Simple Harmonic Motion

The definition of a simple harmonic oscillator is any system which undergoes simple harmonic motion. Simple harmonic motion, in turn, is the term we use to describe any system who's equation of motion is given by the form

The translation: let x be the measurement of some kind of position. The position can be in units of meters, light-years, radians, etc. As long as change in x describes displacement of some kind, the equation above says that the second derivative of x with respect to time must be equal to the negative of some magnitude times x. We say magnitude here to indicate that the constant in the above equation must be positive. The name of the motion, dx/dt, that we apply to solution, x(t), of this equation is simple harmonic motion. The harmonic part comes because one of the natural phenomena that this equation applies to is the generation of sounds from musical instruments. The simple part applies because the description leaves out the effects of friction and other forces that would effect the motion. This is the simplest description of a system that oscillates, i.e. repeats it's motion over and over.

The solution to this equation is one we've already specified. By analogy to the case of the spring and the object in uniform circular motion:

where A is the amplitude, or maximum distance from equilibrium, omega is the angular frequency of the motion and is equal to sqrt[constant] in value, and delta is the phase. The phase gives us information on the initial position, velocity, and acceleration of the motion (at t = 0). The angular frequency, as we saw in the last lecture, is related to the period, T, of the motion by the relationship

For uniform circular motion, we have a ready interpretation for each of these quantities: A is the radius of the circular path followed by the particle, omega is the angular velocity, d(theta)/dt, for the motion, and the phase tells us the initial value of theta. For the spring, A is the maximum extent to which the spring is pushed or pulled from its equilibrium length, omega is given by sqrt[k/m] and is also equal to 2p/T with T being the period, and delta relates to the initial position of the mass m being operated on by the spring. If you want to know why the circular motion expression and spring expression look the same, check here.

Now we need to do something practical with our new found knowledge. It will be awhile before you find out just how universal this simple harmonic equation is, suffice it to say that you will see it again in descriptions of everything from sound, to earthquakes, to optics. For now, let's stick to a few more plebian examples. The goal of problems involving simple harmonic motion is first to show that they are simple harmonic, then find the appropriate relationships between omega, delta, position, velocity, etc. The definition of simple harmonic motion is that the equation of motion for the system must look like

so your first goal is to show this. In the process, you will have to find the appropriate expression for the "constant". Then, you know that omega = sqrt[constant], the amplitude, A, and the phase, delta, can usually be found from the description of the problem, the velocity is dx/dt, etc.

Example 1:

• What if the spring is vertical instead of horizontal? How do you map the motion of a vertical spring onto the vertical component of circular motion?

• Consider the spring hanging vertically. Since it's massless, the spring just hangs at it's equilibrium length. Now attach a mass m to it and gently allow it to lower until you can release it and the spring and mass hang motionless. See the figure below.
  

  Now the mass hangs a distance y₀ below the natural length of the spring. The mass stops moving as you lower it when the weight balances the spring force, so

  -k*y₀ + mg = 0 ==> k*y₀ = mg ==> y₀ = mg/k

  Therefore, y₀ defines the new equilibrium point since, by definition, the equilibrium position is the point where the net force on the mass m equals zero. We can make use of this in the equation of motion since now

  F_net = -k*y + mg = m*a_y = m d²y/dt²

  by noting that we can perform a transformation of coordinates that does not change the physical description. Replace y by y' = y-y₀. Using y', the force exerted by the spring is F(y') = -k(y'+y₀) and we've changed nothing. However, note that dy'/dt = dy/dt and d²y'/dt² = d²y/dt², so our new equation of motion becomes

  Fnet = m d2y/dt2	= -k*(y'+y0) + mg	replace y with y'+y0
  m d2y'/dt2	= -k*(y'+y0) + mg	expand out the rhs
  	= -k*y' - k*y0 + mg
  	= -k*y' - k*(mg/k) + mg	note: last 2 terms cancel
  m d2y'/dt2	= -k*y'	divide through by m
  d2y'/dt2	= -(k/m)*y'

  The last equation is just the condition for simple harmonic motion, only in terms of y' instead of y. That's OK. What it means is that the mass m oscillates around the equilibrium point of y' = 0 or y = y₀ = mg/k. That's fully consistent with our expectation.

Example 2:

• What if more than one spring acts in the horizontal direction? What is the period of the motion if the mass is moved a distance d to the left of the equilibrium position in the figure below. Assume motion to the right is positive.

Solution:

• First note that we need to get the correct expression for the equation of motion. Looking at the figure above, we can see that if the mass m is moved to the left, then the left spring pushes since it is compressed and the right spring pulls since it is stretched. Once we release the mass, the springs push/pull it to the right. The mass overshoots the equilibrium position (it's velocity is non-zero there since it was being accelerated up to that point), and the mass goes to the right of the equilibrium position. On the right, the mass is pushed back to the left by the right spring and pulled to the right by the left spring. In both left and right displacements from equilibrium, we note that the springs apply force in the same directions. Thus, the equation of motion is
  F_net = -k₁x - k₂x ==> m d²x/dt² = -(k₁ + k₂)x ==> d²x/dt² = -(k₁ + k₂)/m * x

  The solution is

  x(t) = A cos(wt + d)

  where A = d, the original displacement from equilibrium, omega = sqrt[(k₁+k₂)/m] ==> T = 2p/d = 2dsqrt[m/(k₁+k₂)].

  To get delta is tricky. If we define the direction of toward the right as positive, then, note that, at t = 0, we want x(t=0) = -d. So, since we already set A = d, we want cos(w*0 + d) = -1 ==> d = p.

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