Download Mass-Spring Systems Mass-Spring Systems

Dulku – Physics 20 – Unit 4 – Topic D
Dulku – Physics 20 – Unit 4 – Topic D
Specific Outcome:
i. I can determine, quantitatively, the relationships among kinetic,
gravitational potential and total mechanical energies of a mass executing
simple harmonic motion.
ii. I can explain, quantitatively, the relationships among displacement,
acceleration, velocity and time for simple harmonic motion, as illustrated
by a frictionless, horizontal mass-spring system or a pendulum, using the
small-angle approximation.
Mass-Spring Systems
Mass-Spring Systems
Mass-Spring Systems
A mass-spring system exhibits simple
harmonic motion
It consists of a spring with one end connected
to a wall and a mass on the other end
Restoring force is the force exerted by the
spring to return the mass to its equilibrium
position
Hooke’s law is used with restoring force
Dulku – Physics 20 – Unit 4 – Topic D
Mass-Spring Systems
Remember that restoring force is given by:
where:
F = restoring force of the spring (N)
F = -kx
k = spring constant (N/m)
x = displacement of the mass or
deformation of the spring (m)
Restoring force may be related to
acceleration by F = ma (Newton’s 2nd law)
Dulku – Physics 20 – Unit 4 – Topic D
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Mass-Spring Systems
Mass-Spring Systems
The maximum speed for a mass-spring system
occurs at the equilibrium position (middle):
force (N)
A graph of force (F) vs. position (x) may be
given, as follows:
where:
vmax = maximum speed (m/s)
position (m)
vmax = A
The average spring constant, k, is given by
the slope (k = ∆F/∆x)
k
m
A = spring amplitude, from
one end to equilibrium (m)
k = spring constant (N/m)
m = mass (kg)
Dulku – Physics 20 – Unit 4 – Topic D
Dulku – Physics 20 – Unit 4 – Topic D
Mass-Spring Systems
LEFT (-)
Mass-Spring Systems
RIGHT (+)
Fnet = 0
m
m
m
m
m
x=0
ex. A spring with a constant of 55.0 N/m is stretched
+2.0 m. Determine the restoring force.
at rest (Equilibrium)
F = -kx = -(55.0 N/m)(2.0 m) = -1.1 x 102 N
F = -max
a = -max
v=0
x=+
F=0
a=0
v = -max
x=0
F = +max
a = +max
v=0
x=-
F=0
a=0
v = +max
x=0
Dulku – Physics 20 – Unit 4 – Topic D
ex. A force of +1.0 N is exerted by a spring to restore it
to equilibrium from a position of -3.0 cm.
Determine the spring constant.
x = -3.0 cm ÷ 100 = -0.030 m
-1.0 N
-F
F = -kx
k=
=
= 33 N/m
-0.030 m
x
Dulku – Physics 20 – Unit 4 – Topic D
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Mass-Spring Systems
ex. A heavy spring (spring constant = 50 N/m) is
attached to a mass of 0.250 kg.
a)Determine the acceleration of the mass-spring
system when the displacement is 0.10 m [right].
F = -kx = ma
a=
-kx -(50 N/m)(0.10 m)
=
m
0.250 kg
= -20 m/s2 = 20 m/s2 [left]
Mass-Spring Systems
ex. A heavy spring (spring constant = 50 N/m) is
attached to a mass of 0.250 kg.
b)Calculate the maximum speed of the mass if the
spring can stretch to a maximum distance of
0.500 m.
k
vmax = A
m
T = 2π
m
T = period of the mass-spring
system (s)
k
m = mass (kg)
0.250 kg
Dulku – Physics 20 – Unit 4 – Topic D
Mass-Spring Systems
where:
50 N/m
= 7.1 m/s
Dulku – Physics 20 – Unit 4 – Topic D
The period of a mass-spring system depends
only on the spring constant and mass:
= (0.500 m)
Mass-Spring Systems
ex. A heavy spring (spring constant = 50 N/m) is
attached to a hanging mass of 0.250 kg. Determine
the period of the mass-spring system.
T = 2π
m
k
= 2π
0.250 kg
50 N/m
= 0.44 s
k = spring constant (N/m)
Dulku – Physics 20 – Unit 4 – Topic D
Dulku – Physics 20 – Unit 4 – Topic D
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Mass-Spring Systems
ex. Find the spring constant if a 17 g mass takes 12
seconds to make 100 oscillations.
m = 17 g ÷ 1000 = 0.017 kg
T=
time
12 s
=
= 0.12 s
# cycles
100
T = 2π
m
k
k=
4π2m
T2
=
4π2(0.017 kg)
(0.12 s)2
= 47 N/m
Dulku – Physics 20 – Unit 4 – Topic D
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