Section 1-10 - Oscillations

Mass on a Spring
Simple Pendulum
Damped Oscillations
Resonance
Correlation to your Textbook

Mass on a Spring

We started by examining a mass on a spring, placed horizontally on a friction less table, so that gravity and the table's normal force cancel, leaving the spring force as the only one to act on the mass.  We know from previous discussion that the spring follows Hooke's Law, F = - kx, the force applied to the mass is proportional to the displacement of the mass from the equilibrium point and is directed opposite to that displacement.  In general, a force which is directed opposite to the displacement, but not necessarily proportional to it, is called a restoring force, since it acts to return the object to its original position.

If we disturb the mass from the equilibrium point, it will oscillate about that point with a well defined frequency, which we find experimentally is independent of the degree of disturbance (i.e., pull the mass a little ways from the eq. pt, or a long ways, and the frequency of oscillation is the same.).  We would like to determine that exact mathematical function that the displacement follows, x(t), and see if there is a way to predict the frequency of the oscillation.  

We can't do this problem directly, so we need to be clever and compare the situation to another one which we do understand.  If we can find a system which moves under the influence of a force F = -Cx for which we know the motion x(t) , then we shall also know the nature of the motion of the mass-spring system.
First, let's try a system which we shall ultimately find doesn't work.  Consider a ball (mass m), bouncing elastically with essentially constant speed between two walls a distance L apart.  We can certainly define the amplitude of oscillation, the farthest displacement from the equilibrium point, to be A = L/2 and w is the angular frequency w = 2pf, whose significance will become apparent later.  The ball then moves back and forth with constant speed.  What is the force which causes this motion?  It is shown in the figure, zero unless the ball is actually in contact with a wall:

Since this F(x) is clearly not of the form 
F(x) = -Cx, 
the motion of the mass on the spring is not back and forth at constant speed.

After trying out a number of systems which didn't work, let's try one which will.  We might well guess that the functional form of the displacement is 
x(t) = A cos(wt + f), 
where A is the amplitude, w is the angular frequency w = 2pf, and f is the phase angle, which simply allows us to change the cosine into a sine.  We need to prove this, however.
Let's consider an identical mass m moving with uniform circular motion about a circle whose radius r we shall make correspond to the amplitude A of the mass/spring system.  We shall also adjust the angular velocity wof the mass around the circle so that it completes one cycle in the same amount of time as the mass/spring system does.

Let us also match their motions by releasing the mass on the spring from x = +A at the instant that the other mass reaches x = +A, or if you prefer, at q = 0.  You might visualize the situation as a toy train moving along a circular track, with its shadow projected on the wall to represent the mass/spring.  At any instant, the location of the mass is given by 
x = r cosq = A cosq
and 
y = A sinq. 
Thinking back to rotational motion, we know that 
q = wt, 
so 
x(t) = A cos(wt). 
This circular motion is caused by a centripetal force, 
FC = mv2/r = mw2r = mw2A. 
Let's consider only the x motion of the mass, which is caused by the x component of the centripetal force: 
(FC)x = mw2A cosq. 
Note however that the force is always directed in the opposite direction as the mass's displacement; that is, when x>0 on the right half of the figure, the force (FC)x points toward the negative direction, and when the mass is on the left half of the figure and x<0, the force points toward the positive direction.  To correct this, let's insert a negative sign: 
(FC)x = - mw2A cosq. 
We already stated that
x = A cosq, 
so we can substitute this back into our result to obtain:

(FC)x = - mw2A cosq =  - mw2(A cosq) =  - mw2(x) = - Cx,

since m and w are constants.

Now, we've shown that motion of the sort 
x(t) = A cos(wt)
is the result of a force of the type 
F = - Cx. 
We can then make the argument in reverse, that a force of the form 
F = - Cx 
causes motion of the form 
x(t) = A cos(wt), 
and since the mass on the spring moves under the influence of just such a force, this function describes that mass's displacement also.

Cool.  Now, we might like to see if we can determine or predict the frequency of oscillation, w.  Remember that we set up the mass/spring system and the circular motion systems so that they had the same amplitudes and frequencies, and we just showed that their motions are identical in all respects.

F = - kx     x(t) = A cos(wt)
F = -(mw2)x     x(t) = A cos(wt)

Now, if the motions of the two masses are to be identical, the masses must have the same accelerations, which implies that they are acted on by forces which, although different in nature, must be identical in terms of magnitude and direction.  This means that the numerical value of the constant C for each system must be the same as for the other!  Otherwise, different accelerations and therefor different motions would result.  So, we write that 
k = mw2, or that 
w = [k/m]1/2. 
Now, it may bother you that the w and the m are the values for the circular motion mass, but remember that we required the two masses and the two frequencies to be the same, so the result we just obtained is valid for the mass on the spring alone.


We noted two things here.  First, the frequency appears to be independent of the amplitude of oscillation, so it doesn't matter how far I pull the mass before I release it, the mass will oscillate with the same frequency.  This type of frequency, at which a system 'prefers' to oscillate, is called the natural frequency.  Second, the frequency depends on the mass and stiffness of the spring, which we then tested qualitatively.

We also took this result and found an expression for the period, P, of the oscillation.  The period is the amount of time necessary to complete one cycle.  We know that the normal frequency f = w/2p represents the number of cycles per second, so the period is clearly
P  =1/f = 1/(w/2p) = 2p/w = 2p[m/k]1/2.

This type of motion,  x(t) = A cos(wt + f), is referred to as simple harmonic motion.

When the spring is horizontal, Newton's second law is
Si Fi  = - kx = ma,
and the motion is as described above.  What if the spring had been oriented vertically, so that gravity were not cancelled?  The new equilibrium point would be where the weight and the spring force balance:
Si Fi  = -mg - kxEQ = 0       xEQ = -mg/k  (negative, because this equilibrium point is below the 'no-gravity' equilibrium point).
Now let the mass oscillate.  Let y' be the displacement from the new equilibrium point, such that x' = x - xEQ.  Substitute.
Si Fi  = -mg - kx = ma
 -mg - k (x' + xEQ) = ma
 -mg - kx' - kxEQ = ma
[-mg - kxEQ] - kx' = ma
The quantity in brackets is known to be zero, so
 - kx' = ma,
which is the same force equation (and therefore the same motion) for the case of no gravity, except that the equilibrium point will be where x' = 0, that is, at x EQ.

Let's see if we can determine any information about the mass's speed and acceleration.  The acceleration is easy; use Newton's Second Law:
F = ma
F = - kx
a = - (k/m) x = - (k/m) A cos(wt + f) = -  w2A cos(wt + f).
The velocity is a bit harder.  Let's consider the energy budget of the mass: as the system oscillates, KE of the mass is converted to PE in the spring:
1/2mv2 + 1/2 kx2 = Etotal.
We can write an expression for Etotal if we look at one extreme of the mass's motion, x = A, at which point the speed is zero and there is no KE.  At that point,
Etotal  = 1/2 kA2.
Then,
1/2mv2 + 1/2 kx2 = 1/2 kA2.
mv2 + kx2 = kA2.
mv2 = kA2 - kx2
mv2 = k(A2 - x2)
v2 = [k/m] [A2 - A2cos2(wt + f)]
v = +/- [k/m]1/2 [A2 - A2cos2(wt + f)]1/2
v = +/- [k/m]1/2 A[1 - cos2(wt + f)]1/2
v = +/- wA [sin2(wt + f)]1/2
v = +/- wA sin(wt + f)
Now, we have to go back to the problem to pick the correct sign.  We have been assuming that x(t) = A cos(wt + f), so that when (wt + f)=0, the object is at its extreme right position and the velocity is zero (so far so good, since sin(wt + f) will be zero).  In the next instant, the mass will be moving to the left, and so we want the velocity function to become negative; this means that we need to use the negative sign.  So,
v(t) = - wA sin(wt + f).
To review:
x(t) = A cos(wt + f)
v(t) = - wA sin(wt + f)
a(t) =  - w2A cos(wt + f)
For fun, what will be the fuctional forms for the jerk, kick, and lurch?
j(t) = w3A sin(wt + f)
k(t) = w4A cos(wt + f)
l(t) = - w5A sin(wt + f)

Simple Pendulum

There are many other systems which exhibit simple harmonic motion (SHM), and even more which are close enough that we can make use of the results above for a reasonably correct approximate solution.  One such system is the simple pendulum, which is a point mass m at the end of a massless string or stick of length l.  Let's look at the free body diagram for such an object.

We are interested in the motion along the circular arc.  Let us describe the bob's position with s = lq, the displacement along the arc which we shall make positive to the right and negative to the left.  Break the forces into tangential and radial components. Radially, we have
T - mgcosq = maC,
where aC is the centripetal acceleration.  We're not really interested in this.
Tangentially, we have
Ft = mg sinq = mg sin(s/l).
We note that the force is really a restoring force, since it points in the opposite direction along the arc as the bob's displacement.  Let's add the minus sign as we did for the mass/spring system to account for the direction of the force:
Ft = -  mg sin(s/l).
Now, this is not the same as the for the mass/spring system, since F is proportional to the sine of the displacement, not to the displacement itself.  However, if an angle is small, the sine of the angle is approximately equal to the angle itself.  So we can make an approximation:
Ft = -  mg (s/l) = -  (mg/l) s.
Now, the restoring force is at least approximately of the form F = - Cx, and we can make use of the result above.  Let's compare the two systems
F = - (k) x, where w = [k/m]1/2
F = - (mg/l) s,
so that it's clear that (mg/l) takes the place of k.  Let's substitute it into the result we already have for w:
w = [(mg/l)/m]1/2  = [mg/lm]1/2 = [g/l]1/2,
which is indeed the correct answer.  In class, we tested qualitatively the dependence of the frequency on the length, and the independence on the mass, and found agreement with the theory; we could not test the dependence on gravity, however.

In fact, you may remember back to the beginning of the year when we discussed dimensional analysis.  We looked at all of the quantities which might possibly affect the frequency of oscillation: mass m: [M], string length l: [L], and gravity g: [L]/[T]2.  We saw that the only combination which could result in the dimensions of the period, [T], is [l / g]1/2, or {[L]/[L]/[T]2}1/2.
In a manner similar to that above, we find that the period of oscillation for a simple pendulum is
P  =1/f = 1/(w/2p) = 2p/w = 2p[ l / g]1/2.

Here is another example for you to work on:
Consider a torsional pendulum, a cylinder hanging from a string.  The torque necessary to twist the string is assumed to be proportional to the angular displacement:
t = - k q.
The cylinder is twisted through some initial angle, then released.  What is the natural frequency of rotational oscillation of this system?


Damped Oscillations

We spoke briefly about damped oscillations.   The equations above suggest that, if one sets the mass/spring system into oscillation, the total energy of the system remains constant and the mass will vibrate forever with the same amplitude.   In fact, we know that the mass will slow a bit on each pass due to friction with the air (usually assumed to be a drag force of the form Ff=-bv) or the table; energy is removed as friction performs negative work on the mass.

The figure shows a lightly damped system (black curve) and an overdamped system (red line), which loses so much energy so quickly that it never oscillates even once.  A good example of the overdamped system is the car shock absorber.  The car (m) is supported by springs (k), so that SHM is possible.  If one were to drive over a bump with faulty shocks, the car would then continue to oscillate at about 1 Hz for several seconds.  Shock absorbers dampen the system so that the ride smooths out without the oscillations.

One caveat: the natural frequency of a lightly damped oscillation is slightly different than the value of an undamped oscillation:
w ' = [(k/m) - b2/4m2]1/2,
where b characterizes the amount of damping.

Resonance

We returned to the notion of the natural frequency of a system.  If one were to disturb the mass/spring system in some way and step back, the system will oscillate with natural frequency wo = [k/m]1/2.  If it's disturbed again in a different manner, the system will still oscillate at that same natural frequency, until its energy is depleted (see damped oscillations).  If we want the system to continue to oscillate, we must replace the energy lost to dissipative forces.  Let's jiggle the other end of the spring, applying a force though a distance (i.e., doing work), at some frequency w, which is then known as the driving frequency.  Let us vary the driving frequency to see the effect on the system.  If we jiggle the spring at a very low frequency, we see that the mass oscillates with the same frequency at which it is driven, but with a small amplitude.  Changing to very high driving frequency, we see once again that the mass oscillates at the driving frequency, but with a very small amplitude.  However, if we excite the system at a driving frequency very near to the natural frequency, we see that the response of the system, as demonstrated by the amplitude of oscillation, increases.  If we plot this response as a function of the driving frequency, we see the curve shown here:

The condition when the system is driven at its natural frequency and delivers its greatest response is called resonance.  Sometimes resonance is desirable, sometimes not.  For example, if one wants to push a small child on a swing, the greatest amount of fun (or terror) is attained when one pushes the swing at its natural frequency.  On the other hand, if the ground shakes at the natural frequency of a skyscraper, the building may respond with an amplitude beyond the limits of structural integrity.  The Tacoma Narrows Bridge collapse, which was featured in an Alpine Stereo advertisement several years ago, occured because the wind passing over the bridge excited one of the span's torsional oscillation modes, resulting in the collapse about three hours later.  The speed of the oscillations was speeded up for the advert, but the amplitude of the oscillations was not.  Are you surprised at the incredible elasticity of steel and concrete?  Only a dog lost its life in the collapse, because the owner left it behind when he abandoned his car on the bridge (Hmm!).  The bridge had exhibited strange effects for the three months it was open.  There are films of the deck of the bridge oscillating in a vibrational mode much like waves in the ocean; cars could actually disappear from view behind the humps which rose and fell in the roadway.  A related system is that of tall skyscrapers.  Once again, if the wind were to gust at the natural frequency of the building, it might cause collapse; modern buildings often have a mechanism to 're-tune' the vibrational modes of the building away from the current driving frequency of the wind.

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