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Vibrations and Oscillations
A mass on a spring will oscillate if the mass is
pushed or pulled from its equilibrium
position. Why?
We saw from the Hooke’s Law experiment
that the force of a spring is related to how
far the spring is pulled or pushed from
equilibrium: Fspring = -kx where the minus
simply indicates that if you push in on the
spring, the spring pushes out.
Springs
The spring constant, k, describes how “stiff”
the spring is. A large k indicates that a
large force is needed to stretch the spring.
But why does the mass oscillate?
From Newton’s Second Law, if we ignore
other forces like friction or gravity:
S F = ma and F = -kx leads to -kx = ma .
Oscillations of Springs
-kx = ma
This says that if x is positive (spring stretched), the
force is negative and hence the acceleration is
negative. This will produce a velocity that, if
initially zero, will become negative, tending to
reduce the positive x. As x approaches zero, the
equilibrium position, the force approaches zero
and the acceleration will approach zero; but this
still leaves a negative velocity!
Oscillations of Springs
Now as it passes the equilibrium position, the mass
has negative velocity so it will pass through the
equilibrium position and end up with a negative
(compressed) position. Here the force
(F=-kx) becomes positive giving a positive
acceleration. This will make the negative velocity
less negative, but still negative. Hence the mass
will go to an even more negative position, with an
even bigger negative force and acceleration that
will continue to slow it down until it reaches zero
speed.
Oscillations of Springs
At this point where the speed is zero, we have a
negative (compressed) position which gives a
positive force and hence positive acceleration.
This acceleration will then cause the speed to
increase to a positive value which will cause the
mass to move back towards equilibrium.
This process continues and we get an
oscillation!
Formulas for Oscillations
Using the calculus, we can solve -kx = ma
(a differential equation: -kx = md2x/dt2) to
get: x = A sin(wt + qo) ,
where w =(k/m) comes from substituting the
solution back into the differential equation .
This should appear reasonable: the sine
function oscillates (goes up and down in
value) just like the mass oscillates! But the
sine function needs an angle to operate on!
Where is the angle in the problem?
Angles: geometric and phase
x = A sin(wt + qo) .
The sine function is operating on the quantity:
(wt + qo). This expression must be an
angle. But what angle?
There is no “geometric” angle in the problem
because the problem is only in one dimension.
Instead, we call this kind of angle a phase
angle. A phase angle simply describes
where in the oscillation the wave is!
Sine and Phase Angles
Sine function
Value of
sine
The crest of the
sine wave is
located at 90o,
1
the trough at 270o
0
-1
and it crosses zero
at 0o, 180o and
starts repeating
at 360o (or 2p radians).
Phase angle in degrees
Series1
Phase angle frequency
x = A sin(q)
where q is a phase angle
in an oscillation; q changes with time
(goes 2p radians in time T) so
q = wt + qo = (2p/T)*t + qo
where
2p/T = w = 2pf ; and
qo is simply the phase angle when t=0
(where the oscillation starts at t=0).
Amplitude
x = A sin(wt + qo) .
The amplitude, A, describes how far up and
how far down x goes. Since sine has a
maximum value of 1 and a minimum value
of -1, A is used to put in units and to give
the amplitude of the oscillation.
What does w depend on?
x = A sin(wt + qo) .
We have seen that w describes how fast the
mass oscillates. But what does this
oscillation (w) depend on?
By substituting our solution into the
differential equation, we get: w = (k/m) .
For stiffer springs and lighter masses,
the frequency of the oscillation increases.
Note: the Amplitude does NOT affect the
frequency!
Amplitude and frequency
w = (k/m)
Why doesn’t the Amplitude affect the frequency?
The more you stretch the spring (bigger
Amplitude), the farther the oscillation has to go.
However, you also have bigger forces which mean
bigger accelerations and bigger speeds. Which
wins, the bigger distance or the faster speeds? The
math tells us that it is a tie! The Amplitude does
NOT matter here!
Oscillations and Energy
For the energy of masses on springs, we have:
Etotal = KE + PE = (1/2)mv2 + (1/2)kx2 .
We also know that x = A sin(wt + qo) .
If we look at how x changes with time, we get
the speed: v = dx/dt = d[A sin(wt+qo)]/dt =
= wA cos(wt + qo) . Therefore, Etotal =
(1/2)mw2A2cos2(wt+qo) + (1/2)kA2sin2 (wt+qo) .
Note: when sine is maximum, cosine is zero;
and when sine is zero, cosine is maximum!
Does Etotal change with time, or is it a constant?
Oscillations and Energy
Etotal =
(1/2)mw2A2cos2(wt+qo) + (1/2)kA2sin2 (wt+qo) .
Recall that w = (k/m) . Thus,
(1/2)mw2A2 = (1/2)kA2 and since sin2q+cos2q = 1,
we see that Etotal = (1/2)mw2A2 = constant .
Note that the energy of oscillation depends on
the frequency squared and the amplitude
squared.
The Pendulum
In the Oscillations Lab in Physics I, we
experimentally developed equations to
describe the oscillations of a pendulum.
In this case, we have a real geometric angle
AND we have a phase angle. The
geometric angle (the angle the pendulum
makes with the vertical) oscillates, so we
might guess a solution:
qpendulum(t) = qmax sin(wt+fo).
The Pendulum
In the lab we found that the period depended
on the square root of the length, but not on
the mass and only slightly on the angle.
If we start from Newton’s Second Law for
rotations: St = Ia, we get:
-mg sin(q) L = mL2 d2q/dt2 .
For small angles, sin(q)  q (where the angle
is measured in radians), so we now have:
- g q = L d2q/dt2 .
The Pendulum
- g q = L d2q/dt2 But this looks just like
-kx = md2x/dt2 and so has a similar
solution:
q(t) = A sin(wt + fo) where
w = [g/L] = 2p/T , where the expression
for w came from substituting our solution
back into the differential equation.
This says that if we don’t have too big an
oscillation, the period does not depend on
the mass or angle, but only on the L !
Waves on a string
If you wiggle one end of a string (or slinky)
that has a tension on it, you will send a
wave down the string. Why?
We can consider this by looking at Newton’s
Second Law:
Waves on a string
If we consider the first part of the wave, we
see that the Tension on the right has a zero
y component, but the Tension on the left is
pulling up. Thus, S Fy = Tleft-y + Tright-y > 0.
This will cause this part of the string to start
moving upwards! [Note: Tleft-x = Tright-x]
Waves on a string
A little later, this part of the string is moving
upwards, but now the Tension on the left
has less of a y component and the Tension
on the right: Tleft-y + Tright-y < 0 . This will
tend to slow this part of the string down!
[Again, Tleft-x = Tright-x ]
Waves on a string
At the top of the pulse, the wave has slowed
to zero, but both the left and right Tensions
are pulling down, so there will continue to
be a negative y force, and this will cause
this part of the string to start moving down.
Waves on a String
By continuing to look at each part of the
string, we can understand how the wave can
move down the string.
Starting with Newton’s Second Law for a
piece of mass on the string, we have:
SFy = +Tsin(qL) - Tsin(qR) = Dmay .
[here T is the tension in the string.]
Note: for small angles, sin(q)  tan(q)  q,
so : sin(qL) - sin(qR) = Dsin(q)  Dtan(q).
Waves on a String
T Dtan(q) = Dm ay .
Further, tan(q) = dy/dx. Next, let’s divide
both sides by Dx, so with Dm/Dx = m we
have: T d2y/dx2 = m ay , or
T 2y/x2 = m 2y/t2 .
The above equation is called the wave
equation for a string. Recall that it comes
directly from Newton’s Second Law.
Waves on a String
T 2y/x2 = m 2y/t2
The solution to this 2nd order partial differential
equation is f(kx  wt + qo), but we’ll choose the
sine function for our function, f. Fourier Series
theory says any periodic function can be expressed
as a series of different sine waves, so our choice of
the sine function really doesn’t limit us.
y(x,t) = A sin(kx  wt + qo) .
Waves on a String
Wave Equation: T 2y/x2 = m 2y/t2
Solution: y(x,t) = A sin(kx  wt + qo) .
– where 2p/T = w (when t=T, wt=2p) and
– where 2p/ = k (when x=, kx=2p).
Note that this k is the wave vector, NOT the
spring constant!
Note: we also have two T’s: the T for tension
and the T for period!
Waves on a String
Tension 2y/x2 = m 2y/t2
y(x,t) = A sin(kx  wt + qo)
If we substitute this solution into our equation,
we get: Tensionk2 = mw2 , or
w/k = [Tension/m] .
Does w/k have any special meaning?
Phase speed of a wave
q(x,t) = (2p/)*x ± (2p/Tp)*t = kx ± wt + qo
If we consider how the crest moves (or any
other particular phase), then q = constant
(= 90o for the crest) and we have for the
position of that phase angle in time:
xq = (q  wt - qo)/k and so the phase speed =
vq = dxq/dt =  w/k . Also,
vq = distance/time = /Tp = f = w/k .
Wave velocity
We saw previously that w/k = [Tension /m] ,
so now we have an expression for the phase
velocity of the wave on a string:
vq = f = w/k = (Ttension)/m)
where m = m/L (mass/length of string).
Note that there are two velocities in this case:
the velocity of a particular phase (such as
a crest, vq=dxq/dt), and the velocity of a
particular part of the string (going up and
down, vy=dy/dt).
Two velocities
The velocity in the relations vq =f and
vq =(Ttension)/m) is the phase velocity
(vq=dxq/dt).
At any particular location (x), the mass at that
point has a y motion of y = A sin(kx-wt+qo)
= -A sin(wt+fo) where the fo=kx+qo=constant
and so has a vy = -Aw cos(wt+fo) .
NOTE: vq and vy are different things!
Phase velocity
Note that the phase velocity is related to the
wavelength () and frequency (f), but it
does not depend on these:
vq = f = w/k .
If you double the wavelength, the frequency will
halve, but the phase velocity remains the same.
The phase speed depends on the material it is
going through (Tension and m):
vq = [Tension /m] .
Waves (in general)
• For nice sine waves:
y(x,t) = A sin(kx  wt + qo)
• For waves in general, such as a square wave
or a triangle wave, we can break any
repeating wave form into component sine
waves; this is called spectral analysis.
Waves (in general)
• sine waves are nice
• other types of waves (such as square waves,
sawtooth waves, etc.) can be formed by a
superposition of sine waves - this is called
Fourier Series . This means that sine
waves can be considered as fundamental.
Power delivered by a wave
Each part of a nice sine wave oscillates just like
a mass on a spring. Hence it’s energy is that of
an oscillating mass on a spring. We saw
before that this depended on w2A2.
The power delivered by a wave is just this
energy per time. The rate of delivery depends
on the phase speed of the wave. Thus the
power in the wave should depend on:
P  w2A2v where  means proportional to.
Reflections of Waves
Do waves “bounce” off of obstacles? In other
words, do waves reflect?
We’ll demonstrate this in class with a slinky.
There are two cases: when waves bounce off
a stiff obstacle, the waves do reflect, but
they change phase by 180o; when waves
bounce off a loose obstacle, the waves do
reflect, but they do NOT change phase.
Interference: Waves on a String
When a wave on a string encounters a fixed end, the
reflected wave must interfere with the incoming wave so
as to produce cancellation. This means the reflected
wave is 180 degrees (or /2) out of phase with the
incoming wave at the fixed end. The image below
shows a picture of the blue incoming wave and the red
reflected wave at one particular instant. At this instant, the
two waves add to zero at the right end, but the two troughs
near the right end add together!
Fixed end
Interference: Waves on a String
Below is the same situation, only this time we
picture the situation at a time (1/4)T later.
Note that the blue incoming wave is at a
trough and the red reflected wave is at a
crest and so the two cancel at the fixed end
as required. Note that at this instant, we get
cancellation everywhere!
Fixed end
Interference: Waves on a String
Below is the same situation, and this time we
picture the situation at a time (1/4)T later
than the previous, or (1/2)T later than the
first. Note that both the blue incident wave
and the red reflected wave are at zero at the
fixed end, but the two crests near the right
end add constructively!
Fixed end
Interference: Waves on a String
When a wave on a string encounters a free
end, the reflected wave does NOT have to
destructively interfere with the incoming
wave since there is no requirement that the
end stays at zero. There is NO phase shift
on this reflection.
Free end
Interference: Waves on a String
Again, we picture the situation at a time
(1/4) T later. The incident wave is at a
trough and the reflected wave is also at a
trough at the free end.
Free end
Interference: Waves on a String
Below is the same situation, and this time we
picture the situation at a time (1/4)T later
than the previous, or (1/2)T later than the
first.
Free end
Below is the situation with two fixed ends spaced 1.625 wavelengths
apart (phase difference of 585o which is the same as 225o).
1. The blue is the incident wave arriving at the right end with a
phase of 180o and is reflected.
2. The red is the first reflected wave from the right end starting
with a phase of 180o+180o = 360o which is the same as 0o. The
red wave reaches the left end with a phase of 225o and is
reflected.
3. The purple reflected wave starts with a phase of 225o+180o =
405o which is the same as 45o. It reaches the right end with a
phase of 45o+225o = 270o and is reflected.
4. The orange reflected wave starts with a phase of 270o+180o =
450o which is the same as 90o.
Fixed end
Fixed end
Interference
Notice how confused the waves become, and when
you start adding them all together, you tend to
get cancellation everywhere all the time.
Fixed end
Fixed end
Standing Waves
However, if the difference between the two ends is
1.5 wavelengths, a stable pattern of constructive
interference persists, and we have what are
called standing waves!
Fixed end
Fixed end
Standing Waves
To create what are called standing waves (we will play
with these in the last lab), we need to create
constructive interference from both ends. This leads
to the following condition: #(/2) = L ,
which says: we need an integer number of half
wavelengths to “fit” on the Length of the string for
standing waves.
We can vary the wavelength by either varying the frequency or
the speed of the wave: recall that phase speed: v =
distance/time = /Tperiod = f . For a wave on a string,
recall that v = f = (Ttension)/m) where m = m/L.
Standing Waves
For stringed instruments (piano, guitar, etc.), the string
vibrates with both ends fixed. However, with
wind instruments (trumpet, trombone, etc.), we can
have the situation where both ends are free and a
different situation where one end is free and one
end is fixed.
1. If both ends are free, we get the same resonance
condition as for both ends fixed: #(/2) = L.
2. If one end is free and the other end is fixed, we
get a different condition: #odd(/4) = L, where
#odd is an odd number (1, 3, 5, etc.).
Interference: Waves on a String
• When a wave is incident on a SLOWER
medium, the reflected wave is 180 degrees
out of phase with the incident wave.
• When a wave is incident on a FASTER
medium, the reflected wave does NOT
undergo a 180 degree phase shift, that is, it
is in phase with the incident wave.