Download Simple Harmonic Motion Serway Chapter 15.1, 15.2

Simple Harmonic Motion
Serway Chapter 15.1, 15.2
Practice: Chapter 15, problems 5, 7, 8, 15, 65
Oscillatory Motion
Motion in the real world may not fit some of our earlier models
(linear or circular motion, uniform acceleration).
Many phenomena are repetitive or oscillatory.
Example: Block and spring
M
Equilibrium: no net force
M
The spring force is always directed back
towards equilibrium. 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.
Simple Harmonic Motion
x(t)
t
In Simple Harmonic Motion (SHM), the displacement is a
sinusoidal function of time, e.g.,:
x ( t ) = A cos ω t
or
x ( t ) = A sin ω t
Question: Is a bouncing ball described by SHM?
x(t)
t
In general,
x ( t ) = A cos( ω t + φ )
Phase
The quantity (ωt + φ) is called the phase, and is measured in radians.
The cosine function traces out one complete cycle when the phase
changes by 2π radians. The phase is not a physical angle!
Three constants specify the motion:
Amplitude, A : maximum displacement from the centre
Angular Frequency, ω
Initial phase (or phase constant), φ
QUIZ
A weight suspended from a spring oscillates up and
down so that its height above the floor varies between
30 cm and 50 cm. The amplitude of the oscillation is:
A)
B)
C)
D)
E)
10 cm
20 cm
30 cm
40 cm
50 cm
The period T of the motion is the time needed to complete one cycle:
x(t) = A cos (ωt+φ) = A cos (ωt+φ+2π)
The motion repeats when the phase (ωt + φ) increases by 2π radians;
x (t + T) = x(t) if ωT = 2π radians (or 360°).
2π
ω=
= 2π f
T
units: radians/second or s-1
ω (“omega”) is called the angular frequency of the oscillation.
The initial phase φ determines where in the motion we set t = 0:
x
x(t) = A cos (ωt )
t
0
x
0
x(t) = A cos (ωt + 45°)
t
x
x(t) = A cos (ωt− 45°)
0
t
Velocity and Acceleration
x (t ) = A cos( ω t + ϕ )
dx
v (t ) =
= − A ω sin( ω t + ϕ )
dt
dv
a (t ) =
= − A ω 2 cos( ω t + ϕ ) = −ω 2 x
dt
Note : v MAX = A ω
a MAX = A ω 2
a(t) =− ω 2 x(t)
Velocity and Acceleration
x(t)
t
v(t)
t
a(t)
t
Example:
SHM can produce very large accelerations if the
frequency is high.
Engine piston at 4000 rpm, amplitude 5 cm:
x = (5.00cm) cosωt
ω = 4000× 2π / 60 radians/sec
= 419 sec−1
aMAX = 0.05 m × (419 s −1 ) 2
= 8770 m/s2
Example
The block is displaced a distance x = 5 cm
from its equilibrium position and released from
rest at time t = 0. Its motion is SHM with
period 2 seconds. Write the function x(t).
Steps:
1) Sketch a graph.
2) Write x = A cos (ωt + φ).
3) Keeping an eye on the graph, evaluate A, ω, and φ.
M
x
Example
The block is at its equilibrium position and
is set in motion by hitting it (and giving it
an initial velocity) at time t = 0. Its motion
is SHM with amplitude 5 cm and period 2
seconds. Write the function x(t).
v0
M
x
QUIZ
The block is at x0 = +5 cm, with positive
velocity v0, at time t = 0. Its motion is SHM
with amplitude 10 cm and period 2
seconds. If x(t) = A cos (ωt + φ), the phase
constant φ should be:
A)
B)
C)
D)
E)
0o
30o
60o
-30o
-60o
M
x0
v0
Quiz
B
µs=0.5
ω=10 s-1
The amplitude of the oscillation gradually increases
until block B starts to slip. At what amplitude A does
this happen?
a)
b)
c)
d)
1cm
5 cm
50 cm
Not known without mB.
Summary
SHM:
x = A cos(ω t + φ )
(get v, a from calculus)
Definitions: amplitude, period, frequency, angular
frequency, phase, phase constant.
The acceleration is proportional to displacement:
a(t) = -ω2 x(t)