Spring-mass oscillators

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Transcript Spring-mass oscillators

SPRING-MASS
OSCILLATORS
AP Physics
Unit 8
Recall Hooke’s Law
• Applied force (Fapplied) stretches or
compresses spring from its natural length
• Restoring force (Fr) acts to return spring to
lowest energy state
Fr   Fapplied
Fr  kx
An energy approach to SHM
• Stretched/compressed spring stores
elastic potential energy: Us = ½kx2
• When released, mass oscillates about its
equilibrium position as PE  KE  etc
– Amplitude of oscillation is xmax
– At x=0, Us = 0 so K is maximized
– At x=A, Us is maximized, so K=0
Us, max = ½kA2
Example #1
• A 2.0 kg block is
attached to an ideal
spring with a force
constant of 500 N/m.
The spring is
stretched 8.0 cm and
released.
• When the block is 4.0
cm from equilibrium
– what is the total
energy of the system?
– what is the velocity of
the block?
Eo  U o  K o
Eo  1 kA2  0  1 (500)(0.08) 2
2
2
Eo  1.6 J
By energy conservation, E=1.6J at
every spring position
1 kx2  1 m v2  E
2
2
1 500(0.04) 2  1 (2.0)v 2  1.6 J
2
2
v  1.1m / s
An energy approach to SHM
Since energy is conserved and
– at x=A, Us is maximized and K=0
– at x=0, Us = 0 so K is maximized
2
it follows that 1 2 kA2  1 2 m vmax
therefore vmax
k
A
m
SHM and the Reference Circle
• Motion of the shadow cast by a particle
moving in a vertical circle mimics SHM
– Amplitude corresponds to the radius of the circle
– Period of the oscillation corresponds to the
period of the UCM
2R 2A
v

T
T
since vmax
or
k
A
m
T
2A
T 
v
2v m
v
k
 2 m
k
Example #2
A 2.0kg block is
attached to a spring
with a force constant
of 300 N/m.
Determine the period
and frequency of the
oscillations.
T  2 m
T  2 2.0
k
300
1
f   1.9 Hz
T
 0.51s
Vertical Spring-Mass Oscillators
• As it turns out, the behavior is the
same regardless of the orientation,
i.e. gravity does not affect the
period or frequency of the
oscillations.
• Sounds improbable? Let’s see why
it is not…
Vertical SMOs
• Consider a spring with constant k on
which a mass m is hung, stretching the
spring some distance x
• The spring is in equilibrium:
Fapplied= Fr or kx=mg
• If the spring is further displaced by some
amount A, the restoring force increases to
k(x+A) while the weight remains mg
Vertical SMOs
The net force on the block is now
F= k(x+A)- mg
But since kx=mg, the force on the
block is F= kA. This is Hooke’s Law!
Instead of oscillating about the natural
length of the spring as happens with a
horizontal SMO, oscillations of a
vertical SMO are about the point at
which the hanging mass is in
equilibrium!
Example #3
• A 1.5 kg block is attached
to the end of a vertical
spring with a constant of
300 N/m. After the block
comes to rest, it is
stretched an additional
2.0 cm and released.
– What is the frequency of
the oscillation?
– What are the maximum &
minimum amounts of
stretch in the spring?
1
2
1
f 
2
f 
k
m
300  2.3Hz
1.5
kx  m g
300x  (1.5)(9.8)
x  0.049m  4.9cm
Since A=2.0 cm, the spring is
stretched a maximum of 6.9 cm
and a minimum of 2.9 cm