1. CHAPTER-15 Oscillations

2. Ch 15-2 Simple Harmonic Motion Simple Harmonic Motion (Oscillatory motion) back and forth periodic motion of a particle about a point (origin of an axis). Frequency f: Number of oscillations completed each second Period of the motion T= 1/f Displacement x of the particle about the origin is a sinusoidal function of time t given by : x(t)=x m cos(  t+  )  = 2  f = 2  /T

3. Ch 15-2 Simple Harmonic Motion

4. The Velocity of SHM V(t)=dx/dt= d/dt[x m cos(  t+  )] v(t)=dx/dt=-  x m sin(  t+  ) v(t)= =-v m sin(  t+  ) v m is maximum value of velocity and is called velocity amplitude The Acceleration of SHM a(t)=dv/dt= d/dt[-  x m sin(  t+  )] a(t) =-  2 x m cos(  t+  )=-  2 x In SHM a(t) =-  2 x a(t)  -x

5. Ch 15-3 Force Law for Simple Harmonic Motion Force Required for SHM F=ma =m(-  2 x)=-(m  2 )x=-kx familiar restoring force of Hook’s law: Spring force with spring constant k= m  2 Block-spring system forms linear simple harmonic oscillator with angular frequency  =  (k/m) ; Oscillation Period T=2  /  = 2  (m/k)

6.  Mechanical Energy E of a Simple Harmonic Oscillator:  E = K(t) +U(t), where K(t) and U(t) are kinetic and potential energies of the oscillator given by:  K(t)=mv 2 /2=[m  2 x 2 m sin 2 (  t+  )]/2  =[kx 2 m sin 2 (  t+  )]/2  U(t)=kx 2 /2=[kx 2 m cos 2 (  t+  )]/2  E=K(t)+U(t) = kx 2 m /2 Ch 15-4 Energy in Simple Harmonic Motion

7. Ch 15-5 An Angular Simple Harmonic Oscillator  Torsion Pendulum Disk of the pendulum oscillates in a horizontal plane with a restoring torque  =-  Then equation T=2  (m/k) modifies to T=2  (I/  ), where I is moment of inertia and  is torsion constant

8. Ch 15-6 Pendulums The Simple Pendulum consists of a particle of mass m (called bob of the pendulum) suspended from one end of an unstretchable, massless string of length L. The bob back and forth motion under a restoring torque  (  =r  F). Then  = -LF g sin  =-Lmgsin  = I  ; For small values of , sin  =  Then I  = =-Lmg  and  = -(Lmg/I)  =-  2  ;  2 = L mg/I T=2  /  = 2  (I/Lmg). For a simple pendulum I=mL 2 ; T=2  (L/g)

9. Ch 15-6 Pendulums  The Physical Pendulum  For a physical pendulum, the period T=2  /  = 2  (I/mgh)  where h is distance of center of mass from pivot point.  For a meterstick pivoted at one end I=ML 2 /3 and h=L/2  T=2  (2L/3g)