Download kx F = The Spring

The Spring
❑Consider a spring, which we apply a force FA to
either stretch it or compress it
FA
x
unstretched
-FA
-x
x=0
FA = kx
k is the spring
constant, units of N/m,
different for different
materials, number of
coils
From Newton’s 3rd Law, the spring exerts a force
that is equal in magnitude, but opposite in
direction
Hooke’s Law for the restoring
force of an ideal spring. (It is
s
a conservative force.)
F = −kx
Chapter 4, problem P29
Five identical springs, each with stiffness 390 N/
m, are attached in parallel (that is side-by-side)
to hold up a heavy weight. If these springs were
replaced by an equivalent single spring, what
should be the stiffness of this single spring?
Oscillatory Motion
❑ We continue our studies of mechanics, but
combine the concepts of translational and rotational
motion.
❑ In particular, we will re-examine the restoring
force of the spring (later its potential energy).
❑ We will consider the motion of a mass, attached
to the spring, about its equilibrium position.
❑ This type of motion is applicable to many other
kinds of situations: pendulum, atoms, planets, ...
Simple Harmonic Motion
❑ If we add a mass m to the end of the (massless)
spring, stretch it to a displacement x0, and release
it. The spring-mass system will oscillate from x0 to
–x0 and back.
Without friction and air
m
resistance, the
oscillation would
k
continue indefinitely
-x0
0
x0
❑ This is Simple Harmonic Motion (SHM)
❑ SHM has a maximum magnitude of |x0|=A,
called the Amplitude
❑ One way to understand SHM is to consider the
circular motion of a particle and rotational
kinematics (The Reference Circle)
❑ The particle travels on a circle of radius r=A with
the line from the center to the particle making an
angle θ with respect to the x-axis at some instant in
time
• Now, project this 2D motion onto a 1D axis
x = A cosθ
but θ = ω t
x=-A
x=0
x=A
x
Therefore,
x = A cos(ω t )
❑ x is the displacement for
SHM, which includes the
motion of a spring
A
θ
A
❑ SHM is also called sinusoidal motion
❑ xmax=A=x0=amplitude of the motion (maximum)
• ω is the angular frequency (speed) in rad/s. It
remains constant during the motion.
❑ ω and the period T are related
• Define the frequency
•
2π
ω=
T
# cycles 1 Units of 1/s = Hertz (Hz)
f =
=
sec
T
2π
Relates frequency and
ω=
= 2π f
angular frequency
T
• As an example, the alternating current (AC) of
electricity in the US has a frequency of 60 Hz
• Now, lets consider the velocity for SHM, again
using the Reference circle
v = -v t sin θ
v t = rω = Aω
v = -Aω sin(ω t )
❑ Amplitude of the velocity
is
v max = A ω
x
vt
v
θ
A
❑ Acceleration of SHM
x
a = −ar cosθ
2
2 2
vt r ω
ar =
=
r
r
a
ar
θ
2
2
2
ar = rω = Aω ⇒ a = − Aω cos(ω t )
❑ The amplitude of the acceleration is
amax = Aω
❑ Summary of SHM
2
xmax = A, ω t = 0, π ,2π ,...
x = A cos(ω t )
π 3π
v = − Aω sin(ω t ) v max = Aω , ω t = , ...
2
2
2
a = − Aω cos(ω t ) a = Aω 2 , ω t = 0, π ,2π ,...
max
x
(A=1)
ωt (π rads)
v
(Aω=1)
ωt (π rads)
a
(Aω2=1)
Example Problem
Given an amplitude of 0.500 m and a frequency of
2.00 Hz for an object undergoing simple harmonic
motion, determine (a) the displacement, (b) the
velocity, and (c) the acceleration at time 0.0500 s.
Solution:
Given: A=0.500 m, f=2.00 Hz, t=0.0500 s.
(a)
ω = 2πf = 2π (2.00 Hz) = 4.00π rad/s
ωt = (4.00π rad/s)(0.0500 s)
= 0.200π rad = 0.628 rad
x = A cos(ωt ) = (0.500 m)cos(0.628 rad)
x = 0.405 m
v = - Aω sin(ω t )
v = −(0.500 m)(4π rad/s) sin(0.628 rad)
v = −3.69 m/s
2
a = Aω cos(ω t )
2
a = −(0.500 m)(4π rad/s) cos(0.628 rad)
2
a = −63.9 m/s
Frequency of Vibration
❑ Apply Newton’s 2nd Law to the spring-mass
system (neglect friction and air resistance)
k
m
0
Fs
x
A
∑F
= ma x
Consider x-direction only
Fs = −kx = ma x Substitute x and a for SHM
2
− k[ A cos(ω t )] = m[− Aω cos(ω t )]
2
k = mω
k Angular frequency of vibration for a
ω=
spring with spring constant k and
m attached mass m. Spring is assumed
x
to be massless.
1
ω = 2π f ⇒ f =
2π
1
m
T = = 2π
=T
f
k
k
m
Frequency of
vibration
Period of
vibration
❑ This last equation can be used to determine k
by measuring T and m
m
2 m
T = 4π
⇒ k = 4π 2
k
T
2
2
❑Note that ω (f or T) does not depend on the
amplitude of the motion A
• Can also arrive at these equations by
considering derivatives
The Simple Pendulum
❑ An application of Simple Harmonic Motion
❑ A mass m at the end of a massless rod of
length L
❑ There is a restoring force which
acts to restore the mass to θ=0
L
θ
F = −mg sin θ
m
• Compare to the spring Fs=-kx
• The pendulum does not
display SHM
T
mgsinθ
mg
❑ But for very small θ (rad), we can make the
approximation (θ<0.5 rad or about 25°)
→
simple pendulum approximation
sin θ ≈ θ ⇒ F = −mgθ This is SHM
since s = rθ = Lθ Arc length
s
mg Looks like spring
F = −mg = −
s force
L
L
mg Like the spring
( Fs = −kx) ⇒ k =
constant
L
❑ Now, consider the angular frequency of the spring
k
mg / L
g
ω=
=
=
=ω
m
m
L
1 g
Simple pendulum
f =
2π L frequency
Simple
pendulum
angular
frequency
• With this ω, the same equations expressing the
displacement x, v, and a for the spring can be used
for the simple pendulum, as long as θ is small
• For θ large, the SHM equations (in terms of sin
and cos) are no longer valid → more complicated
functions are needed (which we will not consider)
• A pendulum does not have to be a point-particle