Download PHYS 342: Modern Physics

XII. Periodic Motion
A. Introduction
1. Oscillations: motions that repeat themselves
a) Swinging chandeliers, boats bobbing at anchor, oscillating guitar strings,
pistons in car engines
2.
Understanding periodic motion essential for later study of
waves, sound, alternating electric currents, and light
3. An object in periodic motion experiences restoring forces or
torques that bring it back toward an equilibrium position
4. Those same forces cause the object to “overshoot” the
equilibrium position
XII.B. Simple Harmonic Motion (SHM)
1.
Definitions
a)
Frequency (f) = number of
oscillations that are completed
each second
[f] = hertz = Hz = 1 oscillation per sec = 1 s–1
b)
Period = time for one complete oscillation (or cycle):
T = 1/f
(XII.B.1)
XII.B. Simple Harmonic Motion (SHM)
2. Displacement x(t):
x(t) = xmcos(wt + f), where
(XII.B.2)
xm = Amplitude of the motion
(wt + f) = Phase of the motion
f = Phase constant (or phase angle) : depends on the initial
displacement and velocity
w = Angular frequency = 2p/T = 2pf (rad/s)
(XII.B.3)
3. Simple harmonic motion = periodic motion is a sinusoidal function of
time (represented by sine or cosine function)
XII.B. Simple Harmonic Motion (SHM)
4. velocity of a particle moving with SHM:
dx(t ) d
v(t ) 
 xm cos(wt  f )  wxm sin( wt  f ) (XII.B.4)
dt
dt
5.The acceleration for SHM:
dv (t ) d
a (t ) 
  w xm sin(wt   )  w 2 xm cos(wt   )  w 2 x(t )
dt
dt
(XII.B.5)
XII.C. Force Law for SHM
1.
From Newton’s 2nd Law:
F = ma = –mw2x
2.
(XII.C.1)
This result (a restoring force that is proportional to the
displacement but opposite in sign) is the same as Hooke’s
Law for a spring:
F = –kx, where k = mw2
k
w
(XII.C.3)
m
(XII.C.2)
m
T  2p / w  2p
k
(XII.C.4)
XII.D. Energy in SHM
1. Elastic potential energy
U = 1/2kx2 = 1/2kxm2cos2(wt + f)
(XII.D.1)
2. Kinetic energy
K = 1/2mv2 = 1/2kxm2sin2(wt + f)
(XII.D.2)
3.
Total mechanical energy = E = U + K
E = 1/2kxm2cos2(wt + f) + 1/2 kxm2sin2(wt + f);
= 1/2kxm2
(XII.D.3)
XII.E. Pendula
1.
A simple pendulum consists of a particle of mass m
(bob) suspended from one end of an unstretchable,
massless string of length L that is fixed at the other end
a) Consider the Forces acting on the bob:
q
L
s
^
q
W = mg
Fq = –mgsinq = mg(s/L); with
sinq = s/L
(XII.E.1)
q
b) If q is small ( 150 or so) then sinq  q:
Fq ~ –mgq = –mgs/L.
(XII.E.2)
XII.E. Pendula
1.
A simple pendulum consists of a particle of mass m
(bob) suspended from one end of an unstretchable,
massless string of length L that is fixed at the other end
c) This equation is the angular equivalent of the
condition for SHM (a = –w2 x), so:
w = (mg/L/ m)1/2 = (g/L)1/2 and
(XII.E.3)
T = 2p(L/g)1/2
(XII.E.4)
Example Problem #12
A pendulum bob swings a total distance of 4.0 cm
from end to end and reaches a speed of 10.0 cm/s at
the midpoint. Find the period of oscillation.
xm = 0.02 m; vm = 0.10 m/s
v(t )  w xm sin(wt   )
vm  (0.10ms 1 )  w xm  w (0.02m).
w  (0.10ms 1 ) / (0.02m)  5.0rad / s.
T  2p / w  2p / (5.0rad / s)  1.3s.