


Origins of Newton's laws of motion

Non-mathematical introduction to relativity


History of mechanical clocks with animations
Understanding a mechanical clock with animations
includes pendulum, balance wheel, and quartz clocks

Water waves, Fourier analysis

Monday, June 11, 2012

Mathematical response to a very short impulse

All postings by author previous: 3.7 More on the impact model up: Contents next: 3.9 Summing impulses using calculus

3.8 Mathematical response to a very short impulse

We started this series of postings discussing continuous excitation of a resonator (rf engineers call this "CW" for continuous wave excitation). We then moved on to showing how we might model sinusoid excitation of a resonator as a series of impulses and how this qualitatively explains the sensitivity of the resonator to the excitation frequency. In this posting and the next, we will use calculus to do a better job of modeling sinusoidal excitation with a series of impulses. In fact, calculus allows us to consider an infinite number of tiny impulses!

Equation (23) of an earlier posting indicates that the impulse response (with damping) of a resonator is given by:

   ,   (80)    in real notation


   ,   (81)    in complex notation.

The decay resonant frequency is given by:

   .   (82)

The complex multiplier s is defined as:

   .   (83)

In the above equations, we have added φ the phase of the amplitude to make the equation a little more general than we used before. In our previous postings on impulse response, we ignored this phase since it wasn't important there but here it is important, so we need to figure it out in terms of our system.

Consider our mass-spring resonator at rest with all parts at rest and in equilibrium. Then suppose the "driver" in Figs. 7 and 8 is made to move up a centimeter very quickly, stay there for 0.01 seconds, then move back to its initial position. The mass would not have time to move significantly in this time, however this action would transmit an impluse   I = f dt = k Δx dt   to the mass. Because of Newton's second law the momentum of the mass will be changed from its equilibrium value of 0 to a value equal to the impulse I:

p = mv0 = I = f dt    .   (84)

See the previous link for a derivation of this equation.

The sudden change in momentum means a sudden change in the velocity of the mass. Solving (84) for the new velocity v0 (immediately after the impulse), we have:

v0 = f dt/m   ,   (85)

where dt is the time duration of the impulse force f.

After the impulse, there is no more intervention of the driver, so the system is described by the normal decay equation, i.e. (81). In order to have an equation that describes this motion completely, we need to adjust A and φ so that the initial velocity (immediately after the impulse) is given by (85).

We can relate (85) to our mass's motion by using the derivative of (81) for the velocity (or at least the real part of our complex velocity):

   ,   (86)

where the impulse is assumed to occur at time t = t'.

Eqn (86) represents one initial condition on our equation at t = t'. The other condition is that the mass does not move a perceptible amount during the impulse, i.e. that at time t = t', x = 0. What we really mean is that the real part of the complex form of x is zero at t'. Using (81) we write:

   .   (87)

In order for Equation (87) to be true, i.e. the left side equaling zero, either the amplitude A must be zero or Re[ei(ωdecayt'+φ)] is zero. Because the impulse starts oscillations we know the amplitude is not zero. Thus the real part of the complex exponential part must be zero. This means that stuff inside the "real part of" sign must be totally imaginary, i.e.  ωdecayt' + φ = ±π/2 . We pick the minus sign because this will make the amplitude A in (87) positive, which is mathematically a little cleaner. Thus  ωdecayt' + φ = −π/2  which means  φ = −ωdecayt' − π/2  and thus:

 e = e−iωdecayt'e/2 = −ie−iωdecayt' .     (88)

We used the fact that  e/2 = -i .

Rewriting (86) we have:

   ,   (89)

which can be solved for the amplitude A:

   ,   (90)

We substitute this and (88) back into (81) to get an equation for the mass's vertical position x versus time after the impulse (remember that Re(x) = 0 before the impulse):


which finally reduces to:

  .  (91)

Fig. 17. The response in the position x of the mass (shown in red as a function of time) of the mass-spring resonator to a very short impulse from the drive. The impulse itself is shown in green (which is a force with different units than x.)

The real representation of this waveform is:

  .  (92)

which we have plotted in Fig 17.
