Transcript Potential Energy - McMaster University

Oscillatory Motion
Serway & Jewett (Chapter 15)
Equilibrium position: no net force
M
The spring force is always directed
back towards equilibrium (hence called
the ‘restoring force’). This leads to an
oscillation of the block about the
equilibrium position.
M
F = -kx
M
x
For an ideal spring, the force is
proportional to displacement. For this
particular force behaviour, the oscillation
is simple harmonic motion.
SHM:
x  A cos(t   )
x(t)
A
T
A = amplitude
t
 = phase constant
 = angular frequency
-A
A is the maximum value of x (x ranges from +A to -A).
 gives the initial position at t=0: x(0) = A cos .
 is related to the period T and the frequency f = 1/T
T (period) is the time for one complete cycle (seconds).
Frequency f (cycles per second or hertz, Hz) is the number of
complete cycles per unit time.
2

 2f
T
units: radians/second or s-1
 (“omega”) is called the angular frequency of the oscillation.
Velocity and Acceleration
x(t )  A cos(t   )
dx
  A sin(t   )
v(t ) 
dt
dv
  A 2 cos(t   )   2 x
a(t ) 
dt
Not e: vMAX  A
aMAX  A 2
a(t)   2 x(t)
Position, Velocity and Acceleration
x(t)
t
v(t)
t
a(t)
t
Question:
Where in the motion is the velocity largest?
Where in the motion is acceleration largest?
Example:
SHM can produce very large accelerations if the
frequency is high.
Engine piston at 4000 rpm, amplitude 5 cm:
x  (5.00cm) cost
  4000 2 / 60 radians/sec
 419 sec 1
aMAX   2 x  0.05 m  (419s 1 ) 2
 8770m/s2
Simple Harmonic Motion
SHM:
x  A cos(t   )
We can differentiate x(t):
dx
dv
v , a
dt
dt
and find that acceleration is proportional to displacement:
a(t) =   2 x(t)
But, how do we know something will obey x=Acos(t) ???
Mass and Spring
Newton’s 2nd Law:
F   kx  ma
F = -kx
M
2
so
d x
k
a 2  x
dt
m
x
This is a 2nd order differential equation for the function x(t). Recall
that for SHM, a   2 x : identical except for the proportionality
constant. So the motion of the mass will be SHM:
x(t) = A cos (t + ), and to make the
equations for acceleration match, we require that
k , or
 
m
2

k
m
(and  = 2 f, etc.).
Note: The frequency is independent of amplitude
Example: Elastic bands and a mass. A mass, m, is attached to two
elastic bands. Each has tension T. The system is on a frictionless
horizontal surface. Will this behave like a SHO?
Quiz
The ball oscillates vertically on a single spring
with period T0 . If two identical springs are
used, the new period will be
A) longer
B) shorter
by a factor of
i) 2
ii) 2
iii) 4
Quiz
The ball oscillates vertically on a single spring
with period T0 . If two identical springs are
used, the period will be
A) longer
B) shorter
by a factor of
i)
ii)
iii)
iv)
2
2
4
1
Quiz
B
μs=0.5
ω=10 s-1
The amplitude of the oscillation gradually increases till block
B starts to slip. At what A does this happen?
(there is no friction between the large block and the surface)
a)
b)
c)
d)
Any A
5 cm
50 cm
Not known without mB.
Solution
Energy in SHM
Look again at the block & spring
M
K  12 mv 2  12 m 2 A2 sin 2 (t   )
U  12 kx2  12 kA2 cos 2 (t   )
 k!
K  U  12 A2 m 2 sin 2 (t   )  k cos 2 (t   ) 
 12 kA2 sin 2 (t   )  cos 2 (t   ) 
 12 kA2  a constant (total mechanical energy)
We could also write E = K+U = ½ m(vmax )2
E
U, K oscillate back and
forth “out of phase” with
each other; the total E is
constant.
K
U
t
T
x
t
v
n.b.! U, K go through
two oscillations while
the position x(t) goes
through one.
Suppose you double the amplitude of the motion:
1) What happens to the maximum speed?
a) Doubles
b) 4 x Larger
c) Doesn’t change
2) What happens to the maximum acceleration?
a) Doubles
b) 4 x Larger
c) Doesn’t change
3) What happens to the the total energy?
a) Doubles
b) 4 x Larger
c) Doesn’t change
Summary
SHM:
x  A cos(t   )
(get v, a with calculus)
Definitions: amplitude, period, frequency, angular
frequency, phase, phase constant.
The acceleration is proportional to displacement:
a(t) = -2 x(t)
Practice problems, Chapter 15:
3, 5, 11, 19, 23
(6th ed – Chapter 13)
1, 3, 5, 9, 19, 23