x 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= - 2x • or in other words x(t)=xm cos(t+)

  8. Solution • (a) Fmax = m amax = m|2xm | (1)+(3) • =2/T = 2/.2 = 10  rads/s • Fmax = (.12 kg)(.085m)(10 s-1)2 =10. N • (b) k= m2 =(.12kg)(10s-1)2 =1.2x102N/m (2)+(3)

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

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

  11. a=v2/r=r2=xm 2 acceleration ax(t)= -2 xmcos(t+)

  12. Uniform Circular Motion • Uniform circular motion • r(t) = x(t) i + y(t) j • x(t)= xm cos(t+), y(t)= xm sin(t+) • r(t) is thesum 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)=xm cos(t+) • x(t) = xm /2 when phase angle = ± 600   phase difference= 1200

  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 x2 = (1/2)k xm2 cos2(t+) • K(t) = (1/2) mv2 = (1/2) m2 xm2 sin2(t+ ) • using k = m 2and cos2 + sin2 =1 • E(total) = U(t)+K(t) =(1/2) k xm2 (constant!) =(1/2) m2 xm2

More Related