Vibrations and waves
Hooke’s Law
The force exerted on an object
attached to a spring is
proportional to the
displacement of the spring.
F = −kx
The constant of
proportionality is the
spring’s constant.
The force acts in opposite
direction to the displacement.
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Acceleration of the Object on a spring
Hooke’s Law gives the force
F = −kx
Newton’s law related the
force with the acceleration
F = ma
Thus
a=
F
k
=− x
m
m
Hooke’s Law: example
A 0.35 Kg object attached to a spring of constant 130 N/m is
free to move on a frictionless surface. If the object is
released from rest at x=0.10 m, find the force on it and
it’s acceleration for
a) x=0.10 m
b) x=0.05 m
c) x=0 m
d) x=-0.05 m
e) Sketch a graph of F(x)
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Simple Harmonic Motion
Simple harmonic motion is the type of motion that results
from a force that is proportional to the displacement but
opposite direction. It is periodic.
Amplitude A: magnitude of the maximum position of the
object relative to its equilibrium position. In the simple
harmonic motion the system will oscillate between +A and –A.
Period T: the time it takes for the object to complete a full
cycle. If the system starts at the maximum x=+A, a full cycle
involves going all the way to x=-A and back to x=+A again.
Frequency f: the number of cycles completed per unit time.
Simple Harmonic motion: question time
Two kids throw a ball to and fro repeating the same path over
and over again.
Is the motion of the ball periodic? (a) yes, (b) no.
Is it a simple harmonic motion? (a) yes, (b) no.
A block on the end of a spring is pulled to position x=A and
released. Through what total distance does it travel in a
full cycle?
a) A/2
b) A
c) 2A
d) 4A
In a simple harmonic motion the acceleration and the
displacement point in the same direction.
(a)True or (b) false
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Energy of the object and spring system
The potential energy
stored in a spring is
proportional to the
square of the
displacement
PE =
1 2
kx
2
a) Potential energy =0
b) Kinetic energy is
being converted into
potential energy
c) Potential energy is
maximum
Velocity in simple harmonic motion
One can determine the
velocity of the object from
energy conservation
1 2 1 2 1 2
kA = mv + kx
2
2
2
v=±
k 2
( A − x2 )
m
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Energy and velocity in SHM: example
A 0.35 Kg object attached to a spring of constant 130 N/m is
free to move on a frictionless surface. The object is
released from rest at x=0.10 m.
a) Calculate the total energy of the system
b) Calculate the maximum speed of the object
c) What is the velocity when the displacement is 0.05?
d) Compute the kinetic energy and the potential energy for
x=0.10 m, x=0.05 m, x=0 m, x=-0.05, x=-0.01.
e) Sketch a graph of KE(x) and PE(x)
Comparing SHM with UCM: derivation
The projection of the uniform circular motion (UCM) in
one of the axes reproduces the pattern of the simple
harmonic motion (SHM). Knowing the definition of the
period in UCM, prove that the period in SHM is given by
T = 2π
Useful relations
T=
2πA
v0
1 2 1
2
kA = mv0
2
2
m
k
One can easily derive the expression
for the angular frequency
ω = 2πf =
k
m
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Period and frequency: question time
Consider a mass M attached to a spring, with a spring
constant K, oscillating with period T on a frictionless
surface with amplitude A. You now replace that mass by
another with 2M and set it into oscillation with amplitude
A. What happens to the period?
a) 4T
b) 2T
c) T/2
d) none are correct
Did the total energy of the system change?
(a) yes, (b) no
Period and frequency: example
A 1300 Kg car is constructed on a frame supported by
four springs, each with k=20000N/m. If two people
riding in the car have a combined mass of 160 Kg, find
the frequency of free vibration of the car when it is
driven over a pothole in the road. How long does it
take to complete 3 periods?
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Position, velocity and acceleration
x = A cos θ
x = A cos(ωt )
θ = ωt
Period and frequency: example
Consider the system represented in
the figure and determine which
of the following statements is
correct.
a) If the amplitude of the oscillation
of the mass is doubled, the total
energy is doubled.
b) If the amplitude is doubled, the
maximum acceleration is
doubled.
c) In order to half the angular
frequency of oscillation, one has
to double the mass.
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Motion of a pendulum: derivation
Consider the forces in a
pendulum. Taking into account
the small angle approximation
derive the expression for the
period of a pendulum:
T = 2π
L
g
small angle approximation
sin θ = θ
The pendulum versus mass and spring system
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Pendulum: example
We want to measure the height of a very tall tower using
a cable attached to the roof. We find that it takes 15 s
for a 30 kg mass attached to the end of the cable to
performed a complete oscillation. Estimate the height
of the tower.
x
Waves
What is a wave?
An oscillation that is
both a function of
time and of
distance.
Snapshot:
behaviour with x
Looking at one point only
behaviour with t
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Waves
A function of time
and of distance.
Types Waves
Transverse: oscillation is
perpendicular to the direction
of propagation.
Longitudinal: oscillation
is in the direction of the
propagation
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Wavelength
λ
The wavelength is the distance
between two maximum points of the wave.
The wave speed is
v=
∆x λ
=
∆t T
λ
x
Wavelength: example
A wave traveling in the position x has frequency 25.0 Hz.
Find the amplitude, the wavelength, the period, and
the speed of the wave.
x
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Interference of waves
x
Interference of waves
Superposition principle:
if two or more traveling waves are moving through a
medium, the resulting wave is found by adding together the
displacements of the individual waves point by point
What happens when wave a) interferes with b) in both cases?
x
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Interference of waves
Constructive interference
Destructive interference
x
Reflection of waves
Rigid boundary
Free boundary
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