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2. Shaken not Stirred |
Resonance
Resonance is a condition associated with oscillations. Resonance occurs when:
The applied frequency is the same as the natural frequency.
The natural frequency is the frequency at which the system oscillates when allowed to do so freely.
The applied (or forced) frequency is the frequency of any outside oscillation that is applied to the system.
Resonance can be useful, for example electrical resonance is essential for any radio receiver. A church bell can only be rung if the bell ringer applies the pull at the same frequency as the bell swings. He cannot make it ring any faster.
It can be a gimmick, such as an opera singer making glasses shatter.
It can be a nuisance, such as the panels on a bus rattling.
It can be dangerous and destructive. For example a helicopter can get into ground resonance (about 1 Hz), and shake itself to pieces. This is what happened to the machine below:
The human body resonates at about 5 - 7 Hz. At best you will feel sick. At worst, an epileptic fit can be induced. This is why news reports often have the warning "This report contains flash photography."
We can investigate resonance with apparatus like this:
If we plot a graph of maximum amplitude against frequency, we see:
At resonant frequency, the amplitude of the waves is very large. Often the slotted mass jumps off.
Simple Harmonic Motion
If you set a simple pendulum swinging, or a mass bouncing up and down on a spring, you will find that its oscillations (complete to-and-fro movements) have a regular periodic motion called harmonic motion. For the motion to show simple harmonic motion (SHM), the following conditions must be met:
The period must be independent of the amplitude;
The restoring force always acts towards the centre point;
The restoring force is proportional to the displacement;
At the equilibrium position the force must be zero (hence the acceleration is zero).
We can write this as a simple equation:
F = -kx
The minus sign is essential; it tells us that the force is towards the centre. If the minus sign were not there, the force would increase with the displacement, which in turn means the greater the distance from the centre, the larger the acceleration.
Consider a
spring with a mass going from side to side.
A mass is mounted on a small railway truck, which is free to move from
side to side, and there is negligible friction in the truck.
The
rest or equilibrium position at O
is where the spring would hold the mass when it is not bouncing.
A is the position where the
spring is stretched the most, and B
is where the spring is squashed most.
At
A there is a large
restoring force because that is where the spring is stretched most.
As a result of this the mass is accelerated. It accelerates towards the rest position.
Its velocity to the
left increases.
The
acceleration decreases as the mass approached the rest position.
Because
of inertia, the mass overshoots the rest position.
Then
the spring is being compressed, and there is a restoring force to the right.
At
B the acceleration is at a
maximum, but this time to the right.
At both A and B, the potential energy is at a maximum; the kinetic energy is zero.
As the mass
passes the equilibrium position, there is zero potential energy, but maximum
kinetic energy because this is the point at which the object has its greatest
velocity (upwards or downwards). Therefore
there is an interchange between potential
and kinetic energy. The
process is never 100 % efficient; some energy is lost as heat and the process is
not indefinite.
We can write down a relationship between the acceleration, a, and the displacement, x.
F = ma and F = kx
Therefore
a = F/m = kx/m
So we are saying that the acceleration is proportional to the displacement from the equilibrium position. However that is not the whole story. Acceleration is a vector, so we must be careful of the direction. The acceleration is towards the equilibrium position.
These relationships are derived by linking SHM to circular motion. Click HERE
Generally
we measure the period, which is the time taken to make a complete
oscillation or cycle. The frequency
is the reciprocal of the period:
f
= 1/T
Acceleration
can be linked to displacement by:
a
= - (2pf )2
x
This
satisfies the condition for SHM that
a = -Kx;
in this case
K = (2pf
)2.
Angular
velocity
is a quantity that is borrowed from circular motion. It is the angle turned per second. In SHM terms, we can consider it as the fraction of a cycle
per second. It can be, of course,
greater than 1:
w
= 2pf
In
some texts you may see the equation for acceleration in SHM written as:
a = -
w2
x
The
velocity at any point in the
oscillation given by:
v2
= (2pf
)2(A2
– x2)
Þ v2 = 4p2f 2(A2 – x2)
Þ v = 2pf Ö(A2 – x2)
In this relationship,
A
is the amplitude and s is the displacement
from the equilibrium position. If
x
= 0,
v has a maximum value; if
x =
A,
v = 0.
The velocity is 0 at each extreme of the oscillation.
The
displacement,
s, is given by:
s =
±
A cos
2pft
The
plus and minus sign here tells us that the motion is forwards and backwards.
Which sign we give for direction is up to the individual.
Generally left to right is forwards.
All these equations are true for any simple harmonic motion. We can show the relationships graphically by showing displacement, velocity, and acceleration against time:
These
graphs are sinusoidal.
The displacement is
p/2
radians (90 o or ¼ cycle)
behind the velocity. The
displacement and acceleration are
p
radians out of phase.
The extension of a spring is directly proportional to the force (Hooke’s Law). Consider a mass, m, put onto a spring of spring constant k so that so that it stretches by an extension l.
The
force on the spring = mg, and the
stretching tension =
kl.
Þ mg = kl
Suppose
the spring is pulled down by a distance x
below the rest position. Now the
stretching force become
k(l + x).
This is also the tension in the spring acting upwards.
So the restoring force,
Fup
= k(l + x) – mg. This is
because
mg is the weight, which always acts downwards.
Since
kl
= mg, we can write:
Fup = kl + kx – kl = kx
We can now apply Newton II to write:
-kx = ma
(The negative
sign tells us that the force is upwards)
We know from SHM that
a
= - (2pf
)2
x, so we can
write:
a = -kx/m = -
(2pf
)2
x
So we can rearrange this to say that
(2pf )2
= k/m.
Since
a = -
(2pf
)2
x we can say
that the condition for SHM is satisfied in this system, as long as Hooke’s Law
is obeyed. Since
T = 1/f,
we can now write down an expression to relate the period with the mass and the
spring constant.
T =
2pÖ(m/k)
This
tells us that if we want to double the period, the mass has to be increased by
four times. If we plot a graph of T2
against m we will get a straight line,
since
T2 = 4p2
(m/k). The gradient will
be
4p2/k
which we can approximate to
40/k,
since p2 »
10.
The relationship of the graph suggests that the line should cut through the origin. However we find that it does not. This is due to the mass of the spring itself; the effective mass of the spring is about 1/3 the actual mass of the spring itself. However if the mass on the spring is very much bigger than the mass of the spring, this effect is negligible.
Worked Example
|
A
light spiral spring is loaded with a mass of 50 g and extends by 10 cm.
What is the period of small vertical oscillations if the
acceleration due to gravity is 10 m/s2? |
|
We
need to work out the spring constant using Hooke’s Law F
= ke k
= F/e =
0.05 kg ´
10 m/s2 = 5.0 N/m
0.1
m |
|
Now
we can use T = 2pÖ(m/k)
to work out the period: 5 N/m |
2. The Simple Pendulum
Consider a small bob of mass m hanging from a very light string, length l , which in turn hangs from a fixed point. If it is pulled to one side through a small angle q, it will swing with a to-and-fro movement in the arc of a circle.
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As weight mg is a vector, we can break it into its two components, mg cos q and mg sin q.
At point A the bob accelerates with an acceleration a due to the force mg sin q. We can apply Newton II to write:
-mg sin
q
= ma
[negative sign as the force is directed to equilibrium position]
If
q
is small and in radians, we can say that sin
q
= q.
This does not work for degrees. We
can measure the chord OA, which is
the displacement. Remember that
displacement is the straight-line distance between two points.
Now we can say that the displacement,
s
= l sin q.
So we can therefore rewrite this as s
= lq.
So now we can write:
-mgq
= -mg(s/l) =
ma
[m
terms cancel out]
Þ a = -gs/l = -(2pf )2s
The relationship a = -(2pf )2s arises because this is a simple harmonic oscillator.
In this case (2pf )2 = g/l because the s terms cancel out.
We know that period is given by T = 1/f.
Þ T = 1 = _2p_ = 2pÖ(l/g)
f Ö(g/l)
So we can write the formula
linking period with length and gravity constant as:
T
=
2pÖ(l/g)
Notice that T is independent of amplitude or mass of the bob. If a pendulum clock were taken to the moon, its time-keeping would be somewhat altered.
If we plot a graph of T2 against l, we get a straight line of which the gradient is 4p2/g (the value of which would be approximately 4). To measure g we divide 4p2 by the gradient we get. We can get a relatively accurate determination of this if we:
Count at least 100 swings
Use a swing angle of less than 10o.
Measure l to the centre of the bob.
Count the oscillations as the bob passes the centre position.
Worked Example
| A simple pendulum has a period of 2.0 s and amplitude of swing of 5.0 cm. What is the maximum velocity of the bob? What is the maximum acceleration? |
| Velocity is at
a maximum when the equilibrium position is reached.
v2 = w2(A2 – x2) x = 0, A = 5 cm. |
| We need to know
w.
w = 2p/T = 2p rad ¸ 2.0 s = p rad/s |
| Now we can work
out the velocity:
v2 = p2(52 – 02) = p2 ´ 25 = 247 cm2 s-2 Þ v = 15.7 cm/s |
| We can use a
= -w2s.
Acceleration is at a maximum when s
= A = 5.0 cm
Þ a = -p2 ´ 5.0 = - 50 cm/s2. The negative sign tells us that the direction of the acceleration is to the rest position. |
Energy in Simple Harmonic
Motion
There is constant interchange between kinetic and potential energy as the pendulum (or other oscillator) swings to-and-fro. If the system does not have to work against restrictive forces, such as friction, the total energy will remain constant.
This is the most likely level you need. Click HERE if you want to know more.
Etot = Ep + Ek
We can show the variation of the energy graphically:
Now we will look at the energy with displacement:
If you are not sure about this, the key points to remember are:
Potential energy is highest when the oscillator is at the maximum amplitude;
Kinetic energy is highest when the oscillator passes the rest position.
In the exam, you will probably be asked simply to mark on a diagram where the maximum potential and kinetic energy occur.
Speed of a compression wave in a material
We can model this as a set of masses connected by springs. The masses represent the molecules, and the springs the bonds between the molecules.
If you push the block at the end, the first spring gets compressed, moving the second block, and so on. A compression pulse propagates down the line. The speed of propagation is:
v = s/t
The spacing between the blocks is s, and t is the time interval taken for the spring to be fully compressed.
The period of the oscillation is T. The time interval for the compression pulse to pass has been found to be the following fraction of a cycle:
t = 1
2p
Therefore:
t = T
2p
We know that for a spring oscillator:
T = 2pÖ(m/k)
If we combine these two:
t = 2pÖ(m/k)
2p
Df = kDe
F = A × kDe
x2
s = kDe
x2
e = De
x
E = s/e = kDe ÷ De
x2 x
Þ E = kDe × x
x2 De
Þ E = k
x
v = xÖ(Ex/m) = Ö(Ex3/m)
v = Ö(E/r)
Now go and have a cup of coffee.
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