![Page 1: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/1.jpg)
Oscillatory Motion
Serway & Jewett (Chapter 15)
![Page 2: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/2.jpg)
M 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
For an ideal spring, the force is proportional to displacement. For this particular force behaviour, the oscillation is simple harmonic motion.
x
F = -kx
![Page 3: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/3.jpg)
)cos( tAxSHM: x(t)
t
A
-A
T
A = amplitude
= phase constant
= angular frequency
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.
![Page 4: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/4.jpg)
2 2
fT
units: radians/second or s-1
(“omega”) is called the angular frequency of the oscillation.
![Page 5: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/5.jpg)
Velocity and Acceleration
2MAX
MAX
22
:Note
)cos()(
)sin()(
)cos( )(
Aa
Av
xtAdt
dvta
tAdt
dxtv
tAtx
a(t) 2 x(t)
![Page 6: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/6.jpg)
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?
![Page 7: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/7.jpg)
Example:
SHM can produce very large accelerations if the frequency is high.
Engine piston at 4000 rpm, amplitude 5 cm:
m/s 8770
)s 419(m 05.0
sec 419
cradians/se 60/24000
cos)cm00.5(
2
212MAX
1
xa
tx
![Page 8: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/8.jpg)
Simple Harmonic Motion
SHM: )cos( tAx
a(t) = 2 x(t)
dtdv
adtdx
v , We can differentiate x(t):
and find that acceleration is proportional to displacement:
But, how do we know something will obey x=Acos(t) ???
![Page 9: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/9.jpg)
Mass and Spring
xmk
dtxd
a 2
2 M
x
F = -kxmakxF Newton’s 2nd Law:
so
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
mk2 , or
mk (and = 2f, etc.).
Note: The frequency is independent of amplitude
![Page 10: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/10.jpg)
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?
![Page 11: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/11.jpg)
Quiz
The ball oscillates vertically on a single spring with period T0 . If two identical springs are used, the new period will be
A) longerB) shorter
by a factor of
i) ii) 2iii) 4
2
![Page 12: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/12.jpg)
Quiz
The ball oscillates vertically on a single spring with period T0 . If two identical springs are used, the period will be
A) longerB) shorter
by a factor of
i) ii) 2iii) 4iv) 1
2
![Page 13: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/13.jpg)
Quiz
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) Any Ab) 5 cmc) 50 cmd) Not known without mB.
Bμs=0.5
ω=10 s-1
![Page 14: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/14.jpg)
Solution
![Page 15: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/15.jpg)
Energy in SHM
MLook again at the block & spring
energy) mechanical (totalconstant a
)(cos)(sin
)(cos)(sin
!
)(cos
)(sin
221
22221
222221
22212
21
222212
21
kA
ttkA
tktmAUK
k
tkAkxU
tAmmvK
We could also write E = K+U = ½ m(vmax )2
![Page 16: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/16.jpg)
U, K oscillate back and forth “out of phase” with each other; the total E is constant.
n.b.! U, K go through two oscillations while the position x(t) goes through one.
U K
E
t
T
x
v
t
![Page 17: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/17.jpg)
1) What happens to the maximum speed?a) Doublesb) 4 x Largerc) Doesn’t change
2) What happens to the maximum acceleration?a) Doublesb) 4 x Largerc) Doesn’t change
3) What happens to the the total energy?a) Doublesb) 4 x Largerc) Doesn’t change
Suppose you double the amplitude of the motion:
![Page 18: Oscillatory Motion Serway & Jewett (Chapter 15). M Equilibrium position: no net force M The spring force is always directed back towards equilibrium (hence](https://reader035.vdocuments.us/reader035/viewer/2022072017/56649efd5503460f94c120cd/html5/thumbnails/18.jpg)
Summary
SHM: )cos( tAx
Definitions: amplitude, period, frequency, angular frequency, phase, phase constant.
(get v, a with calculus)
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