Download Vibrations, springs, and Hooke`s Law

Vibrations and Springs
The force of a spring on a
mass varies with the
displacement, x, as
referenced from the
initial position, x0.
(a) When x is positive
(mass pulled to right for
this figure), the spring is in
tension and the restoring
force, Fs, is to the left.
(b) Spring in equilibrium at
x0. Restoring force, Fs, is
zero.
(c) When x is negative (mass
pushed to left for this figure), the
spring is in compression and the
restoring force, Fs, is to the right.
Vibrations, springs, and Hooke’s Law
The restoring force for
springs was discovered by
Robert Hooke in 1678 and
is know as Hooke’s Law.
Fs = - kx
k is a positive constant know as the spring stiffness
x is the displacement as measure from the equilibrium position
(sign convention: Fs is in the opposite direction of x, therefore “-” sign)
Harmonic Motion
Simple harmonic motion is the study of oscillations. An oscillation is motion of an
object that regularly repeats itself over the same path.
Examples: pendulum, mass vibrating on spring
Here, the restoring force is generated
by gravity trying to put the pendulum
back in equilibrium (vertical position).
Here, Fs = -kx
Harmonic Motion
An object is said to be in simple harmonic motion if the following occurs:
•It moves in a straight line.
•A variable force acts on it.
•The magnitude of force is proportional to the displacement of the mass.
•The force is always opposite in direction to the displacement direction.
•The motion is repetitive and a round trip, back and forth, is always made in
equal time periods.
Definition of terms for simple harmonic motion:
restoring force, Fs– force that restores an object to the equilibrium position
Cycle– one full repetition of periodic motion
Amplitude, A – maximum displacement from the equilibrium position during a cycle
Period, T – time it takes for an object to pass through one complete cycle
Frequency, f– number of cycles per unit of time
Period and frequency have an inverse relationship:
The spring constant, k
For a vertical spring-mass system, the equilibrium
point is when the restoring force, Fs, pulling
upwards is in equilibrium with the weight of the
mass pulling down, W = mg.
Using Hooke’s Law:
Fs = kx = mg
k = mg/x
In this case the acceleration of the mass is due to gravity.
Newton’s Law and Hooke’s Law
The springs we will study here exhibit a linear restoring force, Fs. This spring force is
a conservative force, and is a form of potential energy.
As a object moves in simple harmonic motion, we can use Newton’s 2nd Law (F =
ma) to calculate the acceleration of the mass in motion.
Equating Newton’s and Hooke’s Laws:
-kx = ma
Since amplitude, A, is maximum or minimum displacement, the range in
acceleration over the full harmonic cycle is:
-kA/m= +kA/m
Elastic Potential Energy
Elastic potential energy is Potential energy stored as a result of deformation of an
elastic object, such as the stretching of a spring. It is equal to the work done to stretch
the spring.
Force has the form: Fs = -kx
Work done to stretch the spring a distance x:
Elastic potential energy is proportional
to the spring constant, k, and the
square of the displacement, x2.
The system has zero PE at the
equilibrium position.
Harmonic Motion - Conservation of Energy
In a mass in harmonic motion, three types of energy act on the system:
1) Potential energy from the spring as it tries to restore the system to equilibrium
2) Gravitational potential energy
3) Kinetic energy of mass as it moves
At the maximum displacement (amplitude, A), the initial energy stored in the system:
PEinitial = kA2/2
After the spring is released, the energy of the system in harmonic motion is both the
elastic PE plus the KE:
PE + KE = kx2/2 + mv2/2
For equilibrium:
kA2/2 = kx2/2 + mv2/2
Relationship of velocity to position: v = ±{k(A2 –x2)/m}1/2
Example – 14.4
A 3 kg mass is attached to a spring and pulled out horizontally to a maximum
displacement from equilibrium of 0.5m. What spring constant must the spring
have if the mass is to achieve an acceleration equal to that of gravity?
Example – 14.19
A mass-spring system oscillates with an amplitude of 3.5 cm. If the spring
constant is 250 N/m and the mass is 0.5 kg, determine:
(a) The mechanical energy of the system
Example – 14.19
A mass-spring system oscillates with an amplitude of 3.5 cm. If the spring
constant is 250 N/m and the mass is 0.5 kg, determine:
(b) The maximum speed of the mass
Example – 14.19
A mass-spring system oscillates with an amplitude of 3.5 cm. If the spring
constant is 250 N/m and the mass is 0.5 kg, determine:
(c) The maximum acceleration of the mass
Example – 14.61
A mass of 1.5 kg is attached to a spring of spring constant 1000 n/m. Find the
force on the mass and its acceleration at the positions (equilibrium is at x = 0):
(a) 0.5 m
(b) 0.1 m
(c) 0 m
(d) -0.2m
(e) 0.5m